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.
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