The Fourier–Walsh expansion represents functions on $\Omega_n=\lbrace-1,1\rbrace^n$ in an orthonormal basis of coordinate products. The construction proceeds from a measure space to its Hilbert space $L^2$, then uses orthogonality to derive the coefficients and approximation formulas. On the Boolean cube, the integrals defining the inner product become finite averages.
The construction uses real scalars. Background on measurable functions and integration is given in Abstract measure theory; the geometry of inner products is discussed in Hilbert Space.
Let $\Omega$ be an arbitrary set. The functions $f:\Omega\to\mathbb R$ form a vector space $\mathcal{V}(\Omega)$ under pointwise operations: $(f+g)(x)=f(x)+g(x)$ and $(\alpha f)(x)=\alpha f(x)$ for $x\in\Omega$ and $\alpha\in\mathbb R$. Both operations use the vector space structure of the codomain $\mathbb R$.
If $\Omega$ has $m$ elements, ordering them identifies a function with its vector of values in $\mathbb R^m$, so $\dim\mathcal{V}(\Omega)=m$. For a general domain, an inner product can be constructed by integrating products of functions, after imposing measurability and integrability conditions.
A measure space $(\Omega,\mathcal A,\mu)$ consists of a set $\Omega$, a $\sigma$-algebra $\mathcal A$ of its subsets, and a measure $\mu:\mathcal A\to[0,\infty]$. The $\sigma$-algebra contains $\Omega$ and is closed under complements and countable unions. The measure is countably additive on disjoint measurable sets and satisfies $\mu(\varnothing)=0$.
This structure defines the integral $\int_\Omega f(x)\,\mathrm{d}\mu(x)$ for measurable functions whenever the integral exists. If $\mu(\Omega)=1$, then $\mu$ is a probability measure, and integration gives the expectation of a function under that probability distribution.
Fix a measure space $(\Omega,\mathcal A,\mu)$. The measurable functions $f:\Omega\to\mathbb R$ satisfying $\int_\Omega|f(x)|^2\,\mathrm{d}\mu(x)<\infty$ form a vector space, denoted by $\mathcal V_2(\Omega,\mu)$.
Two such functions are identified when they agree outside a measurable set of measure zero. This is equality almost everywhere with respect to $\mu$. The space $L^2(\Omega,\mu)$ consists of the resulting equivalence classes; a function symbol denotes a representative of its class.
For $f,g\in L^2(\Omega,\mu)$, define
$$\langle f,g\rangle=\int_\Omega f(x)g(x)\,\mathrm{d}\mu(x).$$
The integral form of Cauchy–Schwarz ensures that the product is integrable. The integral defines a symmetric bilinear form on equivalence classes, and $\langle f,f\rangle=0$ holds if and only if $f=0$ almost everywhere. The form is therefore positive definite. Its induced norm is
$$\lVert f\rVert_{L^2(\Omega,\mu)}=\left(\int_\Omega|f(x)|^2\,\mathrm{d}\mu(x)\right)^{1/2}.$$
On the original functions, this expression defines a seminorm whose kernel consists of functions that vanish almost everywhere. Over complex scalars, the corresponding inner product uses $f(x)\overline{g(x)}$ in the integral.
Theorem 2.1. For every measure space $(\Omega,\mathcal A,\mu)$, the space $L^2(\Omega,\mu)$ is complete in the norm defined above. Consequently, $\mathcal{H}=L^2(\Omega,\mu)$ is a Hilbert space.
Completeness is the $p=2$ case of the Riesz–Fischer theorem; a proof is given in Hunter’s notes, Theorem 7.10.
Counting measure on $\lbrace1,\ldots,m\rbrace$ identifies $L^2$ with $\mathbb R^m$. The inner product is $\langle f,g\rangle=\sum_{i=1}^m f(i)g(i)$, so the identification preserves the Euclidean geometry.
Counting measure on $\mathbb N$ gives the sequence space $\ell^2$. A sequence $a=(a_1,a_2,\ldots)$ belongs to this space when $\sum_{j=1}^\infty|a_j|^2<\infty$, and its inner product with $b\in\ell^2$ is $\langle a,b\rangle=\sum_{j=1}^\infty a_jb_j$.
On $[0,1]$ with Lebesgue measure $\lambda$, the same construction gives $L^2([0,1],\lambda)$ with inner product $\langle f,g\rangle=\int_{[0,1]}f(x)g(x)\,\mathrm{d}\lambda(x)$. Here elements are classes of measurable functions whose squared absolute values have finite integrals, with equality understood almost everywhere.
For $1\le p<\infty$, the space $L^p(\Omega,\mu)$ consists of equivalence classes of measurable functions satisfying $\int_\Omega|f(x)|^p\,\mathrm{d}\mu(x)<\infty$. Its norm is
$$\lVert f\rVert_{L^p(\Omega,\mu)}=\left(\int_\Omega|f(x)|^p\,\mathrm{d}\mu(x)\right)^{1/p}.$$
At $p=\infty$, the space $L^\infty(\Omega,\mu)$ consists of classes of essentially bounded measurable functions, and its norm is the essential supremum of $|f|$ on $\Omega$ with respect to $\mu$.
The Riesz–Fischer theorem gives completeness for every $1\le p\le\infty$, so these spaces are Banach spaces. At $p=2$, the identity $\lVert f\rVert_{L^2(\Omega,\mu)}^2=\langle f,f\rangle$ supplies the geometry of orthogonality and projection used in the Fourier construction.
Let $\mathcal{H}$ be a Hilbert space of finite dimension $m$, with inner product $\langle\cdot,\cdot\rangle$. Its induced norm is $\lVert v\rVert_{\mathcal{H}}=\sqrt{\langle v,v\rangle}$.
A family $e_1,\ldots,e_k$ is orthonormal if $\langle e_i,e_j\rangle=\delta_{ij}$, where $\delta_{ij}$ equals $1$ for $i=j$ and $0$ otherwise. Such a family is linearly independent: if $\sum_i a_ie_i=0$, taking the inner product with $e_j$ gives $a_j=0$. An orthonormal family of $m$ vectors therefore forms a basis of $\mathcal{H}$.
Proposition 3.1. If $e_1,\ldots,e_m$ form an orthonormal basis of $\mathcal{H}$, then every $v\in\mathcal{H}$ has the unique expansion $v=\sum_{j=1}^{m}\langle v,e_j\rangle e_j$.
Proof. The basis property gives an expansion $v=\sum_j c_je_j$. Taking the inner product with $e_k$ yields $\langle v,e_k\rangle=c_k$ by orthonormality. Thus every coefficient is uniquely determined.
Proposition 3.2. Let $\mathcal{V}=\operatorname{span}\lbrace e_1,\ldots,e_k\rbrace$, where $e_1,\ldots,e_k$ are orthonormal. For each $v\in\mathcal{H}$, the unique element of $\mathcal{V}$ minimizing $\lVert v-u\rVert_{\mathcal{H}}$ is $P_{\mathcal{V}}v=\sum_{j=1}^{k}\langle v,e_j\rangle e_j$.
Proof. Set $p=P_{\mathcal{V}}v$. For each $j\le k$, orthonormality gives $\langle v-p,e_j\rangle=0$. Hence $v-p$ is orthogonal to $\mathcal{V}$. For any $u\in\mathcal{V}$, the decomposition $v-u=(v-p)+(p-u)$ is orthogonal, and Pythagoras gives
$$\lVert v-u\rVert_{\mathcal{H}}^2=\lVert v-p\rVert_{\mathcal{H}}^2+\lVert p-u\rVert_{\mathcal{H}}^2.$$
The minimum is attained precisely when $u=p$.
Set $\Omega_n=\lbrace-1,1\rbrace^n$, take the full $\sigma$-algebra $\mathcal A=2^{\Omega_n}$, and assign each point probability $\mu(\lbrace x\rbrace)=2^{-n}$. Every function on the cube is measurable and square integrable. Since every point has positive probability, the only set of measure zero is the empty set. Equality almost everywhere is therefore pointwise equality, so $L^2(\Omega_n,\mu)$ is identified with the entire vector space $\mathcal{V}(\Omega_n)$.
In this setting, the integral of any function is a finite average:
$$\int_{\Omega_n}f(x)\,\mathrm{d}\mu(x)=2^{-n}\sum_{x\in\Omega_n}f(x).$$
For the rest of the article, write $\mathcal{H}=L^2(\Omega_n,\mu)$. If $X=(X_1,\ldots,X_n)$ is uniformly distributed on $\Omega_n$, the inner product becomes
$$\langle f,g\rangle=\mathbb E[f(X)g(X)]=2^{-n}\sum_{x\in\Omega_n}f(x)g(x).$$
The notation $\lVert f\rVert_2$ denotes the $L^2(\Omega_n,\mu)$ norm, so $\lVert f\rVert_2=(\mathbb E[f(X)^2])^{1/2}$. The space has dimension $2^n$. The coordinates of $X$ are independent and satisfy $\mathbb E[X_i]=0$ and $\mathbb E[X_i^2]=1$.
It remains to construct $2^n$ orthonormal functions.
For a single coordinate with the uniform distribution on $\lbrace -1,1\rbrace$, the functions $1$ and $x$ are orthonormal: their squared norms equal $1$, and their inner product is $\mathbb E[X]=0$. A product construction extends these functions to $n$ coordinates. At coordinate $i$, choose either $1$ or $x_i$, and multiply the choices. A subset $S\subseteq[n]$, where $[n]=\lbrace1,\ldots,n\rbrace$, records the coordinates contributing $x_i$.
For each $S\subseteq[n]$, define the Walsh function $\chi_S(x)=\prod_{i\in S}x_i$, with the convention $\chi_\varnothing(x)=1$. These functions are also called parity functions: $\chi_S(x)$ equals $1$ when an even number of coordinates indexed by $S$ are negative, and $-1$ when that number is odd.
Theorem 4.1. The family $\lbrace\chi_S:S\subseteq[n]\rbrace$ is an orthonormal basis of $\mathcal{H}$.
Proof. Since $x_i^2=1$, the product of two Walsh functions satisfies $\chi_S(x)\chi_T(x)=\chi_{S\triangle T}(x)$, where $S\triangle T$ is the symmetric difference of the two sets. If $S=T$, this product is identically $1$, so $\langle\chi_S,\chi_T\rangle=1$. If $S\ne T$, independence and the fact that each coordinate has mean zero imply
$$\langle\chi_S,\chi_T\rangle=\mathbb E[\chi_{S\triangle T}(X)]=\prod_{i\in S\triangle T}\mathbb E[X_i]=0.$$
Thus $\langle\chi_S,\chi_T\rangle=\delta_{ST}$. There are $2^n$ subsets of $[n]$, giving $2^n$ orthonormal functions in a space of dimension $2^n$. The family therefore forms a basis.
Theorem 4.2. Every function $f:\Omega_n\to\mathbb R$ has the unique representation
$$f(x)=\sum_{S\subseteq[n]}\widehat f(S)\chi_S(x),$$
where $\widehat f(S)=\langle f,\chi_S\rangle=\mathbb E[f(X)\chi_S(X)]$.
Proof. Apply Proposition 3.1 to the basis in Theorem 4.1.
The sum is finite and holds at every point of $\Omega_n$. Each coordinate occurs at most once in a Walsh function, so the expansion is a multilinear polynomial. It is the unique multilinear polynomial agreeing with $f$ on the cube. The relation $x_i^2=1$ allows distinct polynomials to represent the same function on this domain.
The constant coefficient is $\widehat f(\varnothing)=\mathbb E[f(X)]$. For a singleton set, $\widehat f(\lbrace i\rbrace)=\mathbb E[f(X)X_i]$ measures alignment with coordinate $i$. Larger sets index products involving several coordinates. The degree of $\chi_S$ is $|S|$.
For a Boolean function $f:\Omega_n\to\lbrace-1,1\rbrace$, the product $f(X)\chi_S(X)$ equals $1$ on agreement and $-1$ on disagreement. Consequently,
$$\widehat f(S)=2\mathbb P\bigl(f(X)=\chi_S(X)\bigr)-1.$$
Theorem 4.3. For every $f\in\mathcal{H}$,
$$\lVert f\rVert_2^2=\mathbb E[f(X)^2]=\sum_{S\subseteq[n]}\widehat f(S)^2.$$
Proof. Substituting the Fourier expansion into the inner product and using orthonormality yields
$$\langle f,f\rangle=\sum_{S,T\subseteq[n]}\widehat f(S)\widehat f(T)\langle\chi_S,\chi_T\rangle=\sum_{S\subseteq[n]}\widehat f(S)^2.$$
The same argument gives $\langle f,g\rangle=\sum_{S\subseteq[n]}\widehat f(S)\widehat g(S)$ for $f,g\in\mathcal{H}$. The Fourier transform preserves this inner product when the coefficient space carries the Euclidean inner product.
Since $\widehat f(\varnothing)=\mathbb E[f(X)]$, subtracting the mean removes the constant Walsh term. Applying Parseval’s identity to the remaining expansion gives
$$\operatorname{Var}(f(X))=\sum_{\varnothing\ne S\subseteq[n]}\widehat f(S)^2.$$
For a Boolean function, $f(X)^2=1$, and hence $\sum_{S\subseteq[n]}\widehat f(S)^2=1$.
For an integer $0\le d\le n$, let $\mathcal V_d$ be the subspace of $\mathcal{H}$ spanned by the functions $\chi_S$ with $|S|\le d$. Equivalently, $\mathcal V_d$ consists of functions represented by multilinear polynomials of degree at most $d$. Define the truncated expansion by $f_{\le d}=\sum_{|S|\le d}\widehat f(S)\chi_S$, where the sum runs over subsets of $[n]$.
Proposition 4.4. The function $f_{\le d}$ is the unique minimizer of $\lVert f-g\rVert_2^2=\mathbb E[(f(X)-g(X))^2]$ over $g\in\mathcal V_d$, and its squared approximation error is
$$\lVert f-f_{\le d}\rVert_2^2=\sum_{|S|>d}\widehat f(S)^2.$$
Proof. Proposition 3.2 identifies $f_{\le d}$ as the orthogonal projection onto $\mathcal V_d$. The residual contains precisely the omitted Walsh terms, so Parseval’s identity gives the error formula.
The minimization uses the uniform $L^2$ norm over $\mathcal V_d$, with $\mathbb R$ as the codomain of the approximating function.
[1] O’Donnell, R. (2014). Analysis of Boolean Functions. Cambridge University Press. Chapter 1. A revised text maintained by the author is freely available.
[2] Hunter, J. K. (2011). Measure Theory, Chapter 7: $L^p$ spaces. University of California, Davis.
]]>For any fixed set $X$ denote by $2^X$ the set of all subsets of $X$ and, for any subset $A \subset X$, denote by $A^c := X \setminus A$ its complement.
Let $X$ be a set. A collection $\mathcal{A} \subset 2^X$ of subsets of $X$ is called a $\sigma$-algebra if it satisfies the following axioms:
$\text{(a)} \quad$ $X \in \mathcal{A}.$
$\text{(b)}\quad$ If $A \in \mathcal{A}$, then $A^c \in \mathcal{A}.$
$\text{(c)} \quad$ Every countable union of elements of $\mathcal{A}$ is again an element of $\mathcal{A}$, i.e., if
$A_i \in \mathcal{A}$ for $i = 1, 2, 3, \dots$ then $ \bigcup_{i=1}^{\infty} A_i \in \mathcal{A}. $
A measurable space is a pair $(X, \mathcal{A})$ consisting of a set $X$ and a $\sigma$-algebra $\mathcal{A} \subset 2^X$.
The elements of a $\sigma$-algebra $\mathcal{A}$ are called measurable sets.
Every $\sigma$-algebra $\mathcal{A} \subset 2^X$ satisfies the following:
$\text{(d)} \quad$ $\emptyset \in \mathcal{A}.$
$\text{(e)} \quad$ If $n \in \mathbb{N}$ and $A_1, \dots, A_n \in \mathcal{A}$, then $ \bigcup_{i=1}^{n} A_i \in \mathcal{A}. $
$\text{(f)} \quad$ Every finite or countable intersection of elements of $\mathcal{A}$ is an element of $\mathcal{A}$, i.e., $ \bigcap_{i=1}^{\infty} A_i \in \mathcal{A}. $
$\text{(g)} \quad$ If $A, B \in \mathcal{A}$, then $ A \setminus B \in \mathcal{A}. $
Let $X$ be a set and $\mathcal{E} \subset 2^X$ be any set of subsets of $X$. Then there is a unique smallest $\sigma$-algebra $\mathcal{A} \subset 2^X$ containing $\mathcal{E}$ (i.e., $\mathcal{A}$ is a $\sigma$-algebra, $\mathcal{E} \subset \mathcal{A}$, and if $\mathcal{B}$ is any other $\sigma$-algebra with $\mathcal{E} \subset \mathcal{B}$, then $\mathcal{A} \subset \mathcal{B}$).
Proof.
Uniqueness follows directly from the definition. Namely, if $\mathcal{A}$ and $\mathcal{B}$ are two smallest $\sigma$-algebras containing $\mathcal{E}$, we have both $\mathcal{B} \subset \mathcal{A}$ and $\mathcal{A} \subset \mathcal{B}$ and hence $\mathcal{A} = \mathcal{B}$.
To prove existence, denote by $\mathcal{S} \subset 2^{2^X}$ the collection of all $\sigma$-algebras $\mathcal{B} \subset 2^X$ that contain $\mathcal{E}$ and define $$ \mathcal{A} := \bigcap_{\mathcal{B} \in \mathcal{S}} \mathcal{B} = \left\{ A \subset X \middle|\ \text{if } \mathcal{B} \subset 2^X \text{ is a } \sigma\text{-algebra such that } \mathcal{E} \subset \mathcal{B}, \text{ then } A \in \mathcal{B} \right\}. $$
Thus $\mathcal{A}$ is a $\sigma$-algebra. Moreover, it follows directly from the definition of $\mathcal{A}$ that $\mathcal{E} \subset \mathcal{A}$ and that every $\sigma$-algebra $\mathcal{B}$ that contains $\mathcal{E}$ also contains $\mathcal{A}$. This proves the lemma.
– The letter “$\sigma$” stands for “countable,” and the crucial observation is that axiom (c) allows for countable unions. On the one hand, this is much more general than only allowing for finite unions, which would be the subject of Boolean algebra. On the other hand, it is much more restrictive than allowing for arbitrary unions, which one encounters in topology.
Let $X$ be a set. A collection $\mathcal{U} \subset 2^X$ of subsets of $X$ is called a topology on $X$ if it satisfies the following axioms:
$\text{(a)} \quad$ $\emptyset, X \in \mathcal{U}.$
$\text{(b)} \quad$ If $n \in \mathbb{N}$ and $U_1, \dots, U_n \in \mathcal{U}$, then $ \bigcap_{i=1}^{n} U_i \in \mathcal{U}. $
$\text{(c)} \quad$ If $I$ is any index set and $U_i \in \mathcal{U}$ for $i \in I$, then $ \bigcup_{i \in I} U_i \in \mathcal{U}. $
A topological space is a pair $(X, \mathcal{U})$ consisting of a set $X$ and a topology $\mathcal{U} \subset 2^X$. If $(X, \mathcal{U})$ is a topological space, the elements of $\mathcal{U}$ are called open sets, and a subset $F \subset X$ is called closed if its complement is open, i.e., $F^c \in \mathcal{U}$. Thus finite intersections of open sets are open and arbitrary unions of open sets are open. Likewise, finite unions of closed sets are closed and arbitrary intersections of closed sets are closed.
Let $(X, \mathcal{U})$ be a topological space and let $\mathcal{B} \subset 2^X$ be the smallest $\sigma$-algebra containing $\mathcal{U}$. Then $\mathcal{B}$ is called the Borel $\sigma$-algebra of $(X, \mathcal{U})$ and the elements of $\mathcal{B}$ are called Borel (measurable) sets.
Let $(X, \mathcal{U})$ be a topological space. Then the following holds:
$\text{(i)} \quad$ Every closed subset $F \subset X$ is a Borel set.
$\text{(ii)} \quad$ Every countable union $\bigcup_{i=1}^{\infty} F_i$ of closed subsets $F_i \subset X$ is a Borel set (these are sometimes called $F_\sigma$-sets).
$\text{(iii)} \quad$ Every countable intersection $ \bigcap_{i=1}^{\infty} U_i$ of open subsets $U_i \subset X$ is a Borel set (these are sometimes called $G_\delta$-sets).
For any map $f : X \to Y$ between two sets $X$ and $Y$ and any subset $B \subset Y$, the pre-image $$ f^{-1}(B) := \{ x \in X \mid f(x) \in B \} $$ of $B$ under $f$ is a well-defined subset of $X$, whether or not the map $f$ is bijective, i.e., even if there does not exist any map $f^{-1} : Y \to X$.
The pre-image defines a map from $2^Y$ to $2^X$. It satisfies $$ f^{-1}(Y) = X, \qquad f^{-1}(\emptyset) = \emptyset, $$ and preserves union, intersection, and complement. Thus $$ f^{-1}(Y \setminus B) = X \setminus f^{-1}(B) $$ for every subset $B \subset Y$, and $$ f^{-1}\left( \bigcup_{i \in I} B_i \right) = \bigcup_{i \in I} f^{-1}(B_i), \qquad f^{-1}\left( \bigcap_{i \in I} B_i \right) = \bigcap_{i \in I} f^{-1}(B_i) $$ for every collection of subsets $B_i \subset Y$, indexed by a set $I$.
$\text{(i)} \quad$ Let $(X, \mathcal{A}_X)$ and $(Y, \mathcal{A}_Y)$ be measurable spaces. A map $f : X \to Y$ is called measurable if the pre-image of every measurable subset of $Y$ under $f$ is a measurable subset of $X$, i.e., $$ B \in \mathcal{A}_Y \Rightarrow f^{-1}(B) \in \mathcal{A}_X. $$
$\text{(ii)} \quad$ Let $(X, \mathcal{A}_X)$ be a measurable space. A function $f : X \to \overline{\mathbb{R}}$ is called measurable if it is measurable with respect to the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$ associated to the standard topology.
$\text{(iii)} \quad$ Let $(X, \mathcal{U}_X)$ and $(Y, \mathcal{U}_Y)$ be topological spaces. A map $f : X \to Y$ is called Borel measurable if the pre-image of every Borel measurable subset of $Y$ under $f$ is a Borel measurable subset of $X$.
Let $X$ be a set. The characteristic function of a subset $A \subset X$ is the function $\chi_A : X \to \mathbb{R}$ defined by $$ \chi_A(x) := \begin{cases} 1, & \text{if } x \in A, \\ 0, & \text{if } x \notin A. \end{cases} $$
Now assume $(X, \mathcal{A})$ is a measurable space, consider the Borel $\sigma$-algebra on $\mathbb{R}$, and let $A \subset X$ be any subset. Then $\chi_A$ is a measurable function if and only if $A$ is a measurable set.
Let $(X, \mathcal{A}_X)$, $(Y, \mathcal{A}_Y)$, and $(Z, \mathcal{A}_Z)$ be measurable spaces.
$\text{(i)} \quad$ The identity map $\mathrm{id}_X : X \to X$ is measurable.
$\text{(ii)} \quad$ If $f : X \to Y$ and $g : Y \to Z$ are measurable maps, then so is the composition $g \circ f : X \to Z$.
$\text{(iii)} \quad$ Let $f : X \to Y$ be any map. Then the set $$ f_* \mathcal{A}_X := \left\{ B \subset Y \ \middle| \ f^{-1}(B) \in \mathcal{A}_X \right\} $$ is a $\sigma$-algebra on $Y$, called the pushforward of $\mathcal{A}_X$ under $f$.
$\text{(iv)} \quad$ A map $f : X \to Y$ is measurable if and only if $\mathcal{A}_Y \subset f _* \mathcal{A}_X$.
Let $(X, \mathcal{A}_X)$ and $(Y, \mathcal{A}_Y)$ be measurable spaces. Assume $\mathcal{U}_Y \subset 2^Y$ is a topology on $Y$ such that $\mathcal{A}_Y$ is the Borel $\sigma$-algebra of $(Y, \mathcal{U}_Y)$.
$\text{(i)} \quad$ A map $f : X \to Y$ is measurable if and only if the pre-image of every open subset $V \subset Y$ under $f$ is measurable, i.e., $$ V \in \mathcal{U}_Y \qquad \Longrightarrow \qquad f^{-1}(V) \in \mathcal{A}_X. $$
$\text{(ii)} \quad$ Assume $\mathcal{U}_X \subset 2^X$ is a topology on $X$ such that $\mathcal{A}_X$ is the Borel $\sigma$-algebra of $(X, \mathcal{U}_X)$. Then every continuous map $f : X \to Y$ is (Borel) measurable.
Let $(X, \mathcal{A})$ be a measurable space and let $f : X \to \mathbb{R}$ be any function. Then the following are equivalent:
$\text{(i)} \quad$ $f$ is measurable.
$\text{(ii)} \quad$ $f^{-1}((a, \infty])$ is a measurable subset of $X$ for every $a \in \mathbb{R}$.
$\text{(iii)} \quad$ $f^{-1}([a, \infty])$ is a measurable subset of $X$ for every $a \in \mathbb{R}$.
$\text{(iv)} \quad$ $f^{-1}([-\infty, b))$ is a measurable subset of $X$ for every $b \in \mathbb{R}$.
$\text{(v)} \quad$ $f^{-1}([-\infty, b])$ is a measurable subset of $X$ for every $b \in \mathbb{R}$.
Next, we will show that sums, products, and limits of measurable functions are again measurable. The next two results are useful for the proofs of these fundamental facts.
Let $(X, \mathcal{A})$ be a measurable space and let $f = (f_1, \dots, f_n) : X \to \mathbb{R}^n$ be a function. Then $f$ is measurable if and only if $f_i : X \to \mathbb{R}$ is measurable for each $i$.
Let $(X, \mathcal{A})$ be a measurable space and let $u, v : X \to \mathbb{R}$ be measurable functions. If $\varphi : \mathbb{R}^2 \to \mathbb{R}$ is continuous, then the function $h : X \to \mathbb{R}$ defined by $$ h(x) := \varphi(u(x), v(x)) $$ for $x \in X$ is measurable.
Let $(X, \mathcal{A})$ be a measurable space.
$\text{(i)} \quad$ If $f, g : X \to \mathbb{R}$ are measurable functions, then so are the functions $$f + g,\qquad fg,\qquad \max\{f, g\},\qquad \min{f, g},\qquad |f|.$$
$\text{(ii)} \quad$ Let $f_k : X \to \overline{\mathbb{R}}$, $k = 1, 2, 3, \dots$, be a sequence of measurable functions. Then the following functions from $X$ to $\overline{\mathbb{R}}$ are measurable: $$ \inf_k f_k, \qquad \sup_k f_k, \qquad \limsup_{k \to \infty} f_k, \qquad \liminf_{k \to \infty} f_k. $$
Let $X$ be a set. A function $s : X \to \mathbb{R}$ is called a step function(or simple function) if it takes on only finitely many values, i.e., the image $s(X)$ is a finite subset of $\mathbb{R}$.
– Let $s : X \to \mathbb{R}$ be a step function. Write $ s(X) = \{\alpha_1, \dots, \alpha_\ell\}, \text{with } \alpha_i \neq \alpha_j \text{ for } i \neq j, $ and define $ A_i := s^{-1}(\alpha_i) = \{ x \in X \mid s(x) = \alpha_i \} \text{ for } i = 1, \dots, \ell. $ Then the sets $A_1, \dots, A_\ell$ form a partition of $X$, i.e., $$ X = \bigcup_{i=1}^{\ell} A_i, \quad A_i \cap A_j = \emptyset \text{ for } i \neq j. $$
Moreover, $$ s = \sum_{i=1}^{\ell} \alpha_i \chi_{A_i}, $$ where $\chi_{A_i} : X \to \mathbb{R}$ is the characteristic function of the set $A_i$ for $i = 1,\dots, \ell$. In this situation, $s$ is measurable if and only if each set $A_i \subset X$ is measurable.
Let $(X, \mathcal{A})$ be a measuable space and let $f : X \to [0, \infty]$ be a function. Then $f$ is measurable if and only if there exists a sequence of measurable step functions $s_n : X \to [0, \infty)$ such that $$ 0 \le s_1(x) \le s_2(x) \le \cdots \le f(x), \qquad f(x) = \lim_{n \to \infty} s_n(x) \qquad \text{for all } x \in X. $$
Proof.
If $f$ can be approximated by a sequence of measurable step functions, then $f$ is measurable. Conversely, suppose that $f$ is measurable. For $n \in \mathbb{N}$ define $\varphi_n : [0, \infty] \to \mathbb{R}$ by $$ \varphi_n(t) := \begin{cases} k 2^{-n}, & \text{if } k 2^{-n} \le t < (k+1)2^{-n}, \quad k = 0, 1, \dots, n2^n - 1, \\ n, & \text{if } t \ge n. \end{cases} $$
These functions are Borel measurable and satisfy $\varphi_n(0) = 0$ and $\varphi_n(\infty) = n$ for all $n$ as well as $ t - 2^{-n} \le \varphi_n(t) \le \varphi_{n+1}(t) \le t $ whenever $n \ge t > 0$. Thus $$ \lim_{n \to \infty} \varphi_n(t) = t \qquad \text{for all } t \in [0, \infty]. $$
Hence the functions $s_n := \varphi_n \circ f$ satisfy the requirements of the theorem.
Let $(X, \mathcal{A})$ be a measurable space. A measure on $(X, \mathcal{A})$ is a function $$ \mu : \mathcal{A} \to [0, \infty] $$ satisfying the following axioms:
$\text{(a)}$ $\mu$ is $\sigma$-additive, i.e., if $A_i \in \mathcal{A}$, $i = 1, 2, 3, \dots$, is a sequence of pairwise disjoint measurable sets, then $$ \mu\left( \bigcup_{i=1}^{\infty} A_i \right) = \sum_{i=1}^{\infty} \mu(A_i). $$
$\text{(b)}$ There exists a measurable set $A \in \mathcal{A}$ such that $\mu(A) < \infty$.
A measure space is a triple $(X, \mathcal{A}, \mu)$ consisting of a set $X$, a $\sigma$-algebra $\mathcal{A} \subset 2^X$, and a measure $\mu : \mathcal{A} \to [0, \infty]$.
Let $(X, \mathcal{A}, \mu)$ be a measure space. Then the following holds:
$\text{(i)}$ $\mu(\emptyset) = 0.$
$\text{(ii)}$ If $n \in \mathbb{N}$ and $A_1, \dots, A_n \in \mathcal{A}$ such that $A_i \cap A_j = \emptyset$ for $i \neq j$, then $$ \mu(A_1 \cup \cdots \cup A_n) = \mu(A_1) + \cdots + \mu(A_n). $$
$\text{(iii)}$ If $A, B \in \mathcal{A}$ such that $A \subset B$, then $\mu(A) \le \mu(B)$.
$\text{(iv)}$ Let $A_i \in \mathcal{A}$ be a sequence such that $A_i \subset A_{i+1}$ for all $i$. Then $$ \mu\left( \bigcup_{i=1}^{\infty} A_i \right) = \lim_{i \to \infty} \mu(A_i). $$
$\text{(v)}$ Let $A_i \in \mathcal{A}$ be a sequence such that $A_i \supset A_{i+1}$ for all $i$. Then $$ \mu(A_1) < \infty \qquad \Longrightarrow \qquad \mu\left( \bigcap_{i=1}^{\infty} A_i \right) = \lim_{i \to \infty} \mu(A_i). $$
1. Let $(X, \mathcal{A})$ be a measurable space and fix an element $x_0 \in X$. The Dirac measure at $x_0$ is the measure $\delta_{x_0} : \mathcal{A} \to [0, \infty]$ defined by $$ \delta_{x_0}(A) := \begin{cases} 1, & \text{if } x_0 \in A, \\ 0, & \text{if } x_0 \notin A, \end{cases} \qquad \text{for } A \in \mathcal{A}. $$
2. Let $X$ be an uncountable set and let $\mathcal{A}$ be the $\sigma$-algebra of all subsets of $X$ that are either countable or have countable complements. Then the function $\mu : \mathcal{A} \to [0, 1]$ defined by $\mu(A) := 0$ when $A$ is countable and $\mu(A) := 1$ when $A^c$ is countable is a measure.
3. Let $X = \bigcup\limits_{i \in I} A_i$ be a partition and let $\mathcal{A} \subset 2^X$ be the $\sigma$-algebra generated by this partition. Then any function $I \to [0, \infty]:$ $i \mapsto \alpha_i$ determines a measure $\mu : \mathcal{A} \to [0, \infty]$ via $$ \mu(A_J) := \sum_{j \in J} \alpha_j, $$ for $J \subset I$ and $ A_J = \bigcup\limits_{j \in J} A_j.$
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $E \in \mathcal{A}$ be a measurable set.
$\text{(i)} \quad$ Let $s : X \to [0, \infty)$ be a measurable step function of the form $$ s = \sum_{i=1}^{n} \alpha_i \chi_{A_i} $$ with $\alpha_i \in [0, \infty)$ and $A_i \in \mathcal{A}$ for $i = 1, \dots, n$. The (Lebesgue) integral of $s$ over $E$ is the number $\int_E s \mathrm{d}\mu \in [0, \infty]$ defined by $$ \int_E s \mathrm{d}\mu := \sum_{i=1}^{n} \alpha_i \mu(E \cap A_i). $$
$\text{(ii)} \quad$ Let $f : X \to [0, \infty]$ be a measurable function. The (Lebesgue) integral of $f$ over $E$ is the number $\int_E f \mathrm{d}\mu \in [0, \infty]$ defined by $$ \int_E f \mathrm{d}\mu := \sup_{s \le f} \int_E s \mathrm{d}\mu, $$ where the supremum is taken over all measurable step functions $s : X \to [0, \infty)$ that satisfy $s(x) \le f(x)$ for all $x \in X$.
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $f, g : X \to [0, \infty]$ be measurable functions and let $E \in \mathcal{A}.$ Then the following holds:
$\text{(i)} \quad$ If $f \le g$ on $E$, then $ \int_E f \mathrm{d}\mu \le \int_E g \mathrm{d}\mu. $
$\text{(ii)} \quad$ $ \int_E f \mathrm{d}\mu = \int_X f \chi_E \mathrm{d}\mu. $
$\text{(iii)} \quad$ If $f(x) = 0$ for all $x \in E$, then $ \int_E f \mathrm{d}\mu = 0.$
$\text{(iv)} \quad$ If $\mu(E) = 0$, then $ \int_E f \mathrm{d}\mu = 0. $
$\text{(v)} \quad$ If $A \in \mathcal{A}$ and $E \subset A$, then $ \int_E f \mathrm{d}\mu \le \int_A f \mathrm{d}\mu. $
$\text{(vi)} \quad$ If $c \in [0, \infty)$ then $\int_E cf \mathrm{d}\mu = c \int_E f \mathrm{d} \mu.$
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $s, t : X \to [0, \infty)$ be measurable step functions.
$\text{(i)} \quad$ For every measurable set $E \in \mathcal{A}$, $$ \int_E (s + t) \mathrm{d}\mu = \int_E s \mathrm{d}\mu + \int_E t \mathrm{d}\mu. $$
(ii) If $E_1, E_2, E_3, \dots$ is a sequence of pairwise disjoint measurable sets, then $$ \int_E s \mathrm{d}\mu = \sum_{k=1}^{\infty} \int_{E_k} s \mathrm{d}\mu, \qquad \text{where } E = \bigcup_{k \in \mathbb{N}} E_k. $$
Let $(X, \mathcal{A}, \mu)$ be a measure space.
$\text{(i)} \quad$ If $f, g : X \to [0, \infty]$ are measurable and $E \in \mathcal{A}$, then $$ \int_E (f + g) \mathrm{d}\mu = \int_E f \mathrm{d}\mu + \int_E g \mathrm{d}\mu. $$
$\text{(ii)} \quad$ Let $f_n : X \to [0, \infty]$ be a sequence of measurable functions and define $$ f(x):= \sum_{n=1}^{\infty} f_n(x) \qquad \text{for } x \in X. $$ Then $f : X \to [0, \infty]$ is measurable and, for every $E \in \mathcal{A}$, $$ \int_E f \mathrm{d}\mu = \sum_{n=1}^{\infty} \int_E f_n \mathrm{d}\mu. $$
$\text{(iii)} \quad$ If $f : X \to [0, \infty]$ is measurable and $E_1, E_2, E_3, \dots$ is a sequence of pairwise disjoint measurable sets, then $$ \int_E f \mathrm{d}\mu = \sum_{k=1}^{\infty} \int_{E_k} f \mathrm{d}\mu, \qquad E := \bigcup_{k \in \mathbb{N}} E_k. $$
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $f : X \to [0, \infty]$ be a measurable function. Then the function $\mu_f : \mathcal{A} \to [0, \infty]$, defined by $$ \mu_f(E) := \int_E f \mathrm{d}\mu \qquad \text{for } E \in \mathcal{A} $$ is a measure, and $$ \int_E g \mathrm{d}\mu_f = \int_E f g \mathrm{d}\mu $$ for every measurable function $g : X \to [0, \infty]$ and every $E \in \mathcal{A}$.
Let $(X, \mathcal{A}, \mu)$ be a measure space. A function $f : X \to \mathbb{R}$ is called (Lebesgue) integrable or $\mu$-integrable if $f$ is measurable and $\int_X |f| \mathrm{d}\mu < \infty. $ Denote the set of $\mu$-integrable functions by $$ \mathcal{L}^1(\mu) := \mathcal{L}^1(X, \mathcal{A}, \mu) := \{ f : X \to \mathbb{R} \mid f \text{ is } \mu\text{-integrable} \}. $$
The Lebesgue integral of $f \in \mathcal{L}^1(\mu)$ over a set $E \in \mathcal{A}$ is the real number $$ \int_E f \mathrm{d}\mu := \int_E f^+ \mathrm{d}\mu - \int_E f^- \mathrm{d}\mu $$ where the functions $f^\pm : X \to [0, \infty)$ are defined by $$ f^+(x) := \max\{f(x), 0\}, \qquad f^-(x) := \max\{-f(x), 0\}. $$
Let $(X, \mathcal{A}, \mu)$ be a measure space. Then the following holds:
$\text{(i)} \quad$ The set $\mathcal{L}^1(\mu)$ is a real vector space and, for every $E \in \mathcal{A}$, the function $ \mathcal{L}^1(\mu) \to \mathbb{R}: f \mapsto \int_E f \mathrm{d}\mu $ is linear. That is, if $f, g \in \mathcal{L}^1(\mu)$ and $c \in \mathbb{R}$, then $f + g, cf \in \mathcal{L}^1(\mu)$ and $$ \int_E (f + g) \mathrm{d}\mu = \int_E f \mathrm{d}\mu + \int_E g \mathrm{d}\mu, \qquad \int_E cf \mathrm{d}\mu = c \int_E f \mathrm{d}\mu. $$
$\text{(ii)} \quad$ For all $f, g \in \mathcal{L}^1(\mu)$ and all $E \in \mathcal{A}$, $$ f \le g \text{ on } E \qquad \Longrightarrow\ \qquad \int_E f \mathrm{d}\mu \le \int_E g \mathrm{d}\mu. $$
$\text{(iii)} \quad$ If $f \in \mathcal{L}^1(\mu)$, then $|f| \in \mathcal{L}^1(\mu)$ and, for all $E \in \mathcal{A}$, $$ \left| \int_E f \mathrm{d}\mu \right| \le \int_E |f| \mathrm{d}\mu. $$
$\text{(iv)} \quad$ If $f \in \mathcal{L}^1(\mu)$ and $E_1, E_2, E_3, \dots$ is a sequence of pairwise disjoint measurable sets, then $$ \int_E f \mathrm{d}\mu = \sum_{k=1}^{\infty} \int_{E_k} f \mathrm{d}\mu, \qquad E = \bigcup_{k \in \mathbb{N}} E_k. $$
$\text{(v)} \quad$ For all $E \in \mathcal{A}$ and all $f \in \mathcal{L}^1(\mu)$, $$ \int_E f \mathrm{d}\mu = \int_X f \chi_E \mathrm{d}\mu. $$
$\text{(vi)} \quad$ Let $E \in \mathcal{A}$ and $f \in \mathcal{L}^1(\mu)$. If $\mu(E) = 0$ or $f|_E = 0$, then $ \int_E f \mathrm{d}\mu = 0.$
Let $(X, \mathcal{A}, \mu)$ be a measure space, let $g : X \to [0, \infty)$ be an integrable function, and let $f_n : X \to \mathbb{R}$ be a sequence of integrable functions satisfying $$ |f_n(x)| \le g(x) \quad \text{for all } x \in X \text{ and } n \in \mathbb{N}, $$ and converging pointwise to $f : X \to \mathbb{R}$, i.e., $$ f(x) = \lim_{n \to \infty} f_n(x) \quad \text{for all } x \in X. $$
Then $f$ is integrable and, for every $E \in \mathcal{A}$, $$ \int_E f \mathrm{d}\mu = \lim_{n \to \infty} \int_E f_n \mathrm{d}\mu. $$
Assume throughout this section that $(X, \mathcal{A}, \mu)$ is a measure space. A set of measure zero (or null set) is a measurable set $N \in \mathcal{A}$ such that $\mu(N) = 0$.
Let $\mathscr{P}$ be a name for some property that a point $x \in X$ may have, or not have, depending on $x$. For example, if $f : X \to [0, \infty]$ is a measurable function on $X$, then $\mathscr{P}$ could stand for the condition $f(x) > 0$, or for the condition $f(x) = 0$, or for the condition $f(x) = \infty$. Or if $f_n : X \to \mathbb{R}$ is a sequence of measurable functions, the property $\mathscr{P}$ could stand for the statement “the sequence $f_n(x)$ converges”.
In such a situation, we say that $\mathscr{P}$ holds almost everywhere if there exists a set $N \subset X$ of measure zero such that every element $x \in X \setminus N$ has the property $\mathscr{P}$. It is not required that the set of all elements $x \in X$ that have the property $\mathscr{P}$ is measurable, although that may often be the case.
Let $f : X \to [0, \infty]$ be a measurable function. If $ \int_X f \mathrm{d}\mu < \infty $ then $f < \infty$ almost everywhere.
Proof.
Define $ N := \{x \in X \mid f(x) = \infty\} $ and $ h := \infty \chi_N.$ Then $h \le f$ and so $$ \infty \mu(N) = \int_X h \mathrm{d}\mu \le \int_X f \mathrm{d}\mu < \infty $$ Hence $\mu(N) = 0$.
Assume either that $f, g : X \to [0, \infty]$ are measurable functions that agree almost everywhere, or that $f, g : X \to \mathbb{R}$ are $\mu$-integrable functions that agree almost everywhere. Then $$ \int_A f \mathrm{d}\mu = \int_A g \mathrm{d}\mu \qquad \text{for all } A \in \mathcal{A}. $$
Assume either that $f : X \to [0, \infty]$ is measurable or that $f : X \to \mathbb{R}$ is $\mu$-integrable. Then the following are equivalent:
$\text{(i)}\quad$ $f = 0$ almost everywhere.
$\text{(ii)}\quad$ $\int_A f \mathrm{d}\mu = 0 $ for all $A \in \mathcal{A}$.
$\text{(iii)}\quad$ $ \int_X |f| \mathrm{d}\mu = 0.$
Let $f \in \mathcal{L}^1(\mu)$. Then $$ \left| \int_X f \mathrm{d}\mu \right| = \int_X |f| \mathrm{d}\mu $$ if and only if $f = |f|$ almost everywhere or $f = -|f|$ almost everywhere.
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $f_n : X \to \mathbb{R}$ be a sequence of $\mu$-integrable functions such that $$ \sum_{n=1}^{\infty} \int_X |f_n| \mathrm{d}\mu < \infty. $$
Then there is a set $N$ of measure zero and a function $f \in \mathcal{L}^1(\mu)$ such that $$ \sum_{n=1}^{\infty} |f_n(x)| < \infty \quad \text{and} \quad f(x) = \sum_{n=1}^{\infty} f_n(x) \qquad \text{for all } x \in X \setminus N, \tag{1} $$
and $$ \int_A f \mathrm{d}\mu = \sum_{n=1}^{\infty} \int_A f_n \mathrm{d}\mu \qquad \text{for all } A \in \mathcal{A}, \tag{2} $$
and $$ \lim_{n \to \infty} \int_X \left| f - \sum_{k=1}^{n} f_k \right| \mathrm{d}\mu = 0. \tag{3} $$
Proof.
Define $$ \varphi(x) := \sum_{k=1}^{\infty} |f_k(x)| $$ for $x \in X$. This function is measurable. Moreover, it follows from the Lebesgue Monotone Convergence Theorem that $$ \int_X \varphi \mathrm{d}\mu = \lim_{n \to \infty} \int_X \sum_{k=1}^{n} |f_k| \mathrm{d}\mu = \lim_{n \to \infty} \sum_{k=1}^{n} \int_X |f_k| \mathrm{d}\mu = \sum_{k=1}^{\infty} \int_X |f_k| \mathrm{d}\mu < \infty. $$
Hence the set $ N := \{ x \in X \mid \varphi(x) = \infty \} $ has measure zero, and $ \sum_{k=1}^{\infty} |f_k(x)| < \infty $ for all $x \in X \setminus N$. Define the function $f : X \to \mathbb{R}$ by $ f(x) := 0$ for $x \in N$ and by $$ f(x) := \sum_{k=1}^{\infty} f_k(x) \qquad \text{for } x \in X \setminus N. $$ Then $f$ satisfies (1).
Define the functions $g : X \to \mathbb{R}$ and $g_n : X \to \mathbb{R}$ by $$ g := \varphi \chi_{X \setminus N}, \qquad g_n := \sum_{k=1}^{n} f_k \chi_{X \setminus N}, \quad n \in \mathbb{N}. $$
These functions are measurable. Moreover, $ \int_X g \mathrm{d}\mu = \int_X \varphi \mathrm{d}\mu < \infty .$ Since $|g_n(x)| \le g(x)$ for all $n \in \mathbb{N}$ and $g_n$ converges pointwise to $f$, it follows from the Lebesgue Dominated Convergence Theorem that $f \in \mathcal{L}^1(\mu)$ and, for all $A \in \mathcal{A}$, $$ \int_A f \mathrm{d}\mu = \lim_{n \to \infty} \int_A g_n \mathrm{d}\mu = \lim_{n \to \infty} \int_A \sum_{k=1}^{n} f_k \mathrm{d}\mu = \sum_{n=1}^{\infty} \int_A f_n \mathrm{d}\mu. $$
Here the second step follows from $g_n = \sum_{k=1}^{n} f_k$ almost everywhere. The last step follows by interchanging sum and integral and proves (2).
To prove equation (3), note that $f - \sum_{k=1}^{n} f_k = f - g_n$ almost everywhere, that $f(x) - g_n(x)$ converges to zero for all $x \in X$, and that $ |f - g_n| \le |f| + g, $ where $|f| + g$ is integrable. Hence, by the Lebesgue Dominated Convergence Theorem, $$ \lim_{n \to \infty} \int_X \left| f - \sum_{k=1}^{n} f_k \right| d\mu = \lim_{n \to \infty} \int_X |f - g_n| \mathrm{d}\mu = 0. $$
This proves (3) and the theorem.
Let $(X, \mathcal{A}, \mu)$ be a measure space and let $f_n \in \mathcal{L}^1(\mu)$ be a sequence of integrable functions. Assume $(f_n)$ is a Cauchy sequence with respect to the $L^1$-norm, i.e., for every $\varepsilon > 0$ there exists $n_0 \in \mathbb{N}$ such that, for all $m, n \in \mathbb{N}$, $$ n, m \ge n_0 \qquad \Longrightarrow \qquad \int_X |f_n - f_m| \mathrm{d}\mu < \varepsilon. $$
Then there exists a function $f \in \mathcal{L}^1(\mu)$ such that $$ \lim_{n \to \infty} \int_X |f_n - f| \mathrm{d}\mu = 0. $$
Moreover, there exists a subsequence $(f_{n_i})$ that converges almost everywhere to $f$.
The discussion in Section 5 shows that sets of measure zero are negligible in the sense that the integral of a measurable function remains the same if the function is modified on a set of measure zero. Thus, also subsets of sets of measure zero can be considered negligible. However, such subsets need not be elements of our $\sigma$-algebra $\mathcal{A}$. It is sometimes convenient to form a new $\sigma$-algebra by including all subsets of sets of measure zero. This leads to the notion of a completion of a measure space $(X, \mathcal{A}, \mu)$.
A measure space $(X, \mathcal{A}, \mu)$ is called complete if $$ N \in \mathcal{A}, \quad \mu(N) = 0, \quad E \subset N \qquad \Longrightarrow \qquad E \in \mathcal{A}. $$
Let $(X, \mathcal{A}, \mu)$ be a measure space and define
$$
\mathcal{A}^* :=
\left\{
E \subset X \middle|
\text{there exist measurable sets } A, B \in \mathcal{A}
\text{ such that } A \subset E \subset B \text{ and } \mu(B \setminus A) = 0
\right\}.
$$
Then the following holds:
$\text{(i)} \quad$ $\mathcal{A}^*$ is a $\sigma$-algebra and $\mathcal{A} \subset \mathcal{A}^*$.
$\text{(ii)} \quad$ There exists a unique measure $\mu^* : \mathcal{A}^* \to [0, \infty]$ such that $$ \mu^*|_{\mathcal{A}} = \mu. $$
$\text{(iii)} \quad$ The triple $(X, \mathcal{A}^*, \mu^*)$ is a complete measure space. It is called the completion of $(X, \mathcal{A}, \mu)$.
$\text{(iv)} \quad$ If $f : X \to \mathbb{R}$ is $\mu$-integrable, then $f$ is $\mu^*$-integrable and, for $E \in \mathcal{A}$, $$ \int_E f \mathrm{d}\mu^* = \int_E f \mathrm{d}\mu. $$ This continues to hold for all $\mathcal{A}$-measurable functions $f : X \to [0, \infty]$.
$\text{(v)} \quad$ If $f^* : X \to \overline{\mathbb{R}}$ is $\mathcal{A}^*$-measurable, then there exists an $\mathcal{A}$-measurable function $f : X \to \overline{\mathbb{R}}$ such that the set $$ N^* := \{ x \in X \mid f(x) \ne f^*(x) \} \in \mathcal{A}^* $$ has measure zero, i.e. $ \mu^*(N^*) = 0.$
[1] Salamon, D. A. (n.d.). Measure and integration (preprint). https://people.math.ethz.ch/~salamon/PREPRINTS/measure.pdf
]]>$\quad$ We study representations of functions of inner products via embeddings into Hilbert spaces. While polynomials with nonnegative coefficients admit a single feature map realization, this approach fails when negative coefficients are present. To address this limitation, we introduce a two-map framework. Specifically, for any analytic function $f(x) = \sum_{k=0}^{\infty} a_k x^k$ with global convergence, we construct a Hilbert space $\mathcal{H}$ and transformations $\Phi, \Psi: \mathbb{R}^n \to \mathcal{H}$ such that $$\langle \Phi(u), \Psi(v) \rangle = f(\langle u, v \rangle) \quad \text{for all } u, v \in \mathbb{R}^n.$$
We start with the following problem:
Let $P(x)$ be a polynomial with nonnegative coefficients, then there exists a Hilbert space $\mathcal{H}$ and a transformation $\Phi : \mathbb{R}^n \to \mathcal{H}$ such that: $$\langle \Phi(u), \Phi(v) \rangle = P(\langle u, v \rangle) \quad \text{for all } u, v \in \mathbb{R^n}.$$ But our task can be impossible if $P$ has a negative coefficient since $\langle \Phi(u), \Phi(v) \rangle$ is always nonnegative. Therefore, a natural question is what if we consider two transformations $\Phi, \Psi: \mathbb{R}^n \to \mathcal{H}?$ This leads us to a more general problem about analytic function as follow.
Consider a function $$ f(x) = \sum_{k=0}^{\infty} a_k x^k $$ where the series absolutely converges for all $x \in \mathbb{R}$. There exist a Hilbert space $\mathcal{H}$ and transformations $\Phi, \Psi : \mathbb{R}^n \to \mathcal{H}$ such that $$ \langle \Phi(u), \Psi(v) \rangle = f(\langle u, v \rangle) \quad \text{for all } u, v \in \mathbb{R}^n. $$ Also, for any unit vector $u$, we have $$ \|\Phi(u)\|^2_\mathcal{H} = \|\Psi(u)\|^2_\mathcal{H} =\sum _{k=0}^{\infty} |a_k| .$$ Proof.
For each $k \ge 0$, let $V_k = \mathbb{R}^{n^k}$ and define $$ H = \bigoplus_{k=0}^{\infty} V_k = \left\{ (w_0, w_1, \dots) : w_k \in V_k,\ \sum_{k=0}^{\infty} \|w_k\|^2 < \infty \right\} $$
with inner product $$ \langle w, z \rangle_\mathcal{H} = \sum_{k=0}^{\infty} \langle w_k, z_k \rangle_{V_k}. $$
Consider $s_k = \operatorname{sign}(a_k)$ and write $a_k = s_k |a_k|$. For $u \in \mathbb{R}^n$, define $$ \Phi(u) = (\sqrt{|a_0|} u^{\otimes 0}, \sqrt{|a_1|} u^{\otimes 1}, \dots), $$
$$ \Psi(v) = (s_0 \sqrt{|a_0|} v^{\otimes 0}, s_1 \sqrt{|a_1|} v^{\otimes 1}, \dots). $$
Then we prove that $\Phi(u), \Psi(v) \in \mathcal{H}$. Indeed, we have: $$ \left\|\sqrt{|a_k|} u^{\otimes k}\right\|^2 = |a_k| \left\|u^{\otimes k}\right\|^2 = |a_k| \|u\|^{2k}. $$
Thus, $$ \sum_{k=0}^{\infty}\left\|\sqrt{|a_k|} u^{\otimes k}\right\|^2 = \sum_{k=0}^{\infty} |a_k| \|u\|^{2k} $$ converges, so $\Phi(u) \in \mathcal{H}$. Similarly, $\Psi(v) \in \mathcal{H}$.
On the other hand, $$ \begin{aligned} \langle \Phi(u), \Psi(v) \rangle_\mathcal{H} &= \sum_{k=0}^{\infty} \left\langle \sqrt{|a_k|} u^{\otimes k}, s_k \sqrt{|a_k|} v^{\otimes k} \right\rangle_{V_k} \\ &= \sum_{k=0}^{\infty} s_k |a_k| \langle u^{\otimes k}, v^{\otimes k} \rangle\\ &= \sum_{k=0}^{\infty} a_k \langle u, v \rangle^k \\ &= f(\langle u, v \rangle). \end{aligned} $$
Note. Norm when $u$ is a unit vector: If $\|u\| = 1$, then $$ \|u^{\otimes k}\|^2 = \|u\|^{2k} = 1. $$
Hence, $$\|\Phi(u)\|_\mathcal{H}^2 = \sum _{k=0}^{\infty} |a_k| \| u^{\otimes k} \|^2 = \sum _{k=0}^{\infty} |a_k|.$$
Sine. Let $c > 0$. Then function $ f(x) = \sin(cx) $ is real analytic $$ \sin(cx) = cx - \frac{(cx)^3}{3!} + \frac{(cx)^5}{5!} - \ldots $$
Thus, there exists a Hilbert space $\mathcal{H}$ and transformation $\Phi, \Psi: \mathbb{R}^n \to \mathcal{H}$ such that $$ \langle \Phi(u), \Psi(v) \rangle = \sin(c \langle u, v \rangle)\quad \text{for all } u, v \in \mathbb{R}^n. $$
Also, $\Phi$ and $\Psi$ map unit vectors to unit vectors if $$ 1 = c + \frac{c^3}{3!} + \frac{c^5}{5!} + \ldots = \frac{e^c - e^{-c}}{2}. $$
Hence, $$ c = \ln(1 + \sqrt{2}). $$
Given a function of two variables $ K : \mathcal{X} \times \mathcal{X} \to \mathbb{R},$ on some set sets $\mathcal{X}$, when can we find a Hilbert space $\mathcal{H}$ and a transformation $\Phi : \mathcal{X} \to \mathcal{H}$ so that
1. The kernel trick, which expresses a kernel $K(u, v)$ as an inner product, is widely used in machine learning because it lets us handle non-linear models (determined by $K$) with techniques designed for linear models.$$ \langle \Phi(u), \Phi(v) \rangle = K(u,v)^{\textcolor{gray} 1} \quad \text{for all } u, v \in \mathcal{X}? $$
The answer is given by Mercer theorem2 2. Functions of positive and negative type and their connection with the theory of integral equations (Mercer, 1909). and, more precisely, Moore-Aronszajn theorem:
$-$ The necessary and sufficient condition: $K$ is a positive semidefinite kernel, i.e., for any $u_1, \dots, u_N \in \mathcal{X}$, the matrix $ (K(u_i, u_j))_{i,j=1}^N $ is symmetric and positive semidefinite.
$-$ Transformation $\Phi$ is called a feature map.
$-$ The Hilbert space $\mathcal{H}$ is called a reproducing kernel Hilbert space3 (RKHS). 3. Reproducing Kernel Hilbert Space, Mercer’s Theorem, Eigenfunctions, Nyström Method, and Use of Kernels in Machine Learning: Tutorial and Survey (Ghojogh et al, 2021).
$-$ Popular positive semidefinite kernels in machine learning include the Gaussian and polynomial kernels given by: $$ K(u,v) = \exp \left( -\frac{\|u - v\|^2}{2\sigma^2} \right), \quad K(u,v) = (\langle u, v \rangle + r)^k, \quad u, v \in \mathbb{R}^n. $$ where $\sigma > 0, r > 0,$ and $k \in \mathbb{N}$ are parameters.
[1] Vershynin, R. (2026). High-dimensional probability: An introduction with applications in data science (2nd ed.). Cambridge University Press.
]]>– An inner product space is a vector space $X$ with an inner product defined on $X$. A Hilbert space is a complete inner product space. An inner product on $X$ is a mapping of $X \times X$ into the scalar field $K$ of $X$; that is, with every pair of vectors $x$ and $y$ there is associated a scalar which is written $\langle x,y\rangle$. For all vectors $x, y, z$ and scalar $\alpha,$ we have
$ (\mathrm{IP}1)\quad \langle x+y,z\rangle = \langle x,z\rangle + \langle y,z\rangle $
$ (\mathrm{IP}2)\quad \langle \alpha x,y\rangle = \alpha \langle x,y\rangle $
$ (\mathrm{IP}3)\quad \langle x,y\rangle = \overline{\langle y,x\rangle} $
$ (\mathrm{IP}4)\quad \langle x,x\rangle \ge 0, \qquad \langle x,x\rangle = 0 \iff x = 0 $
An inner product on $X$ defines a norm on $X$ given by $$ \|x\| = \sqrt{\langle x,x\rangle} $$
and a metric on $X$ given by $$ d(x,y) = \|x-y\| = \sqrt{\langle x-y, x-y\rangle} $$
From (IP1) and (IP3), we obtain
$ \text{(a)}\quad \langle \alpha x + \beta y,z\rangle = \alpha \langle x,z\rangle + \beta \langle y,z\rangle $
$ \text{(b)}\quad \langle x,\alpha y\rangle = \overline{\alpha},\langle x,y\rangle $
$ \text{(c)}\quad \langle x,\alpha y + \beta z\rangle = \overline{\alpha},\langle x,y\rangle + \overline{\beta},\langle x,z\rangle $
– An inner product space satisfies the parallelogram equality $$ \|x+y\|^2 + \|x-y\|^2 = 2(\|x\|^2 + \|y\|^2) $$
– An element $x$ of an inner product space $X$ is said to be orthogonal to an element $y \in X$ if $$ \langle x,y\rangle = 0. $$
– We also say that $x$ and $y$ are orthogonal, and we write $x \perp y$. Similarly, for subsets $A, B \subset X$ we write $x \perp A$ if $x \perp a$ for all $a \in A$. $A \perp B$ if $a \perp b$ for all $a \in A$ and $b \in B$.
1. Euclidean space $\mathbb{R}^n$. Hilbert space with inner product: $$ \langle x,y\rangle = \xi_1 \eta_1 + \ldots + \xi_n \eta_n $$ where $x = (\xi_i) = (\xi_1,\ldots,\xi_n); \quad y = (\eta_i) = (\eta_1,\ldots,\eta_n)$. Then $$\|x\| = \langle x,x\rangle^{1/2} = (\xi_1^2 + \ldots + \xi_n^2)^{1/2}$$ with Euclidean metric $$\mathrm{d}(x, y) = \| x - y \|| = \langle x-y, x-y \rangle^{1/2} = [(\xi_1 - \eta_1)^2 + \ldots +(\xi_n - \eta_n)^2]^{1/2}$$
2. Unitary space $\mathbb{C}^n$. Hilbert space with inner product: $$ \langle x,y\rangle = \sum_ {i=1}^n \xi_i \overline{\eta_i} $$ The norm is defined by $$ \|x\| = \left(\sum_{i=1}^n \xi_i \overline{\xi_i}\right)^{1/2} = \left(\sum_{i=1}^n |\xi_i|^2\right)^{1/2} $$
3. Space $L^2[a,b]$. Hilbert space with norm: $$ \|x\| = \left(\int_a^b x(t)^2\ \mathrm{d}t\right)^{1/2} $$ and can be obtained from the inner product defined by $$ \langle x,y\rangle = \int_a^b x(t)y(t)\mathrm{d}t $$ Consider complex-valued functions $\Rightarrow$ complex vector space $\Rightarrow$ inner product space: $$ \langle x,y\rangle = \int_a^b x(t)\overline{y(t)}\mathrm{d}t $$
4. Hilbert sequence space $\ell^2$. Hilbert space with inner product: $$ \langle x,y\rangle = \sum_{j=1}^{\infty} \xi_j \overline{\eta_j} $$ The norm is defined by $$ \|x\| = \langle x,x\rangle^{1/2} = \left(\sum_{j=1}^{\infty} |\xi_j|^2\right)^{1/2} $$
$\text{(a)}$ We have: $$ \|\langle x,y\rangle\| \le \|x\|\ \|y\| $$ The equality sign holds if and only if $\{x,y\}$ is a linearly dependent set.
$\text{(b)}$ The norm also satisfies: $$ \|x+y\| \le \|x\| + \|y\| $$ where the equality sign holds if and only if $y = 0$ or $x = cy$ $(c \in \mathbb{R},\ c \ge 0)$.
If in an inner product space, $x_n \to x$ and $y_n \to y$, then $\langle x_n,y_n\rangle \to \langle x,y\rangle$.
Proof. \begin{align} |\langle x_n,y_n\rangle - \langle x,y\rangle| &= |\langle x_n,y_n\rangle - \langle x_n,y\rangle + \langle x_n,y\rangle - \langle x,y\rangle| \\ & \le |\langle x_n,y_n - y\rangle| + |\langle x_n - x,y\rangle| \\ & \le \|x_n\|\|y_n - y\| + \|x_n - x\|\|y\| \to 0 \end{align}
An isomorphism $T$ of an inner product space $X$ onto an inner product space $\tilde X$ over the same field is a bijective linear operator $T:X \to \tilde X$ which preserves the inner product: $$ \langle Tx,Ty\rangle = \langle x,y\rangle \quad \text{for all } x,y \in X. $$
For any inner product space $X$ there exists a Hilbert space $H$ and an isomorphism $A$ from $X$ onto a dense subspace $W \subset H$. The space $H$ is unique except for isomorphisms.
– $H$ is unique except for isometries, that is, two completions $H$ and $\tilde H$ of $X$ are related by an isometry $T:H \to \tilde H$. Then, $T$ must be an isomorphism of the Hilbert space $H$ onto the Hilbert space $\tilde H$.
– A subspace $Y$ of an inner product space $H$ is defined to be a subspace of $H$, regarded as an inner product space.
$$ !!\quad Y \text{ need not to be a Hilbert space because } Y \text{ may not be complete.} $$
Let $Y$ be a subspace of a Hilbert space $H$. Then:
$\text{(a)} \quad$ $Y$ is complete if and only if $Y$ is closed in $H$.
$\text{(b)} \quad$ If $Y$ is finite dimensional, then $Y$ is complete.
$\text{(c)} \quad$ If $H$ is separable, so is $Y$. More generally, every subset of a separable inner product space is separable.
In a metric space $X$, the distance $\delta$ from an element $x \in X$ to a nonempty subset $M \subset X$ is defined to be: $$ \delta = \inf_{\tilde{y} \in M} d(x,\tilde y). $$
In a normed space this becomes: $$ \delta = \inf_{\tilde y \in M} \|x - \tilde y\|. $$
[1] Kreyszig, E. (1978). Introductory functional analysis with applications. John Wiley & Sons.
]]>$\quad$ Amirali Fatehizadeh & Daniel Yaqubi answered the question when does we have the average of first $n$ Fibonacci numbers are integer. In this post, I will generalize that result.
Given two positive integers $a$ and $m$ such that $a-2$ is not divisible by $4$. Let the sequence $(S_n)_{n \in \mathbb{N}}$ defined by $S_0 = 0, S_1 = 1$ and $S_{n+2} = aS_{n+1} - S_n$ for all $n \in \mathbb{N}.$ Then, for each $m$, there exists a positive integer $N$ has $m$ distinct prime divisors such that $$n \mid S_1 + S_2 + \ldots + S_N$$
We consider the sequence $(T_n)_{n \in \mathbb{N}}$ defined by $T_0 = 2, T_1 = a$ and $T_{n+2} = aT_{n+1}-T_n$. As a result, we have $$ S_n = \dfrac{\alpha^n-\beta^n}{\alpha-\beta} \quad \text{ and } \quad T_n = \alpha^n + \beta^n $$ where $\alpha \ge \beta$ are two roots of equation $x^2-ax+1 = 0.$
As a consequence, we have the following corollary:
Corollary. For a prime $p$ such that $p \mid S_{k}$ then $p^{n+1} \mid S_{kp^n}$ for all $n \in \mathbb{N}.$
Consider the case $m = 1.$ Then, we only need to choose $N = p$ where $p$ is a prime divisor of $\Delta$ then of course, we have $$p \mid S_{p - (\frac{\Delta}{p})} = S_p.$$
Now, we consider the case $m \geq 2.$
Then, we denote $p_1,p_2,\dots, p_m$ are $m$ first primes and we choose $N = 2M$ where $M = (p_1p_2\cdots p_m)^{p_m}$ which give ours the following result: $$ \begin{align*} S_1+\ldots+S_n &= \sum_{i = 1}^n \dfrac{\alpha^n-\beta^n}{\alpha-\beta} \\ &= \dfrac{\dfrac{ \alpha^{n+1}-1}{\alpha-1} - \dfrac{\beta^{n+1}-1}{\beta-1}}{\alpha-\beta} \\ &= \dfrac{(\beta-1)(\alpha^{n+1}-1)-(\alpha-1)(\beta^{n+1}-1)}{(\alpha-\beta)(\alpha-1)(\beta-1)} \\ &= \dfrac{\alpha^n-\beta^n+\alpha-\beta-(\alpha^{n+1}-\beta^{n+1})}{(\alpha-\beta)(2-a)} \\ &= \dfrac{S_{n+1}-1-S_{n}}{a-2} \\ &= \dfrac{S_M(T_{M+1}-T_M)}{a-2}. \end{align*} $$ Note that $T_0 \equiv T_1 \equiv 2 \pmod{a-2}$ so by induction, $T_n \equiv 2 \pmod{a-2}$ for all nonnegative integer $n$. Therefore, $S_1+\ldots+S_n$ is divisible by $n$ is equivalent to $2M \mid S_M$.
Now, consider $p_1 = 2$ then since $a-2$ is not divisible by 4 so $S_2=a$ or $S_3=a^2-1$ is divisible by 4. From this and following lemma 3, we derive that $2^{p_m+1} \mid S_M$. On the other hand, we fix an index $i \ge 2$ then there must exist nonnegative integer numbers $\alpha_{j,i}$ less than $p_m$ for all $j = \overline{1,i}$ such that $$p_i - \left( \frac{\Delta}{p} \right) = p_1^{\alpha_{1,i}} p_2^{\alpha_{2,i}}\cdots p_{i-1}^{\alpha_{i-1,i}} p_i^{\alpha_{i,i}}$$ with note that $\alpha_{i,i} \le 1$. Therefore following from lemma, we see that $$ p_i \mid S_{p - \left(\frac{\Delta}{p}\right)} = S_{p_1^{\alpha_{1,i}}p_2^{\alpha_{2,i}}\cdots p_{i-1}^{\alpha_{i-1,i}}p_i^{\alpha_{i,i}}} $$ so $$ p_i^{p_m} \mid S_{p_1^{\alpha_{1,i}}\cdot p_2^{\alpha_{2,i}}\cdots p_{i-1}^{\alpha_{i-1,i}}p_i^{\alpha_{i,i}+p_m-1}} \mid S_{(p_1p_2\ldots p_m)^{p_m}} = S_K $$
Therefore, we derive that $$2M = 2(p_1p_2\cdots p_m)^{p_m} \mid S_M$$ as desired.
I generalized the reference problem by considering the extension of algebraic number field. On the other hand, on the aforementioned paper, Amirali Fatehizadeh & Daniel Yaqubi also conjectured that for infinitely many odd numbers $n$ also satisfied the problem. This was proved recently by Oisın Flynn-Connolly. Therefore, if I have time, I will consider to extend the result for odd positive integer in a similar way.
[1] Fatehizadeh, A., & Yaqubi, D. (2022). Average of the Fibonacci Numbers. Journal of Integer Sequences, 25(2), Article 22.2.6, 10 pp. Retrieved from https://cs.uwaterloo.ca/journals/JIS/VOL25/Yaqubi/yaq6.html
[2] Flynn-Connolly, O. (2025). On the divisibility of sums of Fibonacci numbers (arXiv:2504.09938). arXiv. https://arxiv.org/abs/2504.09938
]]>A normed space $X$ is a vector space with a norm defined on it. A Banach space is a complete normed space (complete in the metric defined by the norm). Here, a norm on a (real or complex) vector space $X$ is a real-valued function on $X$ whose value at an $x \in X$ is denoted by $\| x \|$ (real “norm of $X$”) and which has the properties:
A norm on $X$ defines a metric $d$ on $X$ by $$ d(x,y) = \|x - y\|, $$ called the metric induced by the norm. The normed space thus defined is denoted by $(X, \| \cdot\|)$, or simply $X$.
1. Euclidean space $\mathbb{R}^n$ and unitary space $\mathbb{C}^n$. Banach space with norm $$ \|x\| = \left( \sum_{i=1}^{n} |x_i|^2 \right)^{\frac{1}{2}} $$ $\mathbb{R}^n$ and $\mathbb{C}^n$ are complete and the induced metric is $$ d(x,y) = \|x - y\| = \left( \sum_{i=1}^{n} |x_i - y_i|^2 \right)^{\frac{1}{2}}. $$
2. Space $\ell^p$.
Banach space with norm
$$ \|x\| = \left( \sum_{i = 1}^{\infty} |x_i|^p \right)^{\frac{1}{p}} $$
and the induced metric is
$$
d(x, y) = \|x - y\| = \left( \sum_{i=1}^{n} |x_i - y_i|^p \right)^{\frac{1}{p}}.
$$
3. Space $\ell^{\infty}$.
Banach space with norm
$$
\|x\|_\infty = \sup _{i}\|x_i\|
$$
4 Space $C[a, b]$.
Banach space with norm
$$
\|x\| = \max _{t \in J}\|x(t)\|
$$
where $J = [a,b]$.
5. Incomplete normed space.
$d(x, y) = \int _{0}^{1} | x(t) - y(t) | \mathrm{d}t$ induce the norm $$\int _{0}^{1} |x(t)| \mathrm{d}t$$
6. Incomplete normed space and its completion $L^p[a,b]$. The Banach space $L^p[a,b]$ with norm $$ \|x\| _p = \left( \int _{a}^{b} |x(t)| \mathrm{d}t \right)^{\frac{1}{p}} $$
7. Space $s$. Metric defined by $$d(x,y) = \sum _{j = 1}^{\infty} \dfrac{1}{2^j} \dfrac{|x_j-y_j|}{1 + |x_j-y_j|}$$ cannot be obtained a norm
A metric $d$ induced by a norm on a normed space $X$ satisfies
$\text{(a)} \quad$ $d(x+a,y+a) = d(x,y)$
$\text{(b)} \quad$ $d(\alpha x, \alpha y) = |\alpha| d(x,y)$
for all $x,y \in X$ and every scalar $\alpha$.
A subspace $Y$ of a Banach space $X$ is complete if and only if $Y$ is closed in $X$.
– Convergence of sequences
Then we write $x_n \to x$ and call $x$ the limit of $(x_n)$.
– Infinite series $(x_k)$ is a sequence in a normed space $X$, we can associate with $(x_k)$ the sequence $(s_n)$ of partial sums $$s_n = x_1 + x_2 + \ldots + x_n.$$
If $(s_n)$ is convergent, $ s_n \to s, $ or $\|s_n - s\| \to 0$ then the infinite series $\sum_{k = 1}^\infty x_k$ is said to converge, $s$ is called the sum of the series and we write $$ s = \sum_{k=1}^{\infty} x_k = x_1 + x_2 + \ldots. $$
If $ \sum_{k=1}^{\infty} \|x_k\| $ converges, then the series is said to be absolutely convergent.
– Basis. A normed space $X$ contains a sequence $(e_n)$ with the property that for every $x \in X$ here is a unique sequence of scalars $(\alpha_n)$ such that $$ \|x - \sum_{k=1}^{n} \alpha_k e_k\| \to 0 \quad \text{ (as } n \to \infty) $$ then $(e_n)$ is called a Schander basis for $X$. The series $\sum_{k=1}^{\infty} \alpha_k e_k$ which has the sum $x$ is then called the expansion of $x$ with to $(e_n)$, and we write $$x = \sum_{k=1}^{\infty} \alpha_k e_k$$
$\ell^p$ has a Schauder, namely $(e_n),$ where $e_n = (\delta_{nj})$, that is, $e_n$ is the sequence whose $n^{\text{th}}$ term is 1 and all other term are zero; thus $$ e_1 = (1,0,0,\dots), \quad e_2 = (0,1,0,\dots), \quad e_3 = (0,0,1,\dots), \ \text{etc.} $$
Let $X = (X, \|\cdot\|)$ be a normed space. Then there exists a Banach space $\tilde{X}$ and an isometry $A$ from $X$ onto a subspace $W$ of $\tilde{X}$ which is dense in $\tilde{X}$. The space $\tilde{X}$ is unique, except for isometris.
Let $\{ x_1, \ldots, x_n \}$ be a linearly independent set of vectors in a normed space $X$ (of any dimension). Then there is a number $c > 0$ such that for every choice of scalars $\alpha_1, \ldots, \alpha_n$ we have
\begin{equation}
\| \alpha_1x_1 + \ldots + \alpha_n x_n \| \ge c(| \alpha_1 | + \ldots + |\alpha_n|) \tag{1}
\end{equation}
Proof.
We write $S = |a_1| + \cdots + |a_n|$. If $S = 0$, all $\alpha_j$ are zero, so (1) holds for any $c$. Let $S > 0$. Then (1) is equivalent to the inequality which we obtain from (1) by dividing $S$ and writing $\beta_j = \dfrac{\alpha_j}{S}$, that is
\begin{equation} \left\| \beta_1 x_1 + \cdots + \beta_n x_n \right\| \ge c \qquad \left( \sum_{j=1}^n |\beta_j| = 1 \right) \tag{2} \end{equation}
Suppose for the contradiction, there exists a sequence $(y_m)$ of vectors
$$ y_m = \beta_1^{(m)} x_1 + \cdots + \beta_n^{(m)} x_n $$
such that $ |y_m| \to 0 \quad \text{as } m \to \infty. $
Since $ \sum_{j=1}^n \left|\beta_j^{(m)}\right| = 1, $ we have $\left|\beta_j^{(m)}\right| \le 1$. Hence for each $j$, the sequence $$ \left(\beta_j^{(m)}\right) = \left(\beta_j^{(1)}, \beta_j^{(2)}, \dots\right) $$ is bounded. Consequently, by the Bolzano-Weierstrass theorem, $\left(\beta_j^{(m)}\right)$ has a convergent subsequence. Let $\beta_1$ denote the limit of that subsequence, let $(y_{1,m})$ denote the corresponding subsequence of $(y_m)$. By the same argument, $(y_{1,m})$ has a subsequence $(y_{2,m})$ for which the corresponding subsequence of scalars $\beta_j^{(m)}$ converges; let $\beta_2$ denote the limit. Continuing in this way, after $n$ steps we obtain a subsequence
$$ (y_{n,m}) = (y_{n,1}, y_{n,2}, \dots) $$
of $(y_m)$ whose terms are of the form
$$ y_{n,m} = \sum_{j=1}^n \gamma_j^{(m)} x_j. $$
with scalars $\gamma_j^{(m)}$ satisfying $ \gamma_j^{(m)} \to \beta_j \text{ as } m \to \infty. $ Hence, as $m \to \infty$, $ y_{n,m} \to y = \sum_{j=1}^n \beta_j x_j $
where $ \sum_{j=1}^n |\beta_j| = 1, $ so that not all $\beta_j$ can be zero. Since $\{x_1,\dots,x_n\}$ is a linearly independent then $y \ne 0$.
On the other hand, $y_{n,m} \to y$ implies $\|y_{n,m}\| \to \|y\|$. Since $\|y_m\| \to 0$ by assumption and $(y_{n,m})$ is a subsequence of $(y_m)$, we must have $\|y_{n,m}\| \to 0$. Hence $\|y\| = 0$ so that $y = 0$ (contradiction).
Every finite dimensional subspace $Y$ of a normed space $X$ is complete. In particular, every finite dimensional normed space is complete.
Proof.
Consider an arbitrary Cauchy sequence $(y_m)$ in $Y$ and show that it is convergent in $Y$; the limit will be denoted by $y$.
Let $\dim Y = n$ and $\{e_1,\ldots,e_n\}$ any basis for $Y$. Then each $y_m$ has a unique representation of the form
$$ y_m = \alpha_1^{(m)} e_1 + \cdots + \alpha_n^{(m)} e_n. $$
Since $(y_m)$ is a Cauchy sequence, for every $\varepsilon > 0$ there is an $N$ such that $\|y_m - y_r\| < \varepsilon$ when $m,r > N$. We have for some $c > 0$
$$ \varepsilon > \|y_m - y_r\| = \left\| \sum_{j=1}^n \left(\alpha_j^{(m)} - \alpha_j^{(r)}\right)e_j \right\| \ge c \sum_{j=1}^n \left|\alpha_j^{(m)} - \alpha_j^{(r)}\right| $$
$$ \Rightarrow \left|a_j^{(m)} - a_j^{(r)}\right| \le \frac{\varepsilon}{c}. $$
This shows that each of the $n$ sequences
$$ \left(\alpha_j^{(m)}\right)=\left(\alpha_j^{(1)},\alpha_j^{(2)},\ldots\right), \qquad j=1,\ldots,n $$
is Cauchy in $\mathbb{R}$ or $\mathbb{C}$. Hence it converges; let $\alpha_j$ denote the limit. Using these $n$ limits $a_1,\ldots,a_n$, we define
$$ y=\alpha_1e_1+\cdots+\alpha_ne_n. $$
Clearly, $y\in Y$ and
$$ \|y_m-y\| = \left\|\sum_{j=1}^n \left(\alpha_j^{(m)}-\alpha_j\right)e_j\right\| \le \sum_{j=1}^n \left|\alpha_j^{(m)}-\alpha_j\right|\|e_j\|. $$
On the right, $\alpha_j^{(m)}\to \alpha_j$. Hence $\|y_m-y\|\to 0$, that is $y_m\to y$.
This shows that $(y_m)$ is convergent in $Y$. Since $(y_m)$ was an arbitrary Cauchy sequence in $Y$, this proves that $Y$ is complete.
Every finite dimensional subspace $Y$ of a normed space $X$ is closed in $X$.
A norm $\|\cdot\|$ on a vector space $X$ is said to be equivalent to a norm $|\cdot|_0$ on $X$ if there are positive numbers $a$ and $b$ such that for all $x\in X$ we have
$$ a\|x\|_0 \le \|x\| \le b\|x\|_0. $$
Equivalent norms on $X$ define the same topology for $X$.
On a finite dimensional vector space $X$ any norm $\|\cdot\|$ is equivalent to any other norm $\|\cdot\|_0$.
Proof.
Let $\dim X=n$ and ${e_1,\ldots,e_n}$ any basis for $X$. Then every $x\in X$ has a unique representation
$$ x=\alpha_1e_1+\cdots+\alpha_ne_n. $$
Then there exists $c>0$ such that
$$ \|x\|\ge c\left(|a_1|+\cdots+|a_n|\right). $$
On the other hand, the triangle inequality gives
$$ \|x\|_0 \le \sum _{j = 1}^n \left|\alpha_j \right|\|e_j\|_0 \le k \sum _{j=1}^n|\alpha_j| \qquad k = \max_j \|e_j\|_0. $$
Hence $a\|x\|_0 \le \|x\|,$ where $ a=\frac{c}{k}>0.$ Interchange of the roles of $\|\cdot\|$ and $\|\cdot\|_0$, we also have $ \|x\|\le b\|x\|_0, $ so these norms are equivalent.
A metric space $X$ is said to be compact if every sequence in $X$ has a convergent subsequence. A subset $M$ of $X$ is said to be compact if $M$ is compact considered as a subspace of $X$, that is, if every sequence in $M$ has a convergent subsequence whose limit is an element of $M$.
A compact subset $M$ of a metric space is closed and bounded.
In a finite dimensional normed space $X$, any subset $M \subset X$ is compact if and only if $M$ is closed and bounded.
Let $Y$ and $Z$ be subspaces of a normed space $X$ (of any dimension), and suppose that $Y$ is closed and is a proper subset of $Z$. Then for every real number $\theta$ in the interval $(0,1)$ there is a $z \in Z$ such that
$$ \|z\| = 1, \qquad \|z - y\| \ge \theta \quad \text{for all } y \in Y. $$
Proof.
We consider any $v \in Z \setminus Y$ and denote its distance from $Y$ by $a$, that is,
$$ a = \inf_{y \in Y} |v - y|. $$
Clearly, $a > 0$ since $Y$ is closed. We now take any $\theta \in (0,1)$. By the definition of an infimum there is a $y_0 \in Y$ such that
$$ a \le |v - y_0| \le \frac{a}{\theta}. $$
Let $ z = c(v - y_0), \text{ where } c = \frac{1}{|v - y_0|}, $ then $\|z\| = 1$, and we show that $\|z - y\| \ge \theta$ for every $y \in Y$. We have
\begin{align*} \|z - y\| &= |c(v - y_0) - y| \\ &= c \|v - y_0 - c^{-1} y\| \\ &= c \| v - y_1 \| \qquad (y_1 = y_0 + c^{-1} y) \end{align*}
The form of $y_1$ show that $y_1 \in Y$. Hence $\|v - y_1\| \ge a$ and we obtain $$ \|z - y\| = c \|v - y_1\| \ge ca = \frac{a}{|v - y_0|} \ge \frac{a}{a/\theta} = \theta. $$
If a normed space $X$ has the property that the closed unit ball $M = \{x \mid \|x\| \le 1\}$ is compact, then $X$ is finite dimensional.
Let $X$ and $Y$ be metric spaces and $T : X \to Y$ a continuous mapping. Then the image of a compact subset $M$ of $X$ under $T$ is compact.
A continuous mapping $T$ of a compact subset $M$ of a metric space $X$ into $\mathbb{R}$ assumes a maximum and minimum at some points of $M$.
A linear operator $T$ is an operator such that
$\text{(i)} \quad$ The domain $\mathcal{D}(T)$ of $T$ is a vector space and the range $\mathcal{R}(T)$ lies in a vector space over the same field.
$\text{(ii)} \quad$ For all $x,y \in \mathcal{D}(T)$ and scalars $\alpha$, $$ T(x+y) = Tx + Ty $$ $$ T(\alpha x) = \alpha Tx $$
Note. $\mathcal{N}(T)$ denotes the null space of $T$, the set of all $x \in \mathcal{D}(T)$ such that $Tx = 0$ (Kernel).
$T$ is a homomorphism of a vector space (its domain) into another space, that is $T$ preserves the two operations of vector space.
1. Identity operator. The identity operator $I_X : X \to X$ is defined by $I_X x = x$ for all $x \in X$. Write simply $I$ for $I_X$, thus $Ix = x$.
2. Differentiation. Let $X$ be the vector space of all polynomials on $[a,b]$. We may define a linear operator $T$ on $X$ by setting $$ T x(t) = x’(t) $$ for every $x \in X$.
3. Integration. A linear operator $T$ from $C[a,b]$ into itself can be defined by $$ T x(t) = \int_a^t x(\tau)\ \mathrm{d}\tau $$
4. Multiplication by $t$. Another linear operator from $C[a,b]$ into itself is defined by $$ T x(t) = t x(t) $$
5. Elementary vector algebra. Cross product with one factor kept fixed defines a linear operator $T_1 : \mathbb{R}^3 \to \mathbb{R}^3$. Similarly, the dot product with one fixed factor defines a linear operator $T_2 : \mathbb{R}^3 \to \mathbb{R}$, say $$ T_2 x = x \cdot a = \xi_1 \alpha_1 + \xi_2 \alpha_2 + \xi_3 \alpha_3 $$ where $\alpha = (\alpha_j) \in \mathbb{R}^3$ is fixed.
6. Matrices. A real matrix $A = (\alpha_{jk})$ with $r$ rows and $n$ columns defines an operator $T : \mathbb{R}^n \to \mathbb{R}^r$ by means of $$ y = Ax $$ where $x = (x_j)$ has $n$ components and $y = (y_j)$ has $r$ components:
$$ \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_r \end{bmatrix} = \begin{bmatrix} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1n} \\ \alpha_{21} & \alpha_{22} & \cdots & \alpha_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{r1} & \alpha_{r2} & \cdots & \alpha_{rn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} $$
Let $T$ be a linear operator, then
$\text{(a)} \quad$ The range $\mathcal{R}(T)$ is a vector space.
$\text{(b)} \quad$ If $\dim \mathcal{D}(T) = n < \infty$, then $\dim \mathcal{R}(T) \le n$.
$\text{(c)} \quad$ The null space $\mathcal{N}(T)$ is a vector space.
Inverse of linear operator. $T : \mathcal{D}(T) \to Y$ is injective or one-to-one, there exists the mapping
\begin{align} T^{-1} : & \mathcal{R}(T) \to \mathcal{D}(T), \\ & \quad y_0 \text{ }\text{ } \mapsto \text{ } \text{ } x_0 \quad (y_0 = Tx_0) \end{align} We have $$ T^{-1}Tx = x, \qquad TT^{-1}y = y $$
Let $X, Y$ be vector spaces, both real or complex. Let $T : \mathcal{D}(T) \to Y$ be a linear operator with domain $\mathcal{D}(T) \subset X$ and $\mathcal{R}(T) \subset Y$. Then:
$\text{(a)} \quad$ The inverse $T^{-1} : \mathcal{R}(T) \to \mathcal{D}(T)$ exists if and only if
$$ Tx = 0 \Rightarrow x = 0 $$
$\text{(b)} \quad$ If $T^{-1}$ exists, it is a linear operator.
$\text{(c)} \quad$ If $\dim \mathcal{D}(T) = n < \infty$ and $T^{-1}$ exists, then $ \dim \mathcal{R}(T) = \dim \mathcal{D}(T) $
Let $T : X \to Y$ and $S : Y \to Z$ be bijective linear operators, where $X, Y, Z$ are vector spaces. Then the inverse $(ST)^{-1} : Z \to X$ of the product $ST$ exists, and
$$ (ST)^{-1} = T^{-1} S^{-1}. $$
Let $X$ and $Y$ be normed spaces and $T : \mathcal{D}(T) \to Y$ a linear operator, where $\mathcal{D}(T) \subset X$. The operator $T$ is said to be bounded if there is a real number $c$ such that for all $x \in \mathcal{D}(T)$,
$$ \|Tx\| \le c \|x\| $$
$\Rightarrow$ $c$ is at least as big as the supremum of the expression on the left taken over $\mathcal{D}(T) - \{0\}$. This supremum is denoted by $\|T\|$; thus
$$ \|T\| = \sup_{\substack{x \in \mathcal{D}(T) \\ x \ne 0}} \frac{\|Tx\|}{\|x\|} $$
$\|T\|$ is called the norm of the operator $T$. If $\mathcal{D}(T) = \{0\}$ we define $\|T\| = 0$. Note that, with $c = \|T\|$,
$$ \|Tx\| \le \|T\| \|x\| $$
Let $T$ be a bounded linear operator. Then, an alternative formula for the norm of $T$ is
$$ \|T\| = \sup_{\substack{x \in \mathcal{D}(T) \\ \|x\| = 1}} \|Tx\| $$
1. Identity operator. The identity operator $I : X \to X$ on a normed space $X \ne \{0\}$ is bounded and has norm $\|I\| = 1$.
2. Differentiation operator. Let $X$ be the normed space of all polynomials on $J = [0,1]$ with norm given $\|x\| = \max |x(t)|$, $t \in J$. A differentiation operator $T$ is defined on $X$ by $$ Tx(t) = x’(t) $$ This operator is linear but not bounded.
3. Integral operator. Define an integral operator $ T: C[0,1] \to C[0,1] $ by $$ y = Tx, \text{ where} \quad y(t)=\int_0^1 k(t,\tau)x(\tau) \mathrm{d}\tau. $$ Here $k$ is a given function, which is called the kernel of $T$, and it is assumed to be continuous on the closed square $ G = J \times J $ in the $t\tau$-plane, where $ J=[0,1]. $
This operator is linear, $T$ is bounded.
4. Matrix. A real matrix $ A=(\alpha_{jk}) $ with $r$ rows and $n$ columns defines an operator $ T:\mathbb{R}^n \to \mathbb{R}^r $ by means of $$ y=Ax. $$ where $ x=(x_i) \text{ and } y=(y_i); \quad y_i=\sum_{k=1}^n \alpha_{jk}x_k \quad (j=\overline{1,r}). $ $T$ is linear and bounded.
If a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded.
Proof. Let $\dim X=n$ and $\{e_1,\dots,e_n\}$ be a basis for $X$. We take any $ x=\sum \xi_i e_i $ and consider any linear operator $T$ on $X$. Since $T$ is linear, \begin{align} \|Tx\|=\left\| \sum \xi_i Te_i \right\| & \le \sum |\xi_i|\ \|Te_i\| \\ & \le \max_k \|Te_k\| \sum |\xi_i| \\ & \le \max_k \|Te_k\| \frac{1}{c}\left\|\sum \xi_i e_i\right\| = \delta \|x\|. \end{align} where $ \delta=\frac{1}{c}\max\limits_k \|Te_k\|. $
Let $ T:\mathcal{D}(T)\to Y $ be a linear operator, where $\mathcal{D}(T)\subset X$ and $X,Y$ are normed spaces. Then:
$\text{(a)} \quad$ $T$ is continuous if and only if $T$ is bounded.
$\text{(b)} \quad$ If $T$ is continuous at a single point, it is continuous.
Let $T$ be a bounded linear operator. Then:
$\text{(a)} \quad$ $x_n\to x$ $( \text{where } x_n,x\in \mathcal{D}(T))$ implies $Tx_n\to Tx$.
$\text{(b)} \quad$ The null space $\mathcal{N}(T)$ is closed.
Let $ T:\mathcal{D}(T)\to Y $ be a bounded linear operator, where $\mathcal{D}(T)$ lies in a normed space $X$ and $Y$ is a Banach space. Then $T$ has an extension $$ \tilde T:\overline{\mathcal{D}(T)}\to Y. $$ where $\tilde T$ is a bounded linear operator with $$ \|\tilde T\|=\|T\|. $$
A linear functional $f$ is a linear operator with domain in a vector space $X$ and range in the scalar field $K$ of $X$, thus $$ f:\mathcal{D}(f)\to K. $$ where $K=\mathbb{R}$ if $X$ is real and $K=\mathbb{C}$ if $X$ is complex.
A bounded linear functional $f$ is a bounded linear operator with range in the scalar field of the normed space $X$ in which the domain $\mathcal{D}(f)$ lies. Thus there exists a real number $c$ such that for all $x\in\mathcal{D}(f)$, $$ |f(x)|\le c\|x\|. $$ Furthermore, the norm of $f$ is $$ \|f\|=\sup_{\substack{x\in\mathcal{D}(f)\\ x\ne 0}}\frac{|f(x)|}{\ |x\|} $$ or $$ \|f\|=\sup_{\substack{x\in\mathcal{D}(f)\\ \|x\|=1}}|f(x)|. $$ $$ \Rightarrow |f(x)|\le \|f\|\|x\|. $$
A linear functional $f$ with domain $\mathcal{D}(f)$ in a normed space is continuous if and only if $f$ is bounded.
1. Norm. The norm $\|\cdot\|:X\to\mathbb{R}$ on a normed space $(X,\|\cdot\|)$ is a functional on $X$ which is not linear.
2. Dot product. The dot product with one factor kept fixed defines a functional
$
f:\mathbb{R}^3\to\mathbb{R}
$
$$
f(x)=x\cdot a=\xi_1 \alpha_1+\xi_2 \alpha_2+\xi_3 \alpha_3.
$$
where $\alpha=(\alpha_j)\in\mathbb{R}^3$ is fixed.
$f$ is linear, bounded. In fact,
$$
|f(x)|=|x\cdot a|\le \|x\|\|a\|
\Rightarrow \|f\|=\|a\|.
$$
3. Definite integral. Consider integral for all functions in a certain function space, $f$ is defined by $$ f(x)=\int_a^b x(t)\ \mathrm{d}t, \qquad x\in C[a,b]. $$ $f$ is linear, bounded and has norm $ \|f\|=b-a. $
4. Space $C[a,b]$. Choose a fixed $t_0\in J=[a,b]$ and set $$ f_t(x)=x(t_0), \qquad x\in C[a,b]. $$ $f_t$ is linear, $f_t$ is bounded and has norm $ \|f_t\|=1. $
5. Space $\ell^2$. We can obtain a linear functional $f$ on the Hilbert space $\ell^2$ by choosing a fixed $ a=(a_j)\in \ell^2 $ and setting $$ f(x)=\sum_{j=1}^{\infty}\xi_j a_j, $$ where $ x=(\xi_j)\in \ell^2. $ This series converges absolutely and $f$ is bounded.
– The set of all linear functionals defined on a vector space $X$ can itself be made into a vector space. This space is denoted by $X^*$ and is called the algebraic dual space of $X$.
– Consider the algebraic dual $(X^{*})^{*}$ of $X^*$, whose elements are the linear functionals defined on $X^*$. Denote $(X^*)^*$ by $X^{* *}$ and call it the second algebraic dual space of $X$.
– To each $x\in X$ there corresponds a $g_x\in X^{**}$. This defines a mapping \begin{align} C:\quad & X \to X^{**} \\ & x \text{ } \mapsto g_x. \end{align}
$C$ is called the canonical mapping of $X$ into $X^{**}$.
Isomorphism. A bijective mapping of $X$ onto $\tilde X$ which preserves the structure.
Accordingly, an isomorphism $T$ of a metric space $X=(X,d)$ onto a metric space $\tilde X=(\tilde X,\tilde d)$ is a bijective mapping which preserves distance, that is, for all $x,y\in X$, $$ \tilde d(Tx,Ty)=d(x,y). $$
$\tilde X$ is then called isomorphic with $X$. $T: X\to \tilde X$ is a bijective linear operator, $$ T(x+y)=Tx+Ty, \qquad T(\alpha x)=\alpha Tx. $$
$X$ and $\tilde X$ are called isomorphic vector spaces.
– Canonical mapping $C$ is injective. Since $C$ is linear, it is an isomorphism of $X$ onto the range $\mathcal{R}(C)\subset X^{**}$.
If $X$ is isomorphic with a subspace of a vector space $Y$, we say that $X$ is embeddable in $Y$. $X$ is embeddable in $X^{**}$ and $C$ is also called the canonical embedding of $X$ into $X^{**}$.
If $C$ is surjective, so that $\mathcal{R}(C)=X^{**}$, then $X$ is said to be algebraically reflexive.
– Let $X$ and $Y$ be finite dimensional vector spaces over the same field and $T:X\to Y$ a linear operator. We choose a basis $ E=\{e_1,\dots,e_n\} \text{ for } X $ and a basis $ B=\{b_1,\dots,b_m\} \text{ for } Y, $ with the vectors arranged in a definite order which we keep fixed. Then every $x\in X$ has a unique representation $$ x=\xi_1 e_1 + \cdots + \xi_n e_n. $$
Since $T$ is linear, $x$ has the image $$ y = Tx = T\left(\sum_{k=1}^n \xi_k e_k\right) = \sum_{k=1}^n \xi_k T e_k. $$
$y$ and $y_k = T e_k$ are in $Y$; they have unique representations of the form $$ y = \sum_{j=1}^r \eta_j b_j, $$ $$ T e_k = \sum_{j=1}^r \tau_{jk} b_j. $$
Then $$ y = \sum_{j=1}^r \eta_j b_j = \sum_{k=1}^n \xi_k T e_k = \sum_{k=1}^n \xi_k \sum_{j=1}^r \tau_{jk} b_j = \sum_{j=1}^r \left(\sum_{k=1}^n \tau_{jk} \xi_k \right) b_j. $$
$$ \Rightarrow \eta_j = \sum_{k=1}^n \tau_{jk} \xi_k, \qquad j=1,\dots,r. $$
The coefficients form a matrix: $ T_{EB} = (\tau_{jk}) $ with $r$ rows and $n$ columns. The matrix $T_{EB}$ represents the operator $T$ with respect to bases.
Let $\tilde{x} = (\xi_k)$ and $\tilde{y} = (\eta_j)$. In matrix notation, $$ \tilde{y} = T_{EB}\tilde{x}, $$ $$ T e = T_{EB} b. $$
– For every functional $f$ and every $x=\sum\limits_{j=1}^n \xi_j e_j \in X$, we have $$ f(x)=f\left(\sum_{j=1}^n \xi_j e_j\right) = \sum_{j=1}^n \xi_j f(e_j) = \sum_{j=1}^n \xi_j \alpha_j, $$ where $\alpha_j = f(e_j)$, and $f$ is uniquely determined by its values $\alpha_j$ at the $n$ basis vectors of $X$. Conversely, every $n$-tuple of scalars $\alpha_1,\dots,\alpha_n$ determines a linear functional on $X$. Take $n$-tuples $$ (1,0,0,\dots,0), \quad (0,1,0,\dots,0), \quad \dots, \quad (0,0,0,\dots,1). $$
This gives $n$ functionals, denote $f_1,\dots,f_n$, with values 1. Kronecker delta $$ f_j(e_k)=\delta_{jk}^{\hspace{0.2cm} \color{gray}{1}}= \begin{cases} 0 & \text{if } j\ne k,\\ 1 & \text{if } j=k, \end{cases} $$
$\{f_1,\dots,f_n\}$ is called the dual basis of the basis $\{e_1,\dots,e_n\}$ of $X$.
Let $X$ be an $n$-dimensional vector space and $E=\{e_1,\dots,e_n\}$ a basis for $X$. Then $ F=\{f_1,\dots,f_n\} $ is a basis for the algebraic dual $X^*$ of $X$, and $ \dim X^* = \dim X = n. $
Let $X$ be a finite dimensional vector space. If $x_0 \in X$ has the property that $ f(x_0)=0 $ for all $f\in X^*$, then $x_0=0$.
A finite dimensional vector space is algebraically reflexive.
– Take any two normed spaces $X$ and $Y$ (both real or complex) and consider the set $B(X,Y)$ consisting of all bounded linear operators from $X$ onto $Y$. $B(X,Y)$ can itself be made into a normed space.
Define $T_1+T_2$ of two operators $T_1,T_2\in B(X,Y)$ by $$ (T_1+T_2)x = T_1x + T_2x. $$
Define the product $\alpha T$ of $T\in B(X,Y)$ and a scalar $\alpha$ by $$ (\alpha T)x = \alpha Tx. $$
Then $B(X,Y)$ becomes a vector space so itself a normed space with norm defined by $$ \|T\| = \sup_{\substack{x\in X \\ x\ne 0}} \frac{\|Tx\|}{\|x\|} = \sup_{\|x\|=1} \|Tx\|. $$
If $Y$ is a Banach space, then $B(X,Y)$ is a Banach space.
Proof.
Consider an arbitrary Cauchy sequence $(T_n)$ in $B(X,Y)$ and show that $(T_n)$ converges to an operator $T\in B(X,Y)$.
Since $(T_n)$ is Cauchy, for every $\varepsilon>0$ there is an $N$ such that $$ \|T_n - T_m\| < \varepsilon \quad (m,n > N). $$
For all $x\in X$ and $m,n>N$ we obtain \begin{equation} \|T_n x - T_m x\| = \|(T_n - T_m)x\| \le \|T_n - T_m\|\|x\| < \varepsilon \|x\| \tag{1}. \end{equation}
$\Rightarrow$ for fixed $x$, $(T_n x)$ is Cauchy in $Y$. Since $Y$ is complete, $(T_n x)$ converges, say $$ T_n x \to y. $$ The limit $y\in Y$ depends on the choice of $x\in X$. Define an operator $$ T: X \to Y, \quad \text{where } y=Tx. $$
The operator $T$ is linear: $$ \lim T_n(\alpha x_1 + \beta x_2) = \lim (\alpha T_n x_1 + \beta T_n x_2) = \alpha \lim T_n x_1 + \beta \lim T_n x_2. $$
We prove that $T$ is bounded and $T_n \to T$, that is, $\|T_n - T\|\to 0$. Since (1) holds for every $m>N$ and $T_m x \to Tx$, let $m\to\infty$. Using the continuity of the norm, for every $n>N$ and all $x\in X$, \begin{equation} \|T_n x - Tx\| = \|T_n x - \lim_{m\to\infty} T_m x\| = \lim_{m\to\infty} \|T_n x - T_m x\| \le \varepsilon \|x\|. \tag{2} \end{equation}
$\Rightarrow (T_n - T)$ with $n>N$ is a bounded linear operator. Since $T_n$ is bounded, $T = T_n - (T_n - T)$ is bounded, hence $T\in B(X,Y)$. In (2), take the supremum over all $x$ of norm $1$, we obtain $$ \|T_n - T\| \le \varepsilon \quad (n>N). $$
Hence, $\|T_n - T\|\to 0$.
Let $X$ be a normed space. Then the set of all bounded linear functionals on $X$ constitutes a normed space with norm defined by
$$ \|f\|=\sup_{\substack{x\in X \\ x\neq 0}}\frac{|f(x)|}{\|x\|}=\sup_{\|x\|=1}|f(x)|. $$
which is called the dual space of $X$ and is denoted by $X’$.
The dual space $X’$ of a normed space $X$ is a Banach space.
An isomorphism of a normed space $X$ onto a normed space $\tilde X$ is a bijective linear operator $T:X\to \tilde X$ which preserves the norm, that is, for all $x\in X$, $ \|Tx\|=\|x\|. $ ($T$ is isometric)
1. Space $\mathbb{R}^n$. The dual space of $\mathbb{R}^n$ is $\mathbb{R}^n$.
Proof.
We have $(\mathbb{R}^{n})^{'}=(\mathbb{R}^{n})^*$ and every $f\in (\mathbb{R}^{n})^*$ has a representation
$$ f(x)=\sum \xi_k \gamma_k,\qquad \gamma_k=f(e_k). $$
By Cauchy–Schwarz inequality, $$ |f(x)|\le \sum |\xi_k \gamma_k| \le \left(\sum \xi_j^2\right)^{1/2}\left(\sum \gamma_k^2\right)^{1/2} =\|x\|\left(\sum \gamma_k^2\right)^{1/2}. $$ Taking the supremum over all $x$ of norm $1$, we obtain $$ \|f\|\le \left(\sum \gamma_k^2\right)^{1/2}. $$ However, since for $x=(x_1,\ldots,x_n)\to$ equality is achieved, we must have
$$ \|f\|=\left(\sum_{k=1}^n \gamma_k^2\right)^{1/2}. $$
This proves that norm of $f$ is the Euclidean norm, $\|f\|=\|c\|$ where $c=(\gamma_k)\in \mathbb{R}^n$.
Hence the mapping of $(\mathbb{R}^{n})^{’}$ onto $\mathbb{R}^n$ defined by $f\mapsto c=(\gamma_k)$, $\gamma_k=f(e_k)$, is norm preserving, it is linear and bijective, it is an isomorphism.
2. Space $\ell^1$. The dual space of $\ell^1$ is $\ell^\infty$.
3. Space $\ell^p$. The dual space of $\ell^p$ is $\ell^q$; here, $1 < p < +\infty$ and $q$ is the conjugate of $p$, that is, $1/p + 1/q = 1.$
[1] Kreyszig, E. (1978). Introductory functional analysis with applications. John Wiley & Sons.
]]>– A metric space is a pair $(X, d)$, where $X$ is a set and $d$ is a metric on $X$ (or distance function on $X$), that is a function defined on $X \times X$ such that for all $x, y, z \in X$, we have
– A subspace $(Y, \tilde{d})$ of $(X, d)$ is obtained if we take a subset $Y_{C} \subset X$ and restrict $d$ to $Y \times Y$; thus the metric on $Y$ is the restriction $$ \tilde{d}= d|_{Y \times Y} $$ $\tilde{d}$ is called the metric induced on $y$ by $d$
1. Euclidean space $\mathbb{R}^{n}$, unitary space $\mathbb{C}^{n}$, complex plan $\mathbb{C}.$
Take the set of all ordered $n$-tuples of real numbers, written $$ x=\left(x_{1}, x_{2}, \ldots, x_{n}\right), \quad y=\left(y_{1}, y_{2}, \ldots, y_{n}\right) $$ and the Euclidean metric defined by $$ d(x, y)=\sqrt{\left(x_{1}-y_{1}\right)^{2}+\left(x_{2}-y_{2}\right)^{2}+\ldots+\left(x_{n}-y_{n}\right)^{2}} $$ $n$ - dimensional unitary space $\mathbb{C}^{n}$ has the metric defined by $$ d(x, y)=\sqrt{\left|x_{1}-y_{1}\right|^{2}+\left|x_{2}-y_{2}\right|^{2}+\ldots+\left|x_{n}-y_{n}\right|^{2}} $$
2. Sequence space $l^{\infty}$.
Take the set of all bounded sequences of complex numbers; every element of $X$ is a complex sequence $$ x=\left(x_{1}, x_{2}, \ldots, x_{n}\right) \text { briefly } x=\left(x_{i}\right) $$ such that for all $j \in \mathbb{N}^{*}$, we have
$$ \left|x_{j}\right| \leqslant c_{x} $$ $c_{x}$ is a real number which may depend on $x$, not on $j$. Metric is defined by $$ d(x, y)=\sup_{j \in \mathbb{N}^{*}}|x_{j}-y_{j}| $$ where $y=\left(y_{j}\right) \in X$.
3. Function space $C[a, b]$.
Take the set of all real-valued function $x, y, \ldots$ and are defined and continuous on a given closed interval $J=[a, b]$ The metric defined by $$ d(x, y)=\max _{t \in J}|x(t)-y(t)| $$
4. Sequence spaces.
The space consists of the set of all (bounded or unbounded) sequences of complex numbers and the metric $d$ defined by where $x=\left(x_{j}\right), y=\left(y_{j}\right)$ $$ d(x, y)=\sum_{j=1}^{\infty} \frac{1}{2^{j}} \frac{\left|x_{j}-y_{j}\right|}{1+\left|x_{j}-y_{j}\right|} $$
Proof.
$f(t)=\frac{t}{1+t}$ is monotone increasing; $|a+b| \leqslant|a|+|b|$ implies $f(|a+b|) \leqslant f(|a|+|b|).$ Then we have: $$ \frac{|a+b|}{1+|a+b|} \leqslant \frac{|a|+|b|}{1+|a|+|b|} \leqslant \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|} $$
Let $a=x_{j}-z_{j}$ and $y=z_{j}-y_{j}$ where $z=\left(z_{j}\right)$. Therefore $a+b=x_{j}-y_{j}$ and $$ \frac{\left|x_{j}-y_{j}\right|}{1+\left|x_{j}-y_{j}\right|} \leqslant \frac{\left|x_{j}-z_{j}\right|}{1+\left|x_{j}-z_{j}\right|}+\frac{\left|z_{j}-y_{j}\right|}{1+\left|z_{j}-y_{j}\right|} $$ Then $$ \sum_{j=1}^{\infty} \frac{1}{2^{j}} \frac{\left|x_{j}-y_{j}\right|}{1+\left|x_{j}-y_{j}\right|} \leqslant \sum_{j=1}^{\infty} \frac{1}{2^{j}} \frac{\left|x_{j}-z_{j}\right|}{1+\left|x_{j}-z_{j}\right|}+\sum_{j=1}^{\infty} \frac{1}{2^{j}} \frac{\left|z_{j}-y_{j}\right|}{1+\left|z_{j}-y_{j}\right|}$$ or $$d(x, y) \leq d(x, z)+d(z, y).$$
5. Space $B(A)$ of bounded function.
Each element $x \in B(A)$ is a function defined and bounded on a given set $A$, the metric is defined: $$ d(x, y)=\sup _{t \in A}|x(t)-y(t)| $$
6. Space $l^{p}$, Hillbert sequence space $l^{2}$, Holder and Minkowski inequalities for sums.
Each element in the space $\ell^{p}$ is a sequence $x=\left(x_{i}\right)=\left(x_{1}, x_{2}, \ldots\right)$ of numbers such that $$ \sum_{j=1}^{\infty}\left|x_{i}\right|^{p}<\infty \quad(p \geqslant 1) \ $$ and the metric is defined by $$ d(x, y)=\left(\sum_{j=1}^{\infty}\left|x_{j}-y_{j}\right|^{p}\right)^{\frac{1}{p}} $$ Proof.
$l^{p}$ is a metric space
$\text{(a)} \quad$ Auxiliary in equality $\frac{1}{p}+\frac{1}{q}=1$ ($p, q$ are called conjugate exponents). Then $$ \alpha \beta \leqslant \frac{\alpha^{p}}{p}+\frac{\beta^{q}}{q} $$
$\text{(b)} \quad$ Holder inequality $$ \sum_{j=1}^{\infty}\left|x_{j} y_{j}\right| \leqslant\left(\sum_{j=1}^{\infty}\left|x_{j}\right|^{p}\right)^{\frac{1}{p}}\left(\sum_{j=1}^{\infty}\left|y_{j}\right|^{q}\right)^{\frac{1}{q}} $$
Let $ \left(\tilde{x_{i}}\right)$ and $\left(\tilde{y_{j}}\right) $ such that
$$ \begin{equation} \sum\left|\tilde{x_{i}}\right|^{p} = 1, \quad \sum\left|\tilde{y_{j}}\right|^{q}=1 \tag{1} \end{equation} $$
Set $\alpha=\left|x_{j}\right|$ and $\beta=\left|y_{j}\right|$, we have $$ \begin{equation} \left|\tilde{x_{j}} \tilde{y_{j}}\right| \leqslant \frac{1}{p}\left|\tilde{x_{j}}\right|^{p}+\frac{1}{q}\left|\tilde{y_{j}}\right|^{q} \tag{2} \end{equation} $$
Then $$ \begin{equation} \sum\left|\tilde{x_{j}}\tilde{y_{j}}\right| \leqslant \frac{1}{p}+\frac{1}{q}=1. \tag{3} \end{equation} $$
Now, take $x=\left(x_{j}\right) \in l^{p}$ and $y=\left(y_{j}\right) \in l^{p}$ and set $$ \tilde{x_{j}}=\dfrac{x_{j}}{\left(\sum\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}}; \quad \tilde{y_{j}}=\dfrac{y_{j}}{\left(\sum\left|y_{j}\right|^{q}\right)^{\frac{1}{q}}} $$
Then (1) sastified, so we apply (2) and (3) to have $$ \sum_{j=1}^{\infty}\left|x_{j} y_{j}\right| \leqslant\left(\sum_{k=1}^{\infty}\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}\left(\sum_{m=1}^{\infty}\left|y_{m}\right|^{q}\right)^{\frac{1}{q}} $$
$\text{(c)} \quad$ Minkowski inequality $$ \begin{equation} \left(\sum_{j=1}^{\infty}\left|x_{j}+y_{j}\right|^{p}\right)^{\frac{1}{p}} \leqslant\left(\sum_{k=1}^{\infty}\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}+\left(\sum_{m=1}^{\infty}\left|y_{m}\right|^{p}\right)^{\frac{1}{p}} \tag{4} \end{equation} $$ where $x=\left(x_{j}\right) \in l^{p}$ and $y=\left(y_{j}\right) \in l^{p}$.
Write $x_{j}+y_{j}=\omega_{j}$, then $$ \begin{aligned} \left|w_{j}\right|^{p} & =\left|x_{j}+y_{j}\right|\left|w_{j}\right|^{p-1} \\ & \leqslant \left( \left|x_{j}\right|+\left|y_{j}\right| \right) \left|w_{j}\right|^{p-1} \end{aligned} $$
Summing over $j$ from 1 to any fixed $n$, we obtain $$ \begin{align} \sum\left|w_{j}\right|^{p} &\leqslant \sum\left|x_{j}\right|\left|w_{j}\right|^{p-1}+\sum\left|y_{j}\right|\left|w_{j}\right|^{p-1} \\ &\leqslant \left(\sum\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}\left(\sum\left|w_{m}\right|^{(p-1)q}\right)^{\frac{1}{q}}+\left(\sum\left|w_{k}\right|^{p}\right)^{\frac{1}{p}}\left(\sum\left|w_{m}\right|^{(p-1) q}\right)^{\frac{1}{q}} \\ &= \left\{\left(\sum\left|x_{k}\right|^{p}\right)^{\frac{1}{p}}+\left(\sum\left|y_{k}\right|^{p}\right)^{\frac{1}{p}}\right\}\left(\sum\left|w_{m}\right|^{p}\right)^{\frac{1}{q}} \quad ((p-1) q=p) \end{align} $$
Dividing by the last factor and noting that $1-\frac{1}{q}=\frac{1}{p}$, we obtain (4).
$\text{(d)} \quad$ From $\text{(c)}$ $$ \begin{align} d(x, y) & =\left(\sum\left|x_{j}-y_{j}\right|^{p}\right)^{\frac{1}{p}} \\ & \leqslant\left(\sum\left(\left|x_{j}-z_{j}\right|+\left|z_{j}-y_{j}\right|^{p}\right)^{\frac{1}{p}}\right. \\ & \leqslant\left(\sum\left|x_{j}-z_{j}\right|^{p}\right)^{\frac{1}{p}}+\left(\sum\left|z_{j}-y_{j}\right|^{p}\right)^{\frac{1}{p}} \\ & =d(x, z)+d(z, y) \end{align} $$
The diameter $\delta(A)$ of a nonempty set $A$ in a metric space $(X, d)$ is defined to be $$ \delta(A)=\sup _{x, y \in A} d(x, y) . $$ $A$ is said to be bounded if $\delta(A)<\infty$
The distance $D(A, B)$ between two nonempty subsets $A$ and $B$ of a metric space $(x, d)$ is defined to be $$ D(A, B)=\inf _{\substack{a \in A, \\ b \in B}} d(a, b) . $$ $D$ does not define a metric on the power set of $X$
The Cartesian product $X=X_{1} \times X_{2}$ of two metric spaces $\left(X_{1}, d_{1}\right)$ and $\left(X_{2}, d_{2}\right)$ can be made into a metric space $(X, d)$.
1. $d(x, y)=d_{1}\left(x_{1}, y_{1}\right)+d_{2}\left(x_{2}, y_{2}\right)$
2. $\tilde{d}(x, y)=\sqrt{d_{1}\left(x_{1}, y_{1}\right)^{2}+d_{2}\left(x_{2}, y_{2}\right)^{2}}$
3. $\tilde{d}(x, y)=\max \left[d_{1}\left(x_{1}, y_{1}\right), d_{2}\left(x_{2}, y_{2}\right)\right]$
Given a point $x_{0} \in X$ and a real number $r>0$, we define three types of sets
$\text{(a)} \quad$ $B\left(x_{0}, r\right)=\left\{x \in X \mid d\left(x, x_{0}\right)<r\right\} \quad$ (Open ball)
$\text{(b)} \quad$ $\tilde{B}\left(x_{0}, r\right)=\left\{x \in X \mid d\left(x, x_{0}\right) \leq r\right\} \quad$ (Closed ball)
$\text{(c)} \quad$ $S\left(x_{0}, r\right)=\left\{x \in X \mid d\left(x, x_{0}\right)=r\right\} \quad$ (Sphere)
$x_{0}$ : center; $r$ : radius. $S\left(x_{0}, r\right)=\tilde{B}\left(x_{0}, r\right)-B\left(x_{0}, r\right)$.
A subset $M$ of a metric space $X$ is said to be open if it contains a ball about each of its points. $A$ subset $K$ of $X$ is said to be closed if its complement (in $X$ ) is open, that is, $K^{c}=X-K$ is open.
– An open ball $B\left(x_{0}, \varepsilon\right)$ of radius $\varepsilon$ is often called an $\varepsilon$-neighborhood of $x_{0}$
– Call $x_{0}$ an interior point of a set $M \subset X$ if $M$ is a neighborhood of $x_{0}$.
– The interior of $M$ is the set of all interior points of $M$, denoted $M^{0}$ or $\operatorname{Int}(M)$.
– The collection of all open subsets of $X$, call it $\mathcal{J}$, has properties:
$T_{1}. \emptyset \in \mathcal{J}, x \in \mathcal{J}$
$T_{2}$. The union of any nembers of $\mathcal{J}$ is a member of $\mathcal{J}$
$T_{3}$. The intersection of finitely many members of $\mathcal{J}$ is a member of $\mathcal{J}$.
– Define a topological space $(x, g)$ to be a set $x$ and a collection $\mathcal{J}$ of subsets of $x$ such that $\mathcal{J}$ satisfies $\left(T_{1}\right)$ and $\left(T_{3}\right)$
– $\mathcal{J}:$ topology for $X$.
Let $X=(x, d)$ and $Y=(y, \tilde{d})$ be metric spaces. A mapping $T: X \rightarrow Y$ is said to be continuous at a point $x_{0} \in X$ if for every $\varepsilon>0$, there is a $\delta>0$ such that $$ \tilde{d}\left(T x, T x_{0}\right)<\varepsilon \quad \text { for all } x \text { satisfying } \quad d\left(x, x_{0}\right)<\delta $$
A mapping $T$ of a metric space $X$ into a metric space $Y$ is continuous if and only if the inverse image of any open subset of $Y$ is an open subset of $X$.
Let $M$ be a subset space $X$. Then a point $x_{0}$ of $X$ is called an accumulation point of $M$ if every neighborhood of $x_{0}$ contains at least one point $y \in M$ distance from $x_{0}$. The set consisting of the points of $M$ and the accumulation points of $M$ is called the closure of $M$ and is denoted by $\bar{M}$.
A subset $M$ of a metric space $X$ is said to be dense in $X$ if $\bar{M}=X$. $X$ is said to be separable if it has countable subset which is dense in $X$
1. Real line $\mathbb{R}$. The real line $\mathbb{R}$ is separable. – The set $\mathbb{Q}$ of all rational numbers is countable and is dense in $\mathbb{R}$.
2. Complex plane $\mathbb{C}$. The complex plane $\mathbb{C}$ is separable – A countable dense subset of $\mathbb{C}$ is the set of all complex numbers whose real and imaginary parts are both rational.
3 Discrete metric space. A discre metric space $X$ is separable if and only if $X$ is countable.
4. Space $l^{\infty}$. The space $l^{\infty}$ is not separable.
5. Space $\ell^{p}$. The space $\ell^{p}$ with $1 \leqslant p<\infty$ is separable
A sequence $\left(x_{n}\right)$ in a metric space $X=(x, d)$ is said to converge or to be convergent if there is an $x \in X$ such that $$ \lim_{n \rightarrow \infty} d\left(x_{n}, x\right)=0 $$ $x$ is called the limit of $\left(x_{n}\right)$ and we write $$ \lim_{n \rightarrow \infty} x_{n}=x $$ or, simply, $$ \lim_{n \rightarrow \infty} x_{n} \to x $$ Hence, if $x_{n} \rightarrow x$, an $\varepsilon>0$ being given, there is an $N=N(\varepsilon)$ such that all $x_{n}$ with $n>N$ lie in the $\varepsilon$ - neighborhood $B(x, \varepsilon)$ of $x$.
Let $(x, d)$ be a metric space. Then:
$\text{(a)} \quad$ A convergent sequence in $x$ is bounded and its limit is unique.
$\text{(b)} \quad$ If $x_{n} \rightarrow x$ and $y_{n} \rightarrow y$ in $x$, then $d\left(x_{n}, y_{n}\right) \rightarrow d(x, y)$.
A sequence $\left(x_{n}\right)$ in a metric space $X=(x, d)$ is said to be cauchy if for every $\varepsilon>0$, there is a $N=N(\varepsilon)$ such that $$ d\left(x_{m}, x_{n}\right)<\varepsilon \quad \forall m, n>N . $$
The space $X$ is said to be complete if every Cauchy sequence in $X$ converges (has a limit which is an element of $x$).
Every convergent sequence in a metric space is a Cauchy sequence.
Let $M$ be a nonempty subset of metric space ($X, d$) and $\bar{M}$ its closure.
$\text{(a)} \quad$ $x \in \bar{M}$ if and only if there is a sequence $\left(x_{n}\right)$ in $M$ such that $x_{n} \rightarrow x$.
$\text{(b)} \quad$ $M$ is closed if and only if the situation $x_{n} \in M, x_{n} \rightarrow x$ implies that $x \in M$.
A subspace $M$ of a complete metric space $X$ is itself complete if and only if the set $M$ is closed in $X$.
A mapping $T: X \rightarrow Y$ of a metric space $(X, d)$ into a metric space $(Y, \tilde{d})$ is continuous at a point $x_{0} \in X$ if and only if $x_{n} \rightarrow x_{0}$ implies $Tx_{n} \rightarrow Tx_{0}$.
1. Completeness of $\mathbb{R}^{n}$ and $\mathbb{C}^{n}$
– Consider $\mathbb{R}^{n}$. The metric on $\mathbb{R}^{n}$ is defined by $$ d(x, y)=\left(\sum_{j=1}^{\infty}\left(x_{j}-y_{j}\right)^{2}\right)^{\frac{1}{2}} $$ where $x=\left(x_{j}\right)$ and $y=\left(y_{j}\right)$. We consider any Cauchy sequence $\left(x_{m}\right)$ in $\mathbb{R}^{n}$, writing $x_{m}=\left(x_{m, 1}, x_{m, 2}, \ldots\right.$, $\left.x_{m, n}\right)$. Since $\left(x_{m}\right)$ is Cauchy, for every $\varepsilon>0$ there exist an $N$ such that
$$ \begin{equation} \quad d\left(x_{m}, x_{r}\right)=\left(\sum_{j=1}^{n}\left(x_{m, j}-x_{r, j}\right)^{2}\right)^{\frac{1}{2}}<\varepsilon \quad (m, r > N) \tag{1} \end{equation} $$
Squaring, we have for $m, r>N$ and $j=1,2, \ldots, n$. $$ \left(x_{m, j}-x_{r, j}\right)^{2}<\varepsilon \quad \text { and } \quad\left|x_{m, j}-x_{r, j}\right|<\varepsilon $$
This show that, for each fixed $j(1 \leqslant j \leqslant n)$, the sequence $\left(x_{1, j}, x_{2,j}, \ldots\right)$ is a Cauchy sequence of real numbers. It converges so $x_{m, j} \rightarrow x_{j}$ as $m \rightarrow \infty$. Using these $n$ limits, we define $x=\left(x_{1}, \ldots, x_{n}\right)$.
Clearly, $x \in \mathbb{R}^{n}$. From (1), with $r \rightarrow \infty$, $$ d\left(x_{m}, x\right)<\varepsilon \quad(m>N) $$
This show that $x$ is the limit of $\left(x_{m}\right)$.
2. Completeness of $l^{\infty}$. The space $l^{\infty}$ is complete.
3. Completeness of $c$. The space $c$ consists of all convergent sequence $x=\left(x_{j}\right)$ of complex numbers, with the metric induced from the space $l^{\infty}$. The space $c$ is complete.
4. Completeness of $l^{p}$. The space $l^{p}$ is complete.
5. Completeness of $C[a, b]$. The function space $C[a, b]$ is complete.
1. Space $\mathbb{Q}$ - set of all rational numbers
2. Polynomials. Let $X$ be the set of all polynomials considered as functions on $t$ on some finite closed interval $T=[a, b]$ and define a metric $d$ on $X$ by $$ d(x, y)=\max_{t \in J}|x(t)-y(t)| $$
3. Continuous functions. Let $x$ be the set of all continuous real-valued function on $J=[0,1]$, and let $$ d(x, y)=\int_{0}^{1}|x(t)-y(t)| \mathrm{d}t $$
Let $X=(x, d)$ and $\tilde{X}=(\tilde{X}, \tilde{d})$ be metric spaces. Then
$\text{(a)} \quad$ A mapping $T$ of $X$ into $\tilde{X}$ is said to be isometric or an isometry if $T$ preserves distances, that is if for all $x, y \in X$ $$ \tilde{d}(T x, T y)=d(x, y) $$ $\text{(b)} \quad$ The space $X$ is said to be isometric with the space $\tilde{X}$ if there exists a bijective isometry of $X$ onto $\tilde{X}.$ The space $X$ and $\tilde{X}$ are then called isometric spaces.
For a metric space $X=(X, d)$ there exists a complete metric space $\hat{X}=(\hat{X}, \hat{d})$ which has a subspace $W$ thas is isometric with $X$ and is dense in $\hat{X}$. This space $\hat{X}$ is unique except for isometries, that is, if $\tilde{X}$ is any complete metric space having a dense subspace $\tilde{W}$ isometric with $X$, then $X$ and $\tilde{X}$ are isometric.
[1] Kreyszig, E. (1978). Introductory functional analysis with applications. John Wiley & Sons.
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