{"author":{"avatar":"https://nguyencungthanh.github.io/animation_avatar%282%29.PNG","name":"Nguyen Cung Thanh","url":"https://nguyencungthanh.github.io/"},"feed_url":"https://nguyencungthanh.github.io/tags/banach-space/feed.json","home_page_url":"https://nguyencungthanh.github.io/tags/banach-space/","items":[{"content_html":"\n  \u003cdiv class='e-content'\u003e\n    \u003ch1 id=\"normed-spaces-and-banach-spaces\"\u003eNormed Spaces and Banach Spaces\u003c/h1\u003e\n\u003ch2 id=\"1-definition-normed-space-banach-space\"\u003e1. Definition (Normed Space, Banach Space)\u003c/h2\u003e\n\u003cp\u003eA 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 \u0026ldquo;norm of $X$\u0026rdquo;) and which has the properties:\u003c/p\u003e\n\u003col\u003e\n\u003cli\u003e$\\|x\\| \\ge 0$\u003c/li\u003e\n\u003cli\u003e$\\|x\\| = 0 \\iff x = 0$\u003c/li\u003e\n\u003cli\u003e$\\|\\alpha x\\| = |\\alpha|\\|x\\|$\u003c/li\u003e\n\u003cli\u003e$\\|x + y\\| \\le \\|x\\| + \\|y\\|$ $\\quad$ (Triangle inequality)\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp\u003eA norm on $X$ defines a metric $d$ on $X$ by\n$$\nd(x,y) = \\|x - y\\|,\n$$\ncalled the \u003cem\u003emetric induced\u003c/em\u003e by the norm. The normed space thus defined is denoted by $(X, \\| \\cdot\\|)$, or simply $X$.\u003c/p\u003e\n\u003ch3 id=\"example\"\u003eExample\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003e1. Euclidean space  $\\mathbb{R}^n$ and unitary space $\\mathbb{C}^n$\u003c/strong\u003e.\nBanach space with norm\n$$\n\\|x\\| = \\left( \\sum_{i=1}^{n} |x_i|^2 \\right)^{\\frac{1}{2}}\n$$\n$\\mathbb{R}^n$ and $\\mathbb{C}^n$ are complete and the induced metric is\n$$\nd(x,y) = \\|x - y\\| = \\left( \\sum_{i=1}^{n} |x_i - y_i|^2 \\right)^{\\frac{1}{2}}.\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e2. Space $\\ell^p$.\u003c/strong\u003e\nBanach space with norm\u003cbr\u003e\n$$ \\|x\\| = \\left( \\sum_{i = 1}^{\\infty} |x_i|^p \\right)^{\\frac{1}{p}} $$\nand the induced metric is\n$$\nd(x, y) = \\|x - y\\| = \\left( \\sum_{i=1}^{n} |x_i - y_i|^p \\right)^{\\frac{1}{p}}.\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e3. Space $\\ell^{\\infty}$.\u003c/strong\u003e\nBanach space with norm\u003cbr\u003e\n$$\n\\|x\\|_\\infty = \\sup _{i}\\|x_i\\|\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e4 Space $C[a, b]$.\u003c/strong\u003e\nBanach space with norm\u003cbr\u003e\n$$\n\\|x\\| = \\max _{t \\in J}\\|x(t)\\|\n$$\nwhere $J = [a,b]$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e5. Incomplete normed space.\u003c/strong\u003e\u003c/p\u003e\n\u003cp\u003e$d(x, y) = \\int _{0}^{1} | x(t) - y(t) | \\mathrm{d}t$ induce the norm\n$$\\int _{0}^{1} |x(t)| \\mathrm{d}t$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e6. Incomplete normed space and its completion $L^p[a,b]$.\u003c/strong\u003e\nThe Banach space $L^p[a,b]$ with norm\n$$ \\|x\\| _p = \\left( \\int _{a}^{b} |x(t)| \\mathrm{d}t  \\right)^{\\frac{1}{p}} $$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e7. Space $s$.\u003c/strong\u003e\nMetric defined by $$d(x,y) = \\sum _{j = 1}^{\\infty} \\dfrac{1}{2^j} \\dfrac{|x_j-y_j|}{1 + |x_j-y_j|}$$\ncannot be obtained a norm\u003c/p\u003e\n\u003ch3 id=\"11-lemma-translation-invariance\"\u003e1.1 Lemma (Translation invariance)\u003c/h3\u003e\n\u003cp\u003eA metric $d$ induced by a norm on a normed space $X$ satisfies\u003c/p\u003e\n\u003cp\u003e$\\text{(a)} \\quad$ $d(x+a,y+a) = d(x,y)$\u003c/p\u003e\n\u003cp\u003e$\\text{(b)} \\quad$ $d(\\alpha x, \\alpha y) = |\\alpha| d(x,y)$\u003c/p\u003e\n\u003cp\u003efor all $x,y \\in X$ and every scalar $\\alpha$.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"2-further-properties-of-normed-space\"\u003e2. Further properties of normed space\u003c/h2\u003e\n\u003ch3 id=\"21-theorem-subspace-of-banach-space\"\u003e2.1 Theorem (Subspace of Banach space)\u003c/h3\u003e\n\u003cp\u003eA subspace $Y$ of a Banach space $X$ is complete if and only if $Y$ is closed in $X$.\u003c/p\u003e\n\u003cp\u003e\u0026ndash; \u003cstrong\u003eConvergence of sequences\u003c/strong\u003e\u003c/p\u003e\n\u003col\u003e\n\u003cli\u003eA sequence $(x_n)$ in a normed space $X$ is convergent if $X$ contains an $x$ such that\n$$\\lim _{n \\to \\infty} \\|x_n - x \\| = 0.$$\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp\u003eThen we write $x_n \\to x$ and call $x$ the limit of $(x_n)$.\u003c/p\u003e\n\u003col start=\"2\"\u003e\n\u003cli\u003eA sequence $(x_n)$ in a normed space $X$ is Cauchy if for every $\\epsilon \u0026gt; 0$ there is an $N$ such that $$\\|x_m - x_n\\| \u0026lt; \\epsilon \\quad \\forall m, n \u0026gt; N$$\u003c/li\u003e\n\u003c/ol\u003e\n\u003cp\u003e\u0026ndash; \u003cstrong\u003eInfinite series\u003c/strong\u003e $(x_k)$ is a sequence in a normed space $X$, we can associate with $(x_k)$ the sequence $(s_n)$ of partial sums $$s_n = x_1 + x_2 + \\ldots + x_n.$$\u003c/p\u003e\n\u003cp\u003eIf $(s_n)$ is convergent,\n$\ns_n \\to s,\n$\nor $\\|s_n - s\\| \\to 0$ 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\n$$\ns = \\sum_{k=1}^{\\infty} x_k = x_1 + x_2 + \\ldots.\n$$\u003c/p\u003e\n\u003cp\u003eIf\n$\n\\sum_{k=1}^{\\infty} \\|x_k\\|\n$\nconverges, then the series is said to be \u003cem\u003eabsolutely convergent\u003c/em\u003e.\u003c/p\u003e\n\u003cp\u003e\u0026ndash; \u003cstrong\u003eBasis.\u003c/strong\u003e 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\n$$\n\\|x - \\sum_{k=1}^{n} \\alpha_k e_k\\| \\to 0 \\quad \\text{ (as } n \\to \\infty)\n$$\nthen $(e_n)$ is called a \u003cstrong\u003eSchander basis\u003c/strong\u003e 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\n$$x = \\sum_{k=1}^{\\infty} \\alpha_k e_k$$\u003c/p\u003e\n\u003ch4 id=\"example-1\"\u003eExample\u003c/h4\u003e\n\u003cp\u003e$\\ell^p$ 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\n$$\ne_1 = (1,0,0,\\dots), \\quad\ne_2 = (0,1,0,\\dots), \\quad\ne_3 = (0,0,1,\\dots), \\ \\text{etc.}\n$$\u003c/p\u003e\n\u003ch3 id=\"21-theorem--completion\"\u003e2.1 Theorem  (Completion)\u003c/h3\u003e\n\u003cp\u003eLet $X = (X, \\|\\cdot\\|)$ 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.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"3-finite-dimensional-normed-spaces-and-subspaces\"\u003e3. Finite Dimensional Normed Spaces and Subspaces.\u003c/h2\u003e\n\u003ch3 id=\"31-lemma-linear-combination\"\u003e3.1 Lemma (Linear combination)\u003c/h3\u003e\n\u003cp\u003eLet $\\{ x_1, \\ldots, x_n \\}$ be a linearly independent set of vectors in a normed space $X$ (of any dimension). Then there is a number $c \u0026gt; 0$ such that for every choice of scalars $\\alpha_1, \\ldots, \\alpha_n$ we have\u003cbr\u003e\n\\begin{equation}\n\\| \\alpha_1x_1 + \\ldots + \\alpha_n x_n \\| \\ge c(| \\alpha_1 | + \\ldots + |\\alpha_n|) \\tag{1}\n\\end{equation}\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eWe write $S = |a_1| + \\cdots + |a_n|$. If $S = 0$, all $\\alpha_j$ are zero, so (1) holds for any $c$. Let $S \u0026gt; 0$. 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\u003c/p\u003e\n\u003cp\u003e\\begin{equation}\n\\left\\| \\beta_1 x_1 + \\cdots + \\beta_n x_n \\right\\| \\ge c\n\\qquad \\left( \\sum_{j=1}^n |\\beta_j| = 1 \\right)  \\tag{2}\n\\end{equation}\u003c/p\u003e\n\u003cp\u003eSuppose for the contradiction, there exists a sequence $(y_m)$ of vectors\u003c/p\u003e\n\u003cp\u003e$$\ny_m = \\beta_1^{(m)} x_1 + \\cdots + \\beta_n^{(m)} x_n\n$$\u003c/p\u003e\n\u003cp\u003esuch that\n$\n|y_m| \\to 0 \\quad \\text{as } m \\to \\infty.\n$\u003c/p\u003e\n\u003cp\u003eSince\n$\n\\sum_{j=1}^n \\left|\\beta_j^{(m)}\\right| = 1,\n$\nwe have $\\left|\\beta_j^{(m)}\\right| \\le 1$. Hence for each $j$, the sequence\n$$\n\\left(\\beta_j^{(m)}\\right) = \\left(\\beta_j^{(1)}, \\beta_j^{(2)}, \\dots\\right)\n$$\nis 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\u003c/p\u003e\n\u003cp\u003e$$\n(y_{n,m}) = (y_{n,1}, y_{n,2}, \\dots)\n$$\u003c/p\u003e\n\u003cp\u003eof $(y_m)$ whose terms are of the form\u003c/p\u003e\n\u003cp\u003e$$\ny_{n,m} = \\sum_{j=1}^n \\gamma_j^{(m)} x_j.\n$$\u003c/p\u003e\n\u003cp\u003ewith scalars $\\gamma_j^{(m)}$ satisfying\n$\n\\gamma_j^{(m)} \\to \\beta_j \\text{ as } m \\to \\infty.\n$\nHence, as $m \\to \\infty$,\n$\ny_{n,m} \\to y = \\sum_{j=1}^n \\beta_j x_j\n$\u003c/p\u003e\n\u003cp\u003ewhere\n$\n\\sum_{j=1}^n |\\beta_j| = 1,\n$\nso that not all $\\beta_j$ can be zero. Since $\\{x_1,\\dots,x_n\\}$ is a linearly independent then $y \\ne 0$.\u003c/p\u003e\n\u003cp\u003eOn 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).\u003c/p\u003e\n\u003ch3 id=\"32-theorem-completeness\"\u003e3.2 Theorem (Completeness)\u003c/h3\u003e\n\u003cp\u003eEvery finite dimensional subspace $Y$ of a normed space $X$ is complete. In particular, every finite dimensional normed space is complete.\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eConsider an arbitrary Cauchy sequence $(y_m)$ in $Y$ and show that it is convergent in $Y$; the limit will be denoted by $y$.\u003c/p\u003e\n\u003cp\u003eLet $\\dim Y = n$ and $\\{e_1,\\ldots,e_n\\}$ any basis for $Y$. Then each $y_m$ has a unique representation of the form\u003c/p\u003e\n\u003cp\u003e$$\ny_m = \\alpha_1^{(m)} e_1 + \\cdots + \\alpha_n^{(m)} e_n.\n$$\u003c/p\u003e\n\u003cp\u003eSince $(y_m)$ is a Cauchy sequence, for every $\\varepsilon \u0026gt; 0$ there is an $N$ such that $\\|y_m - y_r\\| \u0026lt; \\varepsilon$ when $m,r \u0026gt; N$. We have for some $c \u0026gt; 0$\u003c/p\u003e\n\u003cp\u003e$$\n\\varepsilon \u0026gt; \\|y_m - y_r\\|\n= \\left\\| \\sum_{j=1}^n \\left(\\alpha_j^{(m)} - \\alpha_j^{(r)}\\right)e_j \\right\\|\n\\ge c \\sum_{j=1}^n \\left|\\alpha_j^{(m)} - \\alpha_j^{(r)}\\right|\n$$\u003c/p\u003e\n\u003cp\u003e$$\n\\Rightarrow \\left|a_j^{(m)} - a_j^{(r)}\\right| \\le \\frac{\\varepsilon}{c}.\n$$\u003c/p\u003e\n\u003cp\u003eThis shows that each of the $n$ sequences\u003c/p\u003e\n\u003cp\u003e$$\n\\left(\\alpha_j^{(m)}\\right)=\\left(\\alpha_j^{(1)},\\alpha_j^{(2)},\\ldots\\right), \\qquad j=1,\\ldots,n\n$$\u003c/p\u003e\n\u003cp\u003eis 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\u003c/p\u003e\n\u003cp\u003e$$\ny=\\alpha_1e_1+\\cdots+\\alpha_ne_n.\n$$\u003c/p\u003e\n\u003cp\u003eClearly, $y\\in Y$ and\u003c/p\u003e\n\u003cp\u003e$$\n\\|y_m-y\\| =\n\\left\\|\\sum_{j=1}^n \\left(\\alpha_j^{(m)}-\\alpha_j\\right)e_j\\right\\|\n\\le\n\\sum_{j=1}^n \\left|\\alpha_j^{(m)}-\\alpha_j\\right|\\|e_j\\|.\n$$\u003c/p\u003e\n\u003cp\u003eOn the right, $\\alpha_j^{(m)}\\to \\alpha_j$. Hence $\\|y_m-y\\|\\to 0$, that is $y_m\\to y$.\u003c/p\u003e\n\u003cp\u003eThis shows that $(y_m)$ is convergent in $Y$. Since $(y_m)$ was an arbitrary Cauchy sequence in $Y$, this proves that $Y$ is complete.\u003c/p\u003e\n\u003ch3 id=\"33-theorem-closedness\"\u003e3.3 Theorem (Closedness)\u003c/h3\u003e\n\u003cp\u003eEvery finite dimensional subspace $Y$ of a normed space $X$ is closed in $X$.\u003c/p\u003e\n\u003ch3 id=\"34-definition-equivalent-norms\"\u003e3.4 Definition (Equivalent norms)\u003c/h3\u003e\n\u003cp\u003eA 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\u003c/p\u003e\n\u003cp\u003e$$\na\\|x\\|_0 \\le \\|x\\| \\le b\\|x\\|_0.\n$$\u003c/p\u003e\n\u003cp\u003eEquivalent norms on $X$ define the same topology for $X$.\u003c/p\u003e\n\u003ch3 id=\"35-theorem-equivalent-norms\"\u003e3.5 Theorem (Equivalent norms)\u003c/h3\u003e\n\u003cp\u003eOn a finite dimensional vector space $X$ any norm $\\|\\cdot\\|$ is equivalent to any other norm $\\|\\cdot\\|_0$.\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eLet $\\dim X=n$ and ${e_1,\\ldots,e_n}$ any basis for $X$. Then every $x\\in X$ has a unique representation\u003c/p\u003e\n\u003cp\u003e$$\nx=\\alpha_1e_1+\\cdots+\\alpha_ne_n.\n$$\u003c/p\u003e\n\u003cp\u003eThen there exists $c\u0026gt;0$ such that\u003c/p\u003e\n\u003cp\u003e$$\n\\|x\\|\\ge c\\left(|a_1|+\\cdots+|a_n|\\right).\n$$\u003c/p\u003e\n\u003cp\u003eOn the other hand, the triangle inequality gives\u003c/p\u003e\n\u003cp\u003e$$\n\\|x\\|_0 \\le \\sum _{j = 1}^n \\left|\\alpha_j \\right|\\|e_j\\|_0 \\le k \\sum _{j=1}^n|\\alpha_j| \\qquad k = \\max_j \\|e_j\\|_0.\n$$\u003c/p\u003e\n\u003cp\u003eHence $a\\|x\\|_0 \\le \\|x\\|,$ where $ a=\\frac{c}{k}\u0026gt;0.$\nInterchange of the roles of $\\|\\cdot\\|$ and $\\|\\cdot\\|_0$, we also have\n$\n\\|x\\|\\le b\\|x\\|_0,\n$\nso these norms are equivalent.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"4-compactness-and-finite-dimension\"\u003e4. Compactness and Finite Dimension\u003c/h2\u003e\n\u003ch3 id=\"41-definition-compactness\"\u003e4.1 Definition (Compactness)\u003c/h3\u003e\n\u003cp\u003eA 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 $M$.\u003c/p\u003e\n\u003ch3 id=\"42-lemma-compactness\"\u003e4.2 Lemma (Compactness)\u003c/h3\u003e\n\u003cp\u003eA compact subset $M$ of a metric space is closed and bounded.\u003c/p\u003e\n\u003ch3 id=\"43-theorem-compactness\"\u003e4.3 Theorem (Compactness)\u003c/h3\u003e\n\u003cp\u003eIn a finite dimensional normed space $X$, any subset $M \\subset X$ is compact if and only if $M$ is closed and bounded.\u003c/p\u003e\n\u003ch3 id=\"44-f-rieszs-lemma\"\u003e4.4 F. Riesz\u0026rsquo;s Lemma\u003c/h3\u003e\n\u003cp\u003eLet $Y$ 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\u003c/p\u003e\n\u003cp\u003e$$\n\\|z\\| = 1, \\qquad \\|z - y\\| \\ge \\theta \\quad \\text{for all } y \\in Y.\n$$\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eWe consider any $v \\in Z \\setminus Y$ and denote its distance from $Y$ by $a$, that is,\u003c/p\u003e\n\u003cp\u003e$$\na = \\inf_{y \\in Y} |v - y|.\n$$\u003c/p\u003e\n\u003cp\u003eClearly, $a \u0026gt; 0$ 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\u003c/p\u003e\n\u003cp\u003e$$\na \\le |v - y_0| \\le \\frac{a}{\\theta}.\n$$\u003c/p\u003e\n\u003cp\u003eLet\n$\nz = c(v - y_0), \\text{ where } c = \\frac{1}{|v - y_0|},\n$\nthen $\\|z\\| = 1$, and we show that $\\|z - y\\| \\ge \\theta$ for every $y \\in Y$. We have\u003c/p\u003e\n\u003cp\u003e\\begin{align*}\n\\|z - y\\| \u0026amp;= |c(v - y_0) - y| \\\\\n\u0026amp;= c \\|v - y_0 - c^{-1} y\\| \\\\\n\u0026amp;= c \\| v - y_1 \\| \\qquad (y_1 = y_0 + c^{-1} y)\n\\end{align*}\u003c/p\u003e\n\u003cp\u003eThe form of $y_1$ show that $y_1 \\in Y$. Hence $\\|v - y_1\\| \\ge a$ and we obtain\n$$\n\\|z - y\\| = c \\|v - y_1\\| \\ge ca = \\frac{a}{|v - y_0|} \\ge \\frac{a}{a/\\theta} = \\theta.\n$$\u003c/p\u003e\n\u003ch3 id=\"45-theorem-finite-dimension\"\u003e4.5 Theorem (Finite dimension)\u003c/h3\u003e\n\u003cp\u003eIf 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.\u003c/p\u003e\n\u003ch3 id=\"46-theorem-continuous-mapping\"\u003e4.6 Theorem (Continuous mapping)\u003c/h3\u003e\n\u003cp\u003eLet $X$ 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.\u003c/p\u003e\n\u003ch3 id=\"47-corollary-maximum-and-minimum\"\u003e4.7 Corollary (Maximum and Minimum)\u003c/h3\u003e\n\u003cp\u003eA 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 $M$.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"5-linear-operators\"\u003e5. Linear operators\u003c/h2\u003e\n\u003ch3 id=\"51-definition-linear-operator\"\u003e5.1 Definition (Linear operator)\u003c/h3\u003e\n\u003cp\u003eA linear operator $T$ is an operator such that\u003c/p\u003e\n\u003cp\u003e$\\text{(i)} \\quad$ 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.\u003c/p\u003e\n\u003cp\u003e$\\text{(ii)} \\quad$ For all $x,y \\in \\mathcal{D}(T)$ and scalars $\\alpha$,\n$$\nT(x+y) = Tx + Ty\n$$\n$$\nT(\\alpha x) = \\alpha Tx\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eNote.\u003c/strong\u003e $\\mathcal{N}(T)$ denotes the null space of $T$, the set of all $x \\in \\mathcal{D}(T)$ such that $Tx = 0$ (Kernel).\u003c/p\u003e\n\u003cp\u003e$T$ is a homomorphism of a vector space (its domain) into another space, that is $T$ preserves the two operations of vector space.\u003c/p\u003e\n\u003ch3 id=\"example-2\"\u003eExample\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003e1. Identity operator.\u003c/strong\u003e 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 $Ix = x$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e2. Differentiation.\u003c/strong\u003e Let $X$ be the vector space of all polynomials on $[a,b]$. We may define a linear operator $T$ on $X$ by setting\n$$\nT x(t) = x\u0026rsquo;(t)\n$$\nfor every $x \\in X$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e3. Integration.\u003c/strong\u003e A linear operator $T$ from $C[a,b]$ into itself can be defined by\n$$\nT x(t) = \\int_a^t x(\\tau)\\ \\mathrm{d}\\tau\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e4. Multiplication by $t$.\u003c/strong\u003e Another linear operator from $C[a,b]$ into itself is defined by\n$$\nT x(t) = t x(t)\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e5. Elementary vector algebra.\u003c/strong\u003e 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\n$$\nT_2 x = x \\cdot a = \\xi_1 \\alpha_1 + \\xi_2 \\alpha_2 + \\xi_3 \\alpha_3\n$$\nwhere $\\alpha = (\\alpha_j) \\in \\mathbb{R}^3$ is fixed.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e6. Matrices.\u003c/strong\u003e 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\n$$\ny = Ax\n$$\nwhere $x = (x_j)$ has $n$ components and $y = (y_j)$ has $r$ components:\u003c/p\u003e\n\u003cp\u003e$$\n\\begin{bmatrix}\ny_1 \\\\\ny_2 \\\\\n\\vdots \\\\\ny_r\n\\end{bmatrix} =\n\\begin{bmatrix}\n\\alpha_{11} \u0026amp; \\alpha_{12} \u0026amp; \\cdots \u0026amp; \\alpha_{1n} \\\\\n\\alpha_{21} \u0026amp; \\alpha_{22} \u0026amp; \\cdots \u0026amp; \\alpha_{2n} \\\\\n\\vdots \u0026amp; \\vdots \u0026amp; \\ddots \u0026amp; \\vdots \\\\\n\\alpha_{r1} \u0026amp; \\alpha_{r2} \u0026amp; \\cdots \u0026amp; \\alpha_{rn}\n\\end{bmatrix}\n\\begin{bmatrix}\nx_1 \\\\\nx_2 \\\\\n\\vdots \\\\\nx_n\n\\end{bmatrix}\n$$\u003c/p\u003e\n\u003ch3 id=\"52-theorem-range-and-null-space\"\u003e5.2 Theorem (Range and null space)\u003c/h3\u003e\n\u003cp\u003eLet $T$ be a linear operator, then\u003c/p\u003e\n\u003cp\u003e$\\text{(a)} \\quad$ The range $\\mathcal{R}(T)$ is a vector space.\u003c/p\u003e\n\u003cp\u003e$\\text{(b)} \\quad$ If $\\dim \\mathcal{D}(T) = n \u0026lt; \\infty$, then $\\dim \\mathcal{R}(T) \\le n$.\u003c/p\u003e\n\u003cp\u003e$\\text{(c)} \\quad$ The null space $\\mathcal{N}(T)$ is a vector space.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eInverse of linear operator.\u003c/strong\u003e $T : \\mathcal{D}(T) \\to Y$ is injective or one-to-one, there exists the mapping\u003c/p\u003e\n\u003cp\u003e\\begin{align}\nT^{-1} : \u0026amp; \\mathcal{R}(T) \\to \\mathcal{D}(T), \\\\\n\u0026amp; \\quad y_0 \\text{ }\\text{ } \\mapsto \\text{ } \\text{ } x_0 \\quad (y_0 = Tx_0)\n\\end{align}\nWe have\n$$\nT^{-1}Tx = x, \\qquad TT^{-1}y = y\n$$\u003c/p\u003e\n\u003ch3 id=\"53-theorem-inverse-operator\"\u003e5.3 Theorem (Inverse operator)\u003c/h3\u003e\n\u003cp\u003eLet $X, Y$ 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:\u003c/p\u003e\n\u003cp\u003e$\\text{(a)} \\quad$ The inverse $T^{-1} : \\mathcal{R}(T) \\to \\mathcal{D}(T)$ exists if and only if\u003c/p\u003e\n\u003cp\u003e$$\nTx = 0 \\Rightarrow x = 0\n$$\u003c/p\u003e\n\u003cp\u003e$\\text{(b)} \\quad$ If $T^{-1}$ exists, it is a linear operator.\u003c/p\u003e\n\u003cp\u003e$\\text{(c)} \\quad$ If $\\dim \\mathcal{D}(T) = n \u0026lt; \\infty$ and $T^{-1}$ exists, then\n$\n\\dim \\mathcal{R}(T) = \\dim \\mathcal{D}(T)\n$\u003c/p\u003e\n\u003ch3 id=\"54-lemma-inverse-of-product\"\u003e5.4 Lemma (Inverse of product)\u003c/h3\u003e\n\u003cp\u003eLet $T : X \\to Y$ 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\u003c/p\u003e\n\u003cp\u003e$$\n(ST)^{-1} = T^{-1} S^{-1}.\n$$\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"6-bounded-and-continuous-linear-operators\"\u003e6. Bounded and Continuous Linear Operators\u003c/h2\u003e\n\u003ch3 id=\"61-definition-bounded-linear-operator\"\u003e6.1 Definition (Bounded linear operator)\u003c/h3\u003e\n\u003cp\u003eLet $X$ 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 $x \\in \\mathcal{D}(T)$,\u003c/p\u003e\n\u003cp\u003e$$\n\\|Tx\\| \\le c \\|x\\|\n$$\u003c/p\u003e\n\u003cp\u003e$\\Rightarrow$ $c$ 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\u003c/p\u003e\n\u003cp\u003e$$\n\\|T\\| = \\sup_{\\substack{x \\in \\mathcal{D}(T) \\\\ x \\ne 0}} \\frac{\\|Tx\\|}{\\|x\\|}\n$$\u003c/p\u003e\n\u003cp\u003e$\\|T\\|$ is called the norm of the operator $T$. If $\\mathcal{D}(T) = \\{0\\}$ we define $\\|T\\| = 0$. Note that, with $c = \\|T\\|$,\u003c/p\u003e\n\u003cp\u003e$$\n\\|Tx\\| \\le \\|T\\| \\|x\\|\n$$\u003c/p\u003e\n\u003ch3 id=\"62-lemma-norm\"\u003e6.2 Lemma (Norm)\u003c/h3\u003e\n\u003cp\u003eLet $T$ be a bounded linear operator. Then, an alternative formula for the norm of $T$ is\u003c/p\u003e\n\u003cp\u003e$$\n\\|T\\| = \\sup_{\\substack{x \\in \\mathcal{D}(T) \\\\ \\|x\\| = 1}} \\|Tx\\|\n$$\u003c/p\u003e\n\u003ch3 id=\"example-3\"\u003eExample\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003e1. Identity operator.\u003c/strong\u003e The identity operator $I : X \\to X$ on a normed space $X \\ne \\{0\\}$ is bounded and has norm $\\|I\\| = 1$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e2. Differentiation operator.\u003c/strong\u003e 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\n$$\nTx(t) = x\u0026rsquo;(t)\n$$\nThis operator is linear but not bounded.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e3. Integral operator.\u003c/strong\u003e  Define an integral operator\n$\nT: C[0,1] \\to C[0,1]\n$\nby\n$$\ny = Tx, \\text{ where} \\quad  y(t)=\\int_0^1 k(t,\\tau)x(\\tau) \\mathrm{d}\\tau.\n$$\nHere $k$ is a given function, which is called the \u003cem\u003ekernel\u003c/em\u003e of $T$, and it is assumed to be continuous on the closed square\n$\nG = J \\times J\n$\nin the $t\\tau$-plane, where\n$\nJ=[0,1].\n$\u003c/p\u003e\n\u003cp\u003eThis operator is linear, $T$ is bounded.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e4. Matrix.\u003c/strong\u003e A real matrix\n$\nA=(\\alpha_{jk})\n$\nwith $r$ rows and $n$ columns defines an operator\n$\nT:\\mathbb{R}^n \\to \\mathbb{R}^r\n$\nby means of\n$$\ny=Ax.\n$$\nwhere\n$\nx=(x_i) \\text{ and } y=(y_i); \\quad y_i=\\sum_{k=1}^n \\alpha_{jk}x_k \\quad (j=\\overline{1,r}).\n$\n$T$ is linear and bounded.\u003c/p\u003e\n\u003ch3 id=\"63-theorem-finite-dimension\"\u003e6.3 Theorem (Finite dimension)\u003c/h3\u003e\n\u003cp\u003eIf a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded.\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\nLet $\\dim X=n$ and $\\{e_1,\\dots,e_n\\}$ be a basis for $X$. We take any\n$\nx=\\sum \\xi_i e_i\n$\nand consider any linear operator $T$ on $X$. Since $T$ is linear,\n\\begin{align}\n\\|Tx\\|=\\left\\| \\sum \\xi_i Te_i \\right\\| \u0026amp; \\le \\sum |\\xi_i|\\ \\|Te_i\\| \\\\\n\u0026amp; \\le \\max_k \\|Te_k\\| \\sum |\\xi_i| \\\\\n\u0026amp; \\le \\max_k \\|Te_k\\| \\frac{1}{c}\\left\\|\\sum \\xi_i e_i\\right\\|\n= \\delta \\|x\\|.\n\\end{align}\nwhere\n$\n\\delta=\\frac{1}{c}\\max\\limits_k \\|Te_k\\|.\n$\u003c/p\u003e\n\u003ch3 id=\"64-theorem-continuity-and-boundedness\"\u003e6.4 Theorem (Continuity and boundedness)\u003c/h3\u003e\n\u003cp\u003eLet\n$\nT:\\mathcal{D}(T)\\to Y\n$\nbe a linear operator, where $\\mathcal{D}(T)\\subset X$ and $X,Y$ are normed spaces. Then:\u003c/p\u003e\n\u003cp\u003e$\\text{(a)} \\quad$ $T$ is continuous if and only if $T$ is bounded.\u003c/p\u003e\n\u003cp\u003e$\\text{(b)} \\quad$ If $T$ is continuous at a single point, it is continuous.\u003c/p\u003e\n\u003ch3 id=\"65-corollary-continuity-null-space\"\u003e6.5 Corollary (Continuity, null space)\u003c/h3\u003e\n\u003cp\u003eLet $T$ be a bounded linear operator. Then:\u003c/p\u003e\n\u003cp\u003e$\\text{(a)} \\quad$ $x_n\\to x$ $( \\text{where } x_n,x\\in \\mathcal{D}(T))$ implies $Tx_n\\to Tx$.\u003c/p\u003e\n\u003cp\u003e$\\text{(b)} \\quad$ The null space $\\mathcal{N}(T)$ is closed.\u003c/p\u003e\n\u003ch3 id=\"66-theorem-bounded-linear-extension\"\u003e6.6 Theorem (Bounded linear extension)\u003c/h3\u003e\n\u003cp\u003eLet\n$\nT:\\mathcal{D}(T)\\to Y\n$\nbe 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\n$$\n\\tilde T:\\overline{\\mathcal{D}(T)}\\to Y.\n$$\nwhere $\\tilde T$ is a bounded linear operator with\n$$\n\\|\\tilde T\\|=\\|T\\|.\n$$\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"7-linear-functionals\"\u003e7. Linear Functionals\u003c/h2\u003e\n\u003ch3 id=\"71-definition-linear-functional\"\u003e7.1 Definition (Linear functional)\u003c/h3\u003e\n\u003cp\u003eA linear functional $f$ is a linear operator with domain in a vector space $X$ and range in the scalar field $K$ of $X$, thus\n$$\nf:\\mathcal{D}(f)\\to K.\n$$\nwhere $K=\\mathbb{R}$ if $X$ is real and $K=\\mathbb{C}$ if $X$ is complex.\u003c/p\u003e\n\u003ch3 id=\"72-definition-bounded-linear-functional\"\u003e7.2 Definition (Bounded linear functional)\u003c/h3\u003e\n\u003cp\u003eA 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\n$x\\in\\mathcal{D}(f)$,\n$$\n|f(x)|\\le c\\|x\\|.\n$$\nFurthermore, the norm of $f$ is\n$$\n\\|f\\|=\\sup_{\\substack{x\\in\\mathcal{D}(f)\\\\ x\\ne 0}}\\frac{|f(x)|}{\\\n|x\\|}\n$$\nor\n$$\n\\|f\\|=\\sup_{\\substack{x\\in\\mathcal{D}(f)\\\\ \\|x\\|=1}}|f(x)|.\n$$\n$$\n\\Rightarrow |f(x)|\\le \\|f\\|\\|x\\|.\n$$\u003c/p\u003e\n\u003ch3 id=\"73-theorem-continuity-and-boundedness\"\u003e7.3 Theorem (Continuity and boundedness)\u003c/h3\u003e\n\u003cp\u003eA linear functional $f$ with domain $\\mathcal{D}(f)$ in a normed space is continuous if and only if $f$ is bounded.\u003c/p\u003e\n\u003ch3 id=\"examples\"\u003eExamples\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003e1. Norm.\u003c/strong\u003e The norm $\\|\\cdot\\|:X\\to\\mathbb{R}$ on a normed space $(X,\\|\\cdot\\|)$ is a functional on $X$ which is not linear.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e2. Dot product.\u003c/strong\u003e The dot product with one factor kept fixed defines a functional\n$\nf:\\mathbb{R}^3\\to\\mathbb{R}\n$\n$$\nf(x)=x\\cdot a=\\xi_1 \\alpha_1+\\xi_2 \\alpha_2+\\xi_3 \\alpha_3.\n$$\nwhere $\\alpha=(\\alpha_j)\\in\\mathbb{R}^3$ is fixed.\u003cbr\u003e\n$f$ is linear, bounded. In fact,\n$$\n|f(x)|=|x\\cdot a|\\le \\|x\\|\\|a\\|\n\\Rightarrow \\|f\\|=\\|a\\|.\n$$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e3. Definite integral.\u003c/strong\u003e Consider integral for all functions in a certain function space, $f$ is defined by\n$$\nf(x)=\\int_a^b x(t)\\ \\mathrm{d}t, \\qquad x\\in C[a,b].\n$$\n$f$ is linear, bounded and has norm\n$\n\\|f\\|=b-a.\n$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e4. Space $C[a,b]$.\u003c/strong\u003e Choose a fixed $t_0\\in J=[a,b]$ and set\n$$\nf_t(x)=x(t_0), \\qquad x\\in C[a,b].\n$$\n$f_t$ is linear, $f_t$ is bounded and has norm\n$\n\\|f_t\\|=1.\n$\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e5. Space $\\ell^2$.\u003c/strong\u003e We can obtain a linear functional $f$ on the Hilbert space $\\ell^2$ by choosing a fixed\n$\na=(a_j)\\in \\ell^2\n$\nand setting\n$$\nf(x)=\\sum_{j=1}^{\\infty}\\xi_j a_j,\n$$\nwhere\n$\nx=(\\xi_j)\\in \\ell^2.\n$\nThis series converges absolutely and $f$ is bounded.\u003c/p\u003e\n\u003cp\u003e\u0026ndash; 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 \u003cstrong\u003ealgebraic dual space\u003c/strong\u003e of $X$.\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003eThe sum $f_1+f_2$ of two functionals $f_1$ and $f_2$ is the functional $s$ whose value at every $x\\in X$ is\n$$\ns(x)=(f_1+f_2)(x)=f_1(x)+f_2(x).\n$$\u003c/li\u003e\n\u003cli\u003eThe product of a scalar $\\alpha$ and a functional $f$ is the functional $p$ whose value at $x\\in X$ is\n$$\np(x)=(\\alpha f)(x)=\\alpha f(x).\n$$\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003e\u0026ndash; Consider the algebraic dual $(X^{*})^{*}$ of $X^*$, whose elements are the linear functionals defined on $X^*$. Denote $(X^*)^*$ by $X^{* *}$ and call it the \u003cstrong\u003esecond algebraic dual space\u003c/strong\u003e of $X$.\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003eWe can obtain a $g\\in X^{**}$ which is a linear functional defined on $X^*$ by choosing a fixed $x\\in X$ and setting\n$$\ng(f)=g_x(f)=f(x) \\qquad (x\\in X \\text{ fixed}, f\\in X^* \\text{ variable}).\n$$\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003e\u0026ndash; To each $x\\in X$ there corresponds a $g_x\\in X^{**}$. This defines a mapping\n\\begin{align}\nC:\\quad  \u0026amp; X \\to X^{**} \\\\\n\u0026amp; x \\text{ } \\mapsto g_x.\n\\end{align}\u003c/p\u003e\n\u003cp\u003e$C$ is called the \u003cstrong\u003ecanonical mapping\u003c/strong\u003e of $X$ into $X^{**}$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eIsomorphism\u003c/strong\u003e. A bijective mapping of $X$ onto $\\tilde X$ which preserves the structure.\u003c/p\u003e\n\u003cp\u003eAccordingly, 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 $x,y\\in X$,\n$$\n\\tilde d(Tx,Ty)=d(x,y).\n$$\u003c/p\u003e\n\u003cp\u003e$\\tilde X$ is then called \u003cem\u003eisomorphic\u003c/em\u003e with $X$. $T: X\\to \\tilde X$ is a bijective linear operator,\n$$\nT(x+y)=Tx+Ty, \\qquad T(\\alpha x)=\\alpha Tx.\n$$\u003c/p\u003e\n\u003cp\u003e$X$ and $\\tilde X$ are called isomorphic vector spaces.\u003c/p\u003e\n\u003cp\u003e\u0026ndash; Canonical mapping $C$ is injective. Since $C$ is linear, it is an isomorphism of $X$ onto the range $\\mathcal{R}(C)\\subset X^{**}$.\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003e\n\u003cp\u003eIf $X$ is isomorphic with a subspace of a vector space $Y$, we say that $X$ is \u003cem\u003eembeddable\u003c/em\u003e in $Y$. $X$ is embeddable in $X^{**}$ and $C$ is also called the \u003cem\u003ecanonical embedding\u003c/em\u003e of $X$ into $X^{**}$.\u003c/p\u003e\n\u003c/li\u003e\n\u003cli\u003e\n\u003cp\u003eIf $C$ is surjective, so that $\\mathcal{R}(C)=X^{**}$, then $X$ is said to be \u003cem\u003ealgebraically reflexive\u003c/em\u003e.\u003c/p\u003e\n\u003c/li\u003e\n\u003c/ul\u003e\n\u003chr\u003e\n\u003ch2 id=\"8-linear-operators-and-functionals-on-finite-dimensional-spaces\"\u003e8. Linear Operators and Functionals on Finite Dimensional Spaces\u003c/h2\u003e\n\u003cp\u003e\u0026ndash; 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\n$\nE=\\{e_1,\\dots,e_n\\} \\text{ for } X\n$\nand a basis\n$\nB=\\{b_1,\\dots,b_m\\} \\text{ for } Y,\n$\nwith the vectors arranged in a definite order which we keep fixed. Then every $x\\in X$ has a unique representation\n$$\nx=\\xi_1 e_1 + \\cdots + \\xi_n e_n.\n$$\u003c/p\u003e\n\u003cp\u003eSince $T$ is linear, $x$ has the image\n$$\ny = Tx = T\\left(\\sum_{k=1}^n \\xi_k e_k\\right) = \\sum_{k=1}^n \\xi_k T e_k.\n$$\u003c/p\u003e\n\u003cp\u003e$y$ and $y_k = T e_k$ are in $Y$; they have unique representations of the form\n$$\ny = \\sum_{j=1}^r \\eta_j b_j,\n$$\n$$\nT e_k = \\sum_{j=1}^r \\tau_{jk} b_j.\n$$\u003c/p\u003e\n\u003cp\u003eThen\n$$\ny = \\sum_{j=1}^r \\eta_j b_j\n= \\sum_{k=1}^n \\xi_k T e_k\n= \\sum_{k=1}^n \\xi_k \\sum_{j=1}^r \\tau_{jk} b_j\n= \\sum_{j=1}^r \\left(\\sum_{k=1}^n \\tau_{jk} \\xi_k \\right) b_j.\n$$\u003c/p\u003e\n\u003cp\u003e$$\n\\Rightarrow \\eta_j = \\sum_{k=1}^n \\tau_{jk} \\xi_k, \\qquad j=1,\\dots,r.\n$$\u003c/p\u003e\n\u003cp\u003eThe coefficients form a matrix:\n$\nT_{EB} = (\\tau_{jk})\n$\nwith $r$ rows and $n$ columns. The matrix $T_{EB}$ represents the operator $T$ with respect to bases.\u003c/p\u003e\n\u003cp\u003eLet $\\tilde{x} = (\\xi_k)$ and $\\tilde{y} = (\\eta_j)$. In matrix notation,\n$$\n\\tilde{y} = T_{EB}\\tilde{x},\n$$\n$$\nT e = T_{EB} b.\n$$\u003c/p\u003e\n\u003cp\u003e\u0026ndash; For every functional $f$ and every $x=\\sum\\limits_{j=1}^n \\xi_j e_j \\in X$, we have\n$$\nf(x)=f\\left(\\sum_{j=1}^n \\xi_j e_j\\right)\n= \\sum_{j=1}^n \\xi_j f(e_j)\n= \\sum_{j=1}^n \\xi_j \\alpha_j,\n$$\nwhere $\\alpha_j = f(e_j)$, 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 $X$.\nTake $n$-tuples\n$$\n(1,0,0,\\dots,0), \\quad (0,1,0,\\dots,0), \\quad \\dots, \\quad (0,0,0,\\dots,1).\n$$\u003c/p\u003e\n\u003cp\u003eThis gives $n$ functionals, denote $f_1,\\dots,f_n$, with values\n\u003cspan class=\"sidenote\" id=\"note1\"\u003e\n1. Kronecker delta\n\u003c/span\u003e\n$$\nf_j(e_k)=\\delta_{jk}^{\\hspace{0.2cm} \\color{gray}{1}}=\n\\begin{cases}\n0 \u0026amp; \\text{if } j\\ne k,\\\\\n1 \u0026amp; \\text{if } j=k,\n\\end{cases}\n$$\u003c/p\u003e\n\u003cp\u003e$\\{f_1,\\dots,f_n\\}$ is called the \u003cstrong\u003edual basis\u003c/strong\u003e of the basis $\\{e_1,\\dots,e_n\\}$ of $X$.\u003c/p\u003e\n\u003ch3 id=\"81-theorem-dimension-of-x\"\u003e8.1 Theorem (Dimension of $X^*$)\u003c/h3\u003e\n\u003cp\u003eLet $X$ be an $n$-dimensional vector space and $E=\\{e_1,\\dots,e_n\\}$ a basis for $X$. Then\n$\nF=\\{f_1,\\dots,f_n\\}\n$\nis a basis for the algebraic dual $X^*$ of $X$, and\n$\n\\dim X^* = \\dim X = n.\n$\u003c/p\u003e\n\u003ch3 id=\"82-lemma-zero-vector\"\u003e8.2 Lemma (Zero vector)\u003c/h3\u003e\n\u003cp\u003eLet $X$ be a finite dimensional vector space. If $x_0 \\in X$ has the property that\n$\nf(x_0)=0\n$\nfor all $f\\in X^*$, then $x_0=0$.\u003c/p\u003e\n\u003ch3 id=\"83-theorem-algebraic-reflexivity\"\u003e8.3 Theorem (Algebraic reflexivity)\u003c/h3\u003e\n\u003cp\u003eA finite dimensional vector space is algebraically reflexive.\u003c/p\u003e\n\u003chr\u003e\n\u003ch2 id=\"9-normed-spaces-of-operators-dual-space\"\u003e9. Normed Spaces of Operators. Dual Space.\u003c/h2\u003e\n\u003cp\u003e\u0026ndash; 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.\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003e\n\u003cp\u003eDefine $T_1+T_2$ of two operators $T_1,T_2\\in B(X,Y)$ by\n$$\n(T_1+T_2)x = T_1x + T_2x.\n$$\u003c/p\u003e\n\u003c/li\u003e\n\u003cli\u003e\n\u003cp\u003eDefine the product $\\alpha T$ of $T\\in B(X,Y)$ and a scalar $\\alpha$ by\n$$\n(\\alpha T)x = \\alpha Tx.\n$$\u003c/p\u003e\n\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp\u003eThen $B(X,Y)$ becomes a vector space so itself a normed space with norm defined by\n$$\n\\|T\\| = \\sup_{\\substack{x\\in X \\\\ x\\ne 0}} \\frac{\\|Tx\\|}{\\|x\\|}\n= \\sup_{\\|x\\|=1} \\|Tx\\|.\n$$\u003c/p\u003e\n\u003ch3 id=\"91-theorem-completeness\"\u003e9.1 Theorem (Completeness)\u003c/h3\u003e\n\u003cp\u003eIf $Y$ is a Banach space, then $B(X,Y)$ is a Banach space.\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eConsider an arbitrary Cauchy sequence $(T_n)$ in $B(X,Y)$ and show that $(T_n)$ converges to an operator $T\\in B(X,Y)$.\u003c/p\u003e\n\u003cp\u003eSince $(T_n)$ is Cauchy, for every $\\varepsilon\u0026gt;0$ there is an $N$ such that\n$$\n\\|T_n - T_m\\| \u0026lt; \\varepsilon \\quad (m,n \u0026gt; N).\n$$\u003c/p\u003e\n\u003cp\u003eFor all $x\\in X$ and $m,n\u0026gt;N$ we obtain\n\\begin{equation}\n\\|T_n x - T_m x\\| = \\|(T_n - T_m)x\\| \\le \\|T_n - T_m\\|\\|x\\| \u0026lt; \\varepsilon \\|x\\| \\tag{1}.\n\\end{equation}\u003c/p\u003e\n\u003cp\u003e$\\Rightarrow$ for fixed $x$, $(T_n x)$ is Cauchy in $Y$. Since $Y$ is complete, $(T_n x)$ converges, say\n$$\nT_n x \\to y.\n$$\nThe limit $y\\in Y$ depends on the choice of $x\\in X$. Define an operator\n$$\nT: X \\to Y, \\quad \\text{where } y=Tx.\n$$\u003c/p\u003e\n\u003cp\u003eThe operator $T$ is linear:\n$$\n\\lim T_n(\\alpha x_1 + \\beta x_2)\n= \\lim (\\alpha T_n x_1 + \\beta T_n x_2)\n= \\alpha \\lim T_n x_1 + \\beta \\lim T_n x_2.\n$$\u003c/p\u003e\n\u003cp\u003eWe prove that $T$ is bounded and $T_n \\to T$, that is, $\\|T_n - T\\|\\to 0$. Since (1) holds for every $m\u0026gt;N$ and $T_m x \\to Tx$, let $m\\to\\infty$. Using the continuity of the norm, for every $n\u0026gt;N$ and all $x\\in X$,\n\\begin{equation}\n\\|T_n x - Tx\\|\n= \\|T_n x - \\lim_{m\\to\\infty} T_m x\\|\n= \\lim_{m\\to\\infty} \\|T_n x - T_m x\\|\n\\le \\varepsilon \\|x\\|. \\tag{2}\n\\end{equation}\u003c/p\u003e\n\u003cp\u003e$\\Rightarrow (T_n - T)$ with $n\u0026gt;N$ 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\n$$\n\\|T_n - T\\| \\le \\varepsilon \\quad (n\u0026gt;N).\n$$\u003c/p\u003e\n\u003cp\u003eHence, $\\|T_n - T\\|\\to 0$.\u003c/p\u003e\n\u003ch3 id=\"92-definition-dual-space-x\"\u003e9.2 Definition (Dual space $X\u0026rsquo;$)\u003c/h3\u003e\n\u003cp\u003eLet $X$ be a normed space. Then the set of all bounded linear functionals on $X$ constitutes a normed space with norm defined by\u003c/p\u003e\n\u003cp\u003e$$\n\\|f\\|=\\sup_{\\substack{x\\in X \\\\ x\\neq 0}}\\frac{|f(x)|}{\\|x\\|}=\\sup_{\\|x\\|=1}|f(x)|.\n$$\u003c/p\u003e\n\u003cp\u003ewhich is called the dual space of $X$ and is denoted by $X\u0026rsquo;$.\u003c/p\u003e\n\u003ch3 id=\"93-theorem-dual-space\"\u003e9.3 Theorem (Dual Space)\u003c/h3\u003e\n\u003cp\u003eThe dual space $X\u0026rsquo;$ of a normed space $X$ is a Banach space.\u003c/p\u003e\n\u003cp\u003eAn 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 $x\\in X$,\n$\n\\|Tx\\|=\\|x\\|.\n$\n($T$ is isometric)\u003c/p\u003e\n\u003cul\u003e\n\u003cli\u003e$X$ is called isomorphic with $\\tilde X$.\u003c/li\u003e\n\u003cli\u003e$X$ and $\\tilde X$ are called isomorphic normed space.\u003c/li\u003e\n\u003c/ul\u003e\n\u003ch3 id=\"example-4\"\u003eExample\u003c/h3\u003e\n\u003cp\u003e\u003cstrong\u003e1. Space $\\mathbb{R}^n$.\u003c/strong\u003e The dual space of $\\mathbb{R}^n$ is $\\mathbb{R}^n$.\u003c/p\u003e\n\u003cp\u003e\u003cem\u003eProof.\u003c/em\u003e\u003c/p\u003e\n\u003cp\u003eWe have $(\\mathbb{R}^{n})^{'}=(\\mathbb{R}^{n})^*$ and every $f\\in (\\mathbb{R}^{n})^*$ has a representation\u003c/p\u003e\n\u003cp\u003e$$\nf(x)=\\sum \\xi_k \\gamma_k,\\qquad \\gamma_k=f(e_k).\n$$\u003c/p\u003e\n\u003cp\u003eBy Cauchy\u0026ndash;Schwarz inequality,\n$$\n|f(x)|\\le \\sum |\\xi_k \\gamma_k|\n\\le \\left(\\sum \\xi_j^2\\right)^{1/2}\\left(\\sum \\gamma_k^2\\right)^{1/2}\n=\\|x\\|\\left(\\sum \\gamma_k^2\\right)^{1/2}.\n$$\nTaking the supremum over all $x$ of norm $1$, we obtain\n$$\n\\|f\\|\\le \\left(\\sum \\gamma_k^2\\right)^{1/2}.\n$$\nHowever, since for $x=(x_1,\\ldots,x_n)\\to$ equality is achieved, we must have\u003c/p\u003e\n\u003cp\u003e$$\n\\|f\\|=\\left(\\sum_{k=1}^n \\gamma_k^2\\right)^{1/2}.\n$$\u003c/p\u003e\n\u003cp\u003eThis proves that norm of $f$ is the Euclidean norm, $\\|f\\|=\\|c\\|$ where $c=(\\gamma_k)\\in \\mathbb{R}^n$.\u003c/p\u003e\n\u003cp\u003eHence the mapping of $(\\mathbb{R}^{n})^{\u0026rsquo;}$ 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.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e2. Space $\\ell^1$.\u003c/strong\u003e The dual space of $\\ell^1$ is $\\ell^\\infty$.\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003e3. Space $\\ell^p$.\u003c/strong\u003e The dual space of $\\ell^p$ is $\\ell^q$; here, $1 \u0026lt; p \u0026lt; +\\infty$ and $q$ is the conjugate of $p$, that is, $1/p + 1/q = 1.$\u003c/p\u003e\n\u003ch2 id=\"references\"\u003eReferences\u003c/h2\u003e\n\u003cp\u003e[1] Kreyszig, E. (1978). \u003cem\u003eIntroductory functional analysis with applications\u003c/em\u003e. John Wiley \u0026amp; Sons.\u003c/p\u003e\n  \u003c/div\u003e\n\u003cp\u003e\n  \u003ca href=\"mailto:nguyencungthanh11@gmail.com?subject=Re: Banach%20space\" rel='noopener noreferrer'\u003e💬 Reply via e-mail\u003c/a\u003e\n\u003c/p\u003e\n","date_modified":"2025-10-08T00:00:00+00:00","date_published":"2025-10-08T00:00:00+00:00","id":"https://nguyencungthanh.github.io/2025/10/08/banach-space/","summary":"","tags":["Functional Analysis","Banach Space"],"title":"Banach space","url":"https://nguyencungthanh.github.io/2025/10/08/banach-space/"}],"title":"Tag: Banach Space - Nguyen Cung Thanh","version":"https://jsonfeed.org/version/1"}