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Metric Space

Important Equations

Lemma

Assume positive real number \(\alpha\), \(\beta\) satisfies conjugate condition \(\alpha+\beta=1\), then

\[ u^\alpha v^\beta\leq u\cdot\alpha+v\cdot\beta. \]

Since \(f(x)=x^\alpha (0<\alpha<1)\) is concave function, so tangent function at any point on the function curve lies above it. Choose \((1,1)\), we have tangent function

\[ y=\alpha(x-1)+1=\alpha x+1-\alpha=\alpha x+\beta \]

with \(\beta=1-\alpha\in (0,1)\). So

\[ x^\alpha\leq \alpha x + \beta. \]

Let \(x=\frac{u}{v}\), then

\[ \begin{align*} \left(\frac{u}{v}\right)^\alpha &\leq \alpha \frac{u}{v} +\beta\\ u^\alpha\cdot v^{-\alpha}&\leq \alpha u \cdot v^{-1}+\beta\\ u^\alpha\cdot v^{1-\alpha}&\leq \alpha u +\beta v\\ u^\alpha\cdot v^{\beta}&\leq \alpha u +\beta v. \end{align*} \]

If we let \(\alpha=\frac{1}{p}\), \(\beta=\frac{1}{q}\) in inequation, we have

\[ \begin{align} u^{\frac{1}{p}}\cdot v^{\frac{1}{q}}\leq \frac{1}{p}u+\frac{1}{q}v, \quad \frac{1}{p}+\frac{1}{q}=1.\label{conjugate-inequation} \end{align} \]

Hölder Inequation

Assume \(p\), \(q\) satisfies conjugate condition \(\frac{1}{p}+\frac{1}{q}=1\), then

(i) For \(\{\xi_n\}_{n\geq 1}\) and \(\{\eta_n\}_{n\geq 1}\), \(\xi_n,\eta_n\in \mathbb{C}\) we have

\[ \sum_{n=1}^\infty |\xi_n\eta_n|\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\cdot \left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}. \]

(ii) For \(f\), \(g\) satisfies \(\int_E |f(x)|^pdx<\infty\), \(\int_E |g(x)|^qdx<\infty\), we have

\[ \int_E |f(x)g(x)|dx\leq \left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}}\cdot \left(\int_E |g(x)|^qdx\right)^{\frac{1}{q}}. \]

The core is normalization of item to be summed.

In inequation \(\ref{conjugate-inequation}\), for \(n=1,2,\cdots\) let

\[ u=\frac{|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}, \quad v=\frac{|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)}, \]

we have

\[ \frac{|\xi_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}}\cdot \frac{|\eta_n|}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}}\leq \frac{1}{p}\frac{|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}+\frac{1}{q}\frac{|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)} \]

summing up \(n=1,2,\cdots\), we have

\[ \begin{align*} \frac{\sum_{n=1}^\infty|\xi_n\eta_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}} &\leq \frac{1}{p}\frac{\sum_{n=1}^\infty|\xi_n|^p}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)}+\frac{1}{q}\frac{\sum_{n=1}^\infty|\eta_n|^q}{\left(\sum_{n=1}^\infty |\eta_n|^q\right)}\\ \frac{\sum_{n=1}^\infty|\xi_n\eta_n|}{\left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}} &\leq \frac{1}{p}+\frac{1}{q}=1\\ \Rightarrow\quad \sum_{n=1}^\infty|\xi_n\eta_n|& \leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}}\left(\sum_{n=1}^\infty |\eta_n|^q\right)^{\frac{1}{q}}. \end{align*} \]

In inequation \(\ref{conjugate-inequation}\), let

\[ u=\frac{|f(x)|^p}{\left(\int_E |f(x)|^p dx\right)}, \quad v=\frac{|g(x)|^q}{\left(\int_E |g(x)|^q dx\right)}, \]

we have

\[ \frac{|f(x)|}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}}\cdot \frac{|g(x)|}{\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}\leq \frac{1}{p}\frac{|f(x)|^p}{\left(\int_E |f(x)|^p dx\right)}+\frac{1}{q}\frac{|g(x)|^q}{\left(\int_E |g(x)|^q dx\right)} \]

integrate on \(E\) on both sides, we have

\[ \begin{align*} \frac{\int_E |f(x)g(x)|dx}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}&\leq \frac{1}{p}\frac{\int_E |f(x)|^pdx}{\left(\int_E |f(x)|^p dx\right)}+\frac{1}{q}\frac{\int_E |g(x)|^qdx}{\left(\int_E |g(x)|^q dx\right)}\\ \frac{\int_E |f(x)g(x)|dx}{\left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}}&\leq \frac{1}{p}+\frac{1}{q}=1\\ \Rightarrow \quad \int_E |f(x)g(x)|dx &\leq \left(\int_E |f(x)|^p dx\right)^{\frac{1}{p}}\left(\int_E |g(x)|^q dx\right)^{\frac{1}{q}}. \end{align*} \]

Minkowski Inequation

Assume \(p\), \(q\) satisfies conjugate condition \(\frac{1}{p}+\frac{1}{q}=1\), then

(i) For \(\{\xi_n\}_{n\geq 1}\) and \(\{\eta_n\}_{n\geq 1}\), \(\xi_n,\eta_n\in \mathbb{C}\), we have

\[ \left(\sum_{n=1}^\infty |\xi_n+\eta_n|^p\right)^{\frac{1}{p}} \leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^{\frac{1}{p}} + \left(\sum_{n=1}^\infty |\eta_n|^p\right)^{\frac{1}{p}} \]

(ii) For \(f\), \(g\) satisfies \(\int_E |f(x)|^pdx<\infty\), \(\int_E |g(x)|^qdx<\infty\), we have

\[ \left(\int_E |f(x)+g(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_E |f(x)|^pdx\right)^{\frac{1}{p}}+\left(\int_E |g(x)|^pdx\right)^{\frac{1}{p}} \]

The core is to partition one part and use triangle inequation. Conjugate indices would help in the end.

\[ \begin{align*} \left(\sum_{n=1}^\infty |\xi_n+\eta_n|^p\right)&=\sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{p-1}\cdot |\xi_n+\eta_n|\right)\\ &\leq \sum_{n=1}^\infty \left[|\xi_n+\eta_n|^{\frac{p}{q}}\cdot \left(|\xi_n|+|\eta_n|\right)\right]\\ &=\sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{\frac{p}{q}}\cdot |\xi_n|\right) + \sum_{n=1}^\infty \left( |\xi_n+\eta_n|^{\frac{p}{q}}\cdot |\eta_n|\right)\\ &\leq \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^\frac{1}{q}\cdot \left( \sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p} + \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^\frac{1}{q}\cdot \left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \Rightarrow \quad \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{\frac{p}{q}\cdot q} \right)^{1-\frac{1}{q}} &\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p}+\left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \left( \sum_{n=1}^\infty |\xi_n+\eta_n|^{p}\right)^{\frac{1}{p}} &\leq \left(\sum_{n=1}^\infty |\xi_n|^p\right)^\frac{1}{p}+\left(\sum_{n=1}^\infty |\eta_n|^p\right)^\frac{1}{p}\\ \end{align*} \]
\[ \begin{align*} \left(\int_E |f(x)+g(x)|^p dx\right)&=\int_E \left( |f(x)+g(x)|^{p-1}\cdot |f(x)+g(x)|\right) dx\\ &\leq \int_E \left[|f(x)+g(x)|^{\frac{p}{q}}\cdot \left(|f(x)|+|g(x)|\right)\right] dx\\ &=\int_E \left( |f(x)+g(x)|^{\frac{p}{q}}\cdot |f(x)|\right) dx + \int_E \left( |f(x)+g(x)|^{\frac{p}{q}}\cdot |g(x)|\right) dx\\ &\leq \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^\frac{1}{q}\cdot \left( \int_E |f(x)|^pdx\right)^\frac{1}{p} + \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^\frac{1}{q}\cdot \left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \Rightarrow \quad \left( \int_E |f(x)+g(x)|^{\frac{p}{q}\cdot q} dx\right)^{1-\frac{1}{q}} &\leq \left(\int_E |f(x)|^pdx\right)^\frac{1}{p}+\left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \left( \int_E |f(x)+g(x)|^{p}dx\right)^{\frac{1}{p}} &\leq \left(\int_E |f(x)|^pdx\right)^\frac{1}{p}+\left(\int_E |g(x)|^pdx\right)^\frac{1}{p}\\ \end{align*} \]

Basic Concepts

度量空间的定义 | Definition of Metric Space

Assume \(M\) is a set. If there is a function \(\rho: M\times M\mapsto \mathbb{R}\) defined on it, satisfies the following axioms. That is, \(\forall x,y,z \in M\)

  • Positive
\[ \rho(x,y)>0,\quad x\neq y \]
  • Definite
\[ \rho(x, y)=0 \Leftrightarrow x=y \]
  • Symmetry
\[ \rho(x,y) = \rho(y,x) \]
  • Triangle inequality
\[ \rho(x,z) \leq \rho(x,y) +\rho(y,z) \]

then \(M\) is called metric space with metrc \(\rho\), denoted as \((M,\rho)\).

Example. Cases for \((M,\rho)\) with background of Mathematical analysis, show that the following \(\rho\) satisfies axiom of triangle inequation.

(i) \(M=\mathbb{R}^n\), \(\forall x=(\xi_1,\cdots,\xi_2), y=(\eta_1,\cdots,\eta_n)\in \mathbb{R}\), their metric is defined by

\[ \rho(x,y)=\left(\sum_{i=1}^n|\xi_i-\eta_i|^2\right)^\frac{1}{2} \]

(ii) \(M=C[a,b]\), \(\forall f,g\in C[a,b]\), their metric is defined by

\[ \rho(f,g)=\max_{t\in[a,b]}|f(t)-g(t)|. \]

Use Cauchy inequation.

\[ \left(\sum_{i=1}^n a_ib_i\right)^2\leq \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right) \]

So we have

\[ \begin{align*} \sum_{i=1}^n (a_i+b_i)^2&= \sum_{i=1}^n a_i^2 + 2\sum_{i=1}^n a_ib_i +\sum_{i=1}^n b_i^2 \\ &\leq \sum_{i=1}^n a_i^2 + 2 \left(\sum_{i=1}^n a_i^2\right)^\frac{1}{2} \left(\sum_{i=1}^n b_i^2\right)^\frac{1}{2} +\sum_{i=1}^n b_i^2 \\ &=\left[ \left(\sum_{i=1}^n a_i^2\right)^\frac{1}{2}+ \left(\sum_{i=1}^n b_i^2\right)^\frac{1}{2}\right]^2. \end{align*} \]

So we lat \(a_i=\xi_i-\zeta_i, b_i=\zeta_i-\eta_i\), we have the triangle inequation.

By

\[ \begin{align*} |x(t)+y(t)|&\leq |x(t)|+|y(t)|,\quad t\in [a,b]\\ &\leq \max_{t\in [a,b]} |x(t)|+\max_{t\in [a,b]} |y(t)|\\ \Rightarrow \quad \max_{t\in [a,b]} |x(t)+y(t)|&\leq\max_{t\in [a,b]} |x(t)|+\max_{t\in [a,b]} |y(t)|. \end{align*} \]

Example. Cases for \((M,\rho)\) regarding sequence space, show that the following \(\rho\) satisfies axiom of triangle inequation.

(i) \(M=l^p(1\leq p<\infty)\), i.e.

\[ l^p=\left\{ \{\xi_n\}_{n\geq 1}: \sum_{n=1}^\infty |\xi_n|^p<\infty. \right\} \]

\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^p\), their metric is defined by

\[ \rho(x,y)=\left(\sum_{n=1}^\infty |\xi_n-\eta_n|^p\right)^{\frac{1}{p}}. \]

(ii) \(M=l^\infty\), or

\[ l^\infty:=\{\{\xi_n\}_{n\geq 1}: \exists M>0, \text{ s.t } |\xi_n|\leq M,\forall n.\} \]

\(\forall x=\{\xi_n\}_{n\geq 1}, y=\{\eta_n\}_{n\geq 1}\in l^\infty\), their metric is defined by

\[ \rho(x,y)=\sup_{n}|\xi_n-\eta_n|. \]

(i) Using Minkowski Inequation.

(ii) Similar to the previous example.

Example. Cases for \((M,\rho)\) with background of Real analysis, to be more specific, \(L^p\) Space, show that the following \(\rho\) satisfies axiom of triangle inequation. Note that we use "\(=\)" to denote equation almost everywhere.

(i) \(M=L^p(E), (1\leq p<\infty, E\subset \mathbb{R} \text{ is measurable})\), i.e.

\[ L^p=\left\{ \text{measurable } f : \int_E |f(t)|^pdt<\infty. \right\} \]

\(\forall f, g \in L^p(E)\), their metric is defined by

\[ \rho(f,g)=\left( \int_E |f(t)-g(t)|^p dt\right)^{\frac{1}{p}}. \]

(ii) \(M=L^\infty(E)\), or

\[ L^\infty:=\{\text{measurable } f : \exists M>0, \exists E_0\subset E, \text{ s.t } m(E_0)=0, \text{ & } |f(t)|\leq M,\forall t\in E-E_0.\} \]

\(\forall f,g\in L^\infty(E)\), their metric is defined by

\[ \rho(f,g)=\inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|f(t)-g(t)|\right\}. \]

(i) By Minkowski Inequation.

(ii) Here the definition of the metric is to emit the unboundedness on zero-measure set. By infimum, \(\forall a(t),b(t), t\in E\), \(\forall \varepsilon>0\), \(\exists E_1,E_2\subset F\) and \(m(E_1)=m(E_2)=0\), s.t.

\[ \begin{align*} \sup_{t\in E-E_1} |a(t)|\leq \inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|a(t)|\right\} + \frac{\varepsilon}{2},\\ \sup_{t\in E-E_1} |b(t)|\leq \inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|b(t)|\right\} + \frac{\varepsilon}{2}. \end{align*} \]

So since \(m(E_1\cup E_2)=0\),

\[ \begin{align*} \inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|a(t)+b(t)|\right\}&\leq\sup_{t\in E-(E_1+E_2)} |a(t)+b(t)|\\ &\leq \sup_{t\in E-(E_1+E_2)} |a(t)| +\sup_{t\in E-(E_1+E_2)} |b(t)|\\ &\leq \sup_{t\in E-E_1} |a(t)|+\sup_{t\in E-E_2} |b(t)|\\ &\leq \inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|a(t)|\right\} +\inf_{m(E_0)=0,\atop E_0\subset E}\left\{\sup_{t\in E-E_0}|b(t)|\right\} +\varepsilon. \end{align*} \]

Let \(\varepsilon\rightarrow 0\), and we have the result.

Convergence

Readers could compare this definition to convergence in \(L^p\) Space.

Definitions of Convergence

Assume \(\{x_n\}_{n\geq 1}\subset M\). If there exists \(x_0\in M\), such that \(\rho(x_n,x_0)\rightarrow 0(n\rightarrow \infty)\), then we call \(\{x_n\}\) converges to \(x_0\), denoted by

\[ \lim_{n\rightarrow \infty}x_n=x_0 \]

and \(x_0\) is a limit of \(\{x_n\}\).

Properties of convergence

Assume \(\{x_n\}_{n\geq 1}\subset M\) is a convergent sequence, then

(i) Limit of \(\{x_n\}_{n\geq 1}\) exists uniquely.

(ii) \(\forall y_0\in M\), \(\{\rho(x_n,y_0)\}_{n\geq 1}\) is bounded.

(iii) Subsequence. for all subsequence \(\{x_{n_j}\}\) converges to the same limit. Conversely, if for all subsequence \(\{x_{n_j}\}\) converges, then \(\{x_n\}\) also converges.

(i) Assume there are two distinct limits \(x_0,y_0\in M\),

\[ \rho(x_0,y_0)<\rho(x_0,x_n)+\rho(x_n,y_0) \rightarrow 0(n\rightarrow \infty). \]

(ii) easy to see.

(iii) The forward direction is easy. But reversely, if \(\{x_n\}\) does not converge, then there exist at least two distinct subsequential limits \(L_1\) and \(L_2\), \(L_1\neq L_2\), so which contradicts the assumption!

Topology

Definitions

Assume metric space \((M,\rho)\).

(i) If \(x_0\in M\), \(r>0\), Define open ball (neighborhood)

\[ B(x_0, r)=\{x: \rho(x_0,x)<r\}, \]

and closed ball

\[ \overline{B}(x_0, r)=\{x: \rho(x_0,x)\leq r\}. \]

(ii) Assume \(G\subset M\), \(x\in G\), \(x\) is an interior point of \(G\), if \(\exists\) neighborhood \(B(x,r)\subset G\). The kernel of a set \( G \) is a set made up of all its interior points, denoted as \(G^o\). \(G\) is an open set if

\[ \forall x \in G, x \text{ is an interior point of } G. \]

Define \(\varnothing\) is an open set.

Properties of open sets

(i) \(M\) is an open set.

(ii) \(\{X_\lambda\}_{\lambda\in \Lambda}\), \(\bigcup_{\lambda\in \Lambda} X_\lambda\) is an open set.

(iii) \(\{X_n\}_{1\leq n\leq N}\), \(\bigcap_{n=1}^N X_n\) is an open set.

Definitions 2

Assume metric space \((M,\rho)\). \(G\subset M\), \(x_0\in M\).

(i) If \(\forall \varepsilon>0\),

\[ B(x_0, \varepsilon)\cap (G-\{x_0\})\neq \varnothing, \]

then \(G\) is an accumulation point.

(ii) The Derived Set of \( G \) is defined as the set composed of all accumulation points of \( G \), denoted as \( G' \).

(iii) The Closure of \( G \) is the union of \( G \) and its derived set, denoted as:

\[ \overline{G} = G\cup G'. \]

Properties of Closure

Assume metric space \((M,\rho)\), \(A,B\subset X\), then

(i) \(A\subset \overline{A}\), (ii) \(\overline{\overline{A}}=\overline{A}\), (iii) \(\overline{A\cup B}=\overline{A}\cup \overline{B}\), (iv) \(\overline{\varnothing}=\varnothing\).

Density & Separability

Definition of Density

Assume \(X\) is a metric space, \(A,B\subset X\). If \(\overline{B}\supset A\), then \(B\) is dense in \(A\).

Equivalent proposition of Density

Assume \(X\) is a metric space, \(A,B\subset X\). Then the following statements are equivalent.

(i) \(B\) is dense in \(A\),

(ii) \(\forall x\in A\), \(\forall \varepsilon>0\), \(\exists y\in B\), such that \(x\in B(y,\varepsilon)\) (or \(\rho(x,y)<\varepsilon\)).

(iii) \(\forall x\in A\), \(\exists \{x_n\}\in B\), such that \(\lim\limits_{n\rightarrow\infty}x_n=x\).

(iv) \(\forall \varepsilon>0\), \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\).

  • (i) \(\Rightarrow\) (ii)

\(\forall x\in A\), since \(A\subset \overline{B}\), then \(x\in \overline{B}\), so \(\forall \varepsilon>0\), \(B(x,\varepsilon)\cap B\neq\varnothing\). So \(\exists y\in B(x,\varepsilon)\cap B\), which satisfies \(\rho(x,y)<\varepsilon\).

  • (ii) \(\Rightarrow\) (iii)

\(\forall x\in A\), \(\forall \varepsilon_k=\frac{1}{k}>0\), \(\exists y_k\in B\), such that \(\rho(x,y_k)<\varepsilon_k\). So \(\lim\limits_{k\rightarrow \infty}y_k=x\).

  • (iii) \(\Rightarrow\) (iv)

\(\forall x\in A\), \(\exists \{x_n\}\in B\), such that \(\lim\limits_{n\rightarrow\infty}x_n=x\). So \(\forall \varepsilon>0\), \(\exists N>0\), \(\rho(x,x_N)<\varepsilon\), which means \(x\in B(x_N,\varepsilon)\). So \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\).

  • (iv) \(\Rightarrow\) (i)

\(\forall \varepsilon>0\), \(A\subset \bigcup\limits_{x\in B}B(x, \varepsilon)\), so \(\forall x\in A\), \(\exists y\in B\), \(x\in B(y,\varepsilon)\), which means \(B(x,\varepsilon)\cap B=\neq \varnothing\) (at least we have \(y\)), so \(x\in \overline{B}\).

Definition of Separability

Assume \(X\) is a metric space. If there exists denumerable set \(B\subset X\), \(B\) is dense in \(X\), then \(X\) is separable. This is an essential difference between finite-dimension and infinite-dimension space.

Example. Prove \(L^\infty[a,b]\) is not separable in metric

\[ \inf_{F_0\subset [a,b]\atop m(F_0)=0}\sup_{t\in [a,b]-F_0 }|x(t)-y(t)|. \]

Actually, \(F=[a,b]\), which means we have a non-denumerable set, we could find a set with the distance of all its elements not really small.

Choose element

\[ f_s(t)=\chi_{[a,s]},\quad s\in [a,b] \]

So if \(t\neq s\in [a,b]\), we have

\[ \rho(f_t,f_s)=1. \]

We construct set \(A\) to be all the functions above.

We prove the proposition by contradiction. If it is separable, then there exists a dense denumberable subset \(A_0\) of \(L^\infty [a,b]\), such that \(\forall \varepsilon>0\), \(L^\infty [a,b]\subset \bigcup\limits_{x\in A_0}B(x,\varepsilon)\). Choose \(\varepsilon=\frac{1}{3}\).

Since \(A\) has non-denumerable elements, so there exists 2 elements \(f_t,f_s \in A\), \(\exists g_0\in A_0\), such that \(f_t,f_s\in B(g_0,\frac{1}{3})\), so

\[ \rho(f_t,f_s)\leq \rho(f_t,g_0)+\rho(g_0,f_s)<\frac{2}{3}<1, \]

which contradicts!

Continuous Mapping

Mapping & Continuous Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If \(\forall x\in X\), \(\exists ! y\in Y\), such that \(Tx=y\), then we call \(T\) is a mapping from \(X\) to \(Y\).

If mapping \(T\) satisfies that for a fixed point \(x_0\in X\), \(\forall \varepsilon>0\), \(\exists \delta>0\), such that \(\rho_1(Tx,Tx_0)<\varepsilon\) whenever \(\rho(x,x_0)<\delta\), then we call \(T\) is a continuous mapping at \(x_0\).

If the above property holds for all \(x_0\in X\), then we call \(T\) continuous mapping on \(X\).

Usually we call a mapping from metric space \(X\) to \(\mathbb{R}\) functional.

Example. (i) Assume \(X\) is a metric space, \(x_0\in X\), mapping \(T(x)=\rho(x,x_0), (x\in X)\) is a continuous functional from \(X\) to \(\mathbb{R}\).

(ii) Assume \(X\) is a metric space, \(F\subset X\), mapping \(T(x)=\inf\limits_{y\in F}\rho(x,y), (x\in X)\) is continuous (actually uniformly continuous) functional from \(X\) to \(\mathbb{R}\).

\(\forall \varepsilon>0\), \(\forall x\in X\), choose \(\delta=\varepsilon\), \(\forall y\in X\) such that \(\rho(x,y)<\delta\), we have

\[ |T(x)-T(y)|=|\rho(x,x_0)-\rho(y,x_0)|\leq \rho(x,y)<\varepsilon. \]

\(\square\)

\(\forall y\in F\), \(\forall x, z\in X\),

\[ \begin{align*} \rho(x,y)&\leq \rho(x,z)+\rho(z,y),\\ \rho(z,y)&\leq \rho(z,x)+\rho(x,y).\\ \Rightarrow \quad T(x)&\leq \rho(x,z)+T(z)\quad \text{take } \inf_{y\in F}\\ \Rightarrow \quad T(z)&\leq \rho(x,z)+T(x)\quad \text{take } \inf_{y\in F}\\ \Rightarrow \quad |T(x)-T(z)|&\leq \rho(x,z). \end{align*} \]

\(\forall \varepsilon>0\), choose \(\delta=\varepsilon\), whenever \(\rho(x,z)<\delta\),

\[ |T(x)-T(z)|\leq \rho(x,z)<\varepsilon. \]

\(\square\)

Equivalent definition for continuous mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively.

(i) Choose a point \(x_0\in X\). A mapping \(T\) from \(X\) to \(Y\) is continuous at \(x_0\), iff

\[ \forall \{x_n\}_{n\geq 1}\subset X, \lim\limits_{n\rightarrow \infty}x_n=x_0, s.t. \lim\limits_{n\rightarrow \infty}Tx_n=Tx. \]

(ii) A mapping \(T\) from \(X\) to \(Y\) is continuous, iff one of the following two condition holds

\[ \forall \text{ open set } G\in Y, T^{-1}(G) \text{ is an open set.} \]
\[ \forall \text{ closed set } F\in Y, T^{-1}(F) \text{ is a closed set.} \]
  • Necessary. Here we have to find a \(N\) for \(\rho(Tx_n,Tx_0)<\varepsilon\) whenever \(n>N\).

Actually, since \(T\) is continuous, then \(\forall \varepsilon>0\), \(\exists \delta>0\), such that \(\forall \rho(x, x_0)<\delta\), we have \(\rho_1(Tx, Tx_0)<\varepsilon\). Assume \(\{x_n\}\) converges to \(x_0\). For the above \(\delta\), \(\exists N>0\), such that \(\rho(x_n, x_0)<\delta\) whenever \(n>N\). So for \(n>N\),

\[ \rho_1(T(x_0), T(x_n))<\frac{1}{n}. \]
  • Sufficient. Here we have to find \(\delta\), such that \(\rho(Tx,Tx_0)<\varepsilon\) whenever \(\rho(x,x_0)<\delta\). This is a little hard, we prove by contradiction. Assume \(T\) is not continuous, then \(\exists \varepsilon_0\), \(\forall \delta_n=\frac{1}{n}\), \(\exists x_n\) such that \(\rho(x_n,x_0)<\delta_n\), we have \(\rho_1(Tx_n,Tx_0)\geq \varepsilon_0\). So we have a sequence \(x_n\rightarrow x_0\) but \(\rho_1(Tx_n, Tx_0)\not\rightarrow 0\), which contrsdicts!

\(\square\)

  • Necessary.

Assume \(T\) is continuous. we have to prove, for any open set \(G\in Y\), \(T^{-1}(G)\) is also an open set, which means \(\forall x_0\in T^{-1}(G)\), \(Tx_0=y_0\), we have to find a neighborhood of \(x_0\) contained in \(X\).

Actually, since \(T\) is continuous, \(\forall \varepsilon>0\), \(\exists \delta>0\), \(\rho_1(Tx,Tx_0)<\varepsilon\) whenever \(\rho(x,x_0)<\delta\). By \(G\) is a open set, we choose a neighborhood of radius \(\varepsilon\), then \(B_\varepsilon(y_0)\subset G\). By the above assumption, \(\exists \delta>0\), \(\forall x\in B_\delta(x_0)\subset X\), such that \(\rho_1(Tx, Tx_0)=\rho_1(y,y_1)<\varepsilon\). which means \(Tx\in B_\varepsilon(y_0)\subset G\), so \(x\in T^{-1}(G)\). By arbitrariness of \(x\), we know \(B_\delta(x_0, x)\subset T^{-1}(G)\), which is exactly the definition of open set.

  • Sufficient.

Assume the original image of open set \(G\) is open set under \(T\). \(\forall \varepsilon>0\), \(\forall x_0\in X\), \(y_0=Tx_0\). Choose a specific open set \(G=B_\varepsilon(y_0)\), then \(T^{-1}(G)\) is an open set. By its definition, \(\exists \delta>0\), such that \(B_\delta(x_0)\subset T^{-1}(G)\). So \(\forall x\in B_\delta(x_0)\), we have \(Tx\in B_\varepsilon(y_0)\), which is exactly definition of continuous mapping at \(x_0\). By arbitrariness of \(x_0\), we know \(T\) is continuous on \(X\).

\(\square\)

Definition of inverse mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\) is injective and surjective, or bijective, then \(\forall y\in Y\), \(\exists ! x\in X\), such that

\[ Tx=y. \]

here we have a new mapping, called Inverse Mapping, deonted as \(T^{-1}\).

Definition of Homeomorphic Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\) is bijective, and \(T\), \(T^{-1}\) are both continuous on \(X\), \(Y\), respectively, then we call \(T\) is a Homeomorphic Mapping.

If there exists a homeomorphic mapping \(T\) from \(X\) to \(Y\), then we call \(X\) and \(Y\) are homeomorphic.

Definition of Isometric Mapping

Assume \(X,Y\) are metric space, with its distance \(\rho\), \(\rho_1\), respectively. If a mapping \(T\) from \(X\) to \(Y\), satisfies that \(\forall x,y\in X\), \(\rho(x,y)=\rho_1(Tx,Ty)\), then we call \(T\) is an isometric mapping.

If there exists a bijective and isometric mapping \(T\) from \(X\) to \(Y\), then we call \(X\) and \(Y\) are isometric.

When Two metric space are homeomorphic or isometric, we view them as the same.

Completeness

Definition of Basic sequence (Cauchy Sequence)

Assume \(X\) is a metric space with metric \(\rho\). Sequence \(\{x_n\}_{n\geq 1}\subset X\) is said to be a Cauchy Sequence, if \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n,m>N\), such that

\[ \rho(x_n,x_m)<\varepsilon. \]

The core problem of functional analysis, lies in formulating a complete space, i.e. every Cauchy sequence of \(X\) converge to the point in the space \(X\) itself.

Example. Basic space \(\mathbb{R}\) and \(C[a,b]\) are complete.

For \(\mathbb{R}\), it is done in mathematical analysis.

For \(C[a,b]\), we have prove it actually in series of functions. By definition of metric, \(\forall\text{ Cauchy's Seq } \{x_n\}\subset C[a,b]\), \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n,m>N\),

\[ \sup_{t\in [a,b]}|x_m(t)-x_n(t)|<\varepsilon. \]

So

\[ \begin{align} |x_m(t)-x_n(t)|<\varepsilon,\quad \forall t\in [a,b].\label{uniform-convergence} \end{align} \]

which means for fixed \(t\in [a,b]\), \(x_m(t), x_n(t)\) are cauchy sequence on \(\mathbb{R}\), so by completeness of \(\mathbb{R}\), we have their convergent function \(x(t)\). Let \(n,t\) fixed and let \(m\rightarrow \infty\) in inequation \(\ref{uniform-convergence}\), we have

\[ |x_n(t)-x(t)|\leq \varepsilon, \quad \forall t\in [a,b]. \]

So \(x_n(t)\) converges to \(x(t)\) uniformly, so by conclusion of mathematical analysis, we have \(x(t)\) is continuous and thus \(x(t)\in C[a,b]\). So from above, we take a supremum, which gives \(\rho(x_n, x)\leq \varepsilon\). So \(x_n\) converges to an element in \(C[a,b]\).

Example. Assume \(S\) is a space composed by functions defined on a finite-measure set \(F\), which is finite-valued almost everywhere. Define metric

\[ \rho(x,y)=\int_F \frac{|x(t)-y(t)|}{1+|x(t)+y(t)|}dt,\quad x,y\in S. \]

Prove \(S\) is complete.

  • Show that \(\rho\) is a metric, using
\[ \begin{align*} \frac{|a+b|}{1+|a+b|}&\leq \frac{|a|+|b|}{1+|a+b|}\\ &=\frac{|a|}{1+|a+b|}+\frac{|b|}{1+|a+b|}\\ &\leq \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}. \end{align*} \]
  • We have to prove that convergence in this metric is equivalent to convergence in measure.

Assume \(\{x_n\}\rightarrow x\) in this metric, then \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n>N\), and cccording to the relationship, \(\forall \delta>0\),

\[ \begin{align*} \varepsilon&>\int_F \frac{|x_n(t)-x(t)|}{1+|x_n(t)+x(t)|}dt\\ &\geq \int_{F(|x_n-x|\geq \delta)} \frac{|x_n(t)-x(t)|}{1+|x_n(t)+x(t)|}dt\\ &\geq \frac{\delta}{1+\delta} m[F(|x_n-x|\geq \delta)] \end{align*} \]

which means \(\lim_{n\rightarrow \infty}m[F(|x_n-x|\geq \delta)]=0\).

on the other hand, assume \(\{x_n\}\rightarrow x\) in measure, then \(\forall \varepsilon>0, \delta>0\), \(\exists N>0\), \(\forall n>N\), we have

\[ m[F(|x_n-x|\geq \delta)]<\frac{\varepsilon}{2}. \]

While according to relationship

\[ \begin{align*} \int_F \frac{|x_n(t)-x(t)|}{1+|x_n(t)+x(t)|}dt& =\left[\int_{F(|x_n-x|\geq \delta)} + \int_{F(|x_n-x|< \delta)}\right]\frac{|x_n(t)-x(t)|}{1+|x_n(t)+x(t)|}dt\\ &\leq \frac{\delta}{1+\delta} m[F(|x_n-x|<\delta)] + \int_{F(|x_n-x|\geq \delta)} 1dt\\ &=\delta m[F(|x_n-x|<\delta)] + m[F(|x_n-x|\geq \delta)] \end{align*} \]

here we have to choose \(\delta\) such that \(\delta<\frac{\varepsilon}{2m[F(|x_n-x|<\delta)]}\), so the above estimate is less then \(\varepsilon\).

  • We have to prove \(\forall \text{ Cauchy sequence } \{x_n(t)\}\subset S\), \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n,m>N\) such that
\[ \begin{align*} \varepsilon&>\int_F \frac{|x_n(t)-x_m(t)|}{1+|x_n(t)+x_m(t)|}dt.\\ &>\int_F \frac{\delta}{1+\delta}m[F(|x_n-x_m|>\varepsilon)] \end{align*} \]

which means \(x_n\) is a cauchy sequence in a sense of measure. By Riesz Theorem, we have a subsequence \(x_{n_k}\) which is Cauchy sequence a.e. So it converges to a finite-valued measurable function \(x(t)\) a.e. So \(x\in S\). Then for \(n_k>N\), we have

\[ \int_F \frac{|x_n(t)-x_{n_k}(t)|}{1+|x_n(t)+x_{n_k}(t)|}dt<\varepsilon. \]

Since \(\left|\frac{|x_n(t)-x_{n_k}(t)|}{1+|x_n(t)+x_{n_k}(t)|}\right|<1\), \(\int_F 1dt<\infty\), so by Lebesgue's Dominated Convergence Theorem.

\[ \begin{align*} \rho(x,x_n)&=\int_F \lim_{k\rightarrow\infty} \frac{|x_n(t)-x_{n_k}(t)|}{1+|x_n(t)+x_{n_k}(t)|}dt\\ &=\lim_{k\rightarrow \infty}\int_F \frac{|x_n(t)-x_{n_k}(t)|}{1+|x_n(t)+x_{n_k}(t)|}dt \leq \varepsilon. \end{align*} \]

Example. Assume \(s\) is a space composed by sequences \(x=\{\xi_1,\cdots,\xi_n,\cdots\}\). Define metric

\[ \rho(x,y)=\sum_{k=1}^\infty \frac{1}{2^k}\frac{|\xi_k-\eta_k|}{1+|\xi_k+\eta_k|} \]

Prove \(s\) is complete.

  • Similar idea to prove it is a metric.

  • Prove that convergence in the above metric is equivalent to convergence for each coordinate.

Assume \(x_n\rightarrow x\) in the above metric, i.e. \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n>N\),

\[ \begin{align*} \sum_{k=1}^\infty \frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi_k|}{1+|\xi^{(n)}_k+\xi_k|}&<\varepsilon\\ \Rightarrow \quad \frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi_k|}{1+|\xi^{(n)}_k+\xi_k|}&<\varepsilon\\ \end{align*} \]

For fixed \(k\), let \(\varepsilon 2^k<1\), and \(|\xi^{(n)}_k-\xi_k|<\frac{\varepsilon}{1-\varepsilon}<\varepsilon\). By arbitrariness of \(\varepsilon<\frac{1}{2^k}\), we have its convergence for each coordinate.

Assume \(\varepsilon>0\), \(\exists K\) such that \(\frac{1}{2^K}<\frac{\varepsilon}{2}\). Still the above \(\varepsilon\), for each coordinate \(k\), \(\exists N_k>0\), \(|\xi^{(n)}_k-\xi_k|<\varepsilon/K\) whenever \(n>N_k\). For \(k=1:K\), choose \(N=\max\{N_1,\cdots,N_k\}\), let \(n>N\), we have

\[ \begin{align*} \sum_{k=1}^\infty \frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi_k|}{1+|\xi^{(n)}_k+\xi_k|}&=\left[\sum_{k=1}^K + \sum_{k=K+1}^\infty\right]\frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi_k|}{1+|\xi^{(n)}_k+\xi_k|}\\ &\leq \sum_{k=1}^K \frac{|\xi^{(n)}_k-\xi_k|}{1+|\xi^{(n)}_k+\xi_k|} + \sum_{k=K+1}^\infty \frac{1}{2^k}\\ &\leq \sum_{k=1}^K |\xi^{(n)}_k-\xi_k| + \frac{1}{2^K}<\varepsilon. \end{align*} \]
  • Prove its completeness.

Similar to the above proof, a Cauchy sequence \(x_n\) in the above metric is also a Cauchy sequence for each coordinate, which is in \(\mathbb{R}\). So by completeness of \(\mathbb{R}\), we have \(\xi_k\) \((k=1,2,\cdots)\) as its convergence. So define \(x=\{\xi_1,\cdots,\xi_k\}\in s\) and prove that \(x_n\) converges to \(x\) in the above metric. This is apparent because we could change the order of limit for absolute convergent series.

\[ \begin{align*} \rho(x_n,x)&=\sum_{k=1}^\infty\lim_{m\rightarrow \infty} \frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi^{(m)}_k|}{1+|\xi^{(n)}_k+\xi^{(m)}_k|}\\ &=\lim_{m\rightarrow \infty}\sum_{k=1}^\infty \frac{1}{2^k}\frac{|\xi^{(n)}_k-\xi^{(m)}_k|}{1+|\xi^{(n)}_k+\xi^{(m)}_k|}\leq \varepsilon \end{align*} \]

\(\square\)

Meager sets

Definition of nowhere dense sets

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be Nowhere Dense Set, if \(\forall \text{ open set }G\subset X\), \(A\) is not dense in \(G\).

Equivalent definition of nowhere dense sets

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be Nowhere Dense Set, iff

\[ \forall \text{ open ball }S(x_0,r), \exists \text{ open ball }S(y_0,r')\subset S(x_0,r), s.t. A\cap S(y_0,r')=\varnothing. \]
  • Necessary.

Assume \(A\) is a nowhere dense set, since \(\forall \text{ open ball } S(x,r)\subset X\), \(A\) is not dense in \(S(x,r)\), so \(\exists \varepsilon_0>0\), \(\exists x_0\in S(x,r)\), \(\forall y\in A\), \(\rho(x_0,y)\geq \varepsilon_0\), so \(S(x_0,\varepsilon)\cap A=\varnothing\).

or by definition of density, \(S(x, r)-\overline{A}\) is not empty and still open, so we have a small open ball in the rest which has no common element with \(A\).

  • Sufficient.

\(\forall \text{ open set }G \subset X\), \(\forall x\in G\), \(\exists r>0\), such that \(S(x,r)\subset G\). By condition, \(\exists \text{ open ball }S(y_0,r')\subset S(x_0,r)\) s.t. \(A\cap S(y_0,r')=\varnothing\), which means \(\exists r'>0\), \(\exists y\in S(y_0,r')\subset G\), \(\rho(y,z)>r'\) for all \(z\in A\).

\(\square\)

Set of the first & second category

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is said to be a set of the first category, if there exists a sequence of nowhere dense sets \(\{F_n\}_{n\geq 1}\), such that

\[ A=\bigcup_{n=1}^\infty F_n. \]

If not, we call \(A\) a set of the second category.

Theorem of Nested Closed Balls

Metric Space \(X\) is complete, iff \(\forall\) nested closed ball \(\{K_n=\overline{S}(x_n,r_n)\}_{n\geq 1}\), which satisfies

\[ \overline{S}(x_1,r_1)\supset \overline{S}(x_2,r_2)\supset \cdots, \]

and \(\lim\limits_{n\rightarrow \infty}r_n=0\), then there exists a unique point \(x_0\in X\), such that \(x_0\in \bigcap_{n=1}^\infty \overline{S}(x_n,r_n)\).

  • Necessary.

Assume \(X\) is complete. Choose a sequence \(\{x_n\}\), just the center of each closed ball, then for each \(n\geq 1\), \(\forall m>n\),

\[ \begin{align} \rho(x_n, x_{m})<r_n.\label{nested} \end{align} \]

\(\forall \varepsilon>0\), \(\exists N>0\), such that \(\forall n>N\), \(r_n<\varepsilon\). Choose \(\forall m>n>N\), \(\rho(x_n, x_{m})<\varepsilon\).

So \(x_n\) is a Cauchy sequence. By completeness of \(X\), there exists a point \(x\in X\), such that \(x_n\rightarrow n\). Because for each \(n\geq 1\), \(x_k\subset \overline{S}(x_n,r_n)\) for \(k\geq n\), so \(x\in K_n\). So \(x=\bigcup_{n=1}^\infty K_n\). In inequation \(\ref{nested}\), fix \(n\) and let \(m\rightarrow \infty\), by continuity of metric, we have

\[ \rho(x_n, n)\leq r_n, \]

which means \(x\subset K_n\) for each n.

To prove the uniqueness, we assume there exists another point \(y\in \bigcup_{n=1}^\infty K_n\). If \(x\neq y\), then \(\rho(x,y)>0\), which contradicts \(r_n\rightarrow 0\)!

\(\square\)

  • Sufficient.

Assume \(\{x_n\}\subset X\) is a Cauchy sequence, choose a subsequence which converges much faster. Since \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n,m>N\), we have \(\rho(x_n,x_m)<\varepsilon\). Let \(\varepsilon_1=\frac{1}{2}\), \(\exists N_1>0\), choose \(n_1>N_1\), and \(\forall m_1>N_1\), we have

\[ \rho(x_{n_1}, x_{m_1})<\frac{1}{2}. \]

Choose \(\varepsilon_2=\frac{1}{2^2}\), \(\exists N_2>0\), choose \(n_2>\max\{N_1,N_2\}\), and \(\forall m_2\), we have

\[ \rho(x_{n_2}, x_{m_2})<\frac{1}{2^2}. \]

Since \(n_2>N_1\), so we have

\[ \rho(x_{n_2},x_{n_1})<\frac{1}{2}. \]

Follow the same routine, we choose \(\varepsilon_k=\frac{1}{2^k}\), then \(\exists N_k>0\), choose \(n_k>\max\{N_k,N_{k-1},\cdots, N_1\}\), and we have

\[ \rho(x_{n_k},x_{n_{k-1}})<\frac{1}{2^{k-1}}. \]

Construct nested closed ball \(K_k=\overline{S}(x_{n_k}, \frac{1}{2^{k}})\). \(\forall x\in K_{k+1}\), we have

\[ \rho(x_{k},x)\leq \rho(x_k,x_{k+1}) +\rho(x_{k+1}, x)<\frac{1}{2^{k}}+\frac{1}{2^{k}}=\frac{1}{2^{k-1}}. \]

So \(x\in K_k\). By condition of the proposition, we have a unique point \(x\in \bigcup_{k=1}^\infty K_k\). Since \(\forall \varepsilon>0\), \(\exists N>0\), s.t. \(\frac{1}{2^{N-1}}<\varepsilon\), and for \(n, n_k>N\),

\[ \rho(x_n, x)\leq \rho(x_n, x_{n_k})+\rho(x_{n_k}, x)<\frac{1}{2^{N}}+\frac{1}{2^{N}}<\varepsilon. \]

So \(x_n\rightarrow x\in X\).

Baire Theorem

Every complete space is a set of the second category.

We prove by contradiction. Assume \(X\) is a complete metric space, which is a set of the first category. Then \(X=\bigcup_{n=1}^\infty F_n\) where \(F_n\) is nowhere dense set. We use the nested closed ball theorem.

Since \(F_1\) is nowhere dense set, then choose a closed ball \(\overline{S}(x_0,r_0)\subset X\), exists another smaller closed ball \(\overline{S}(x_1,r_1)\subset \overline{S}(x_0,r_0)\), such that \(\overline{S}(x_1,r_1)\cap F_1=\varepsilon\). If the ball is a little large, we take a smaller part of it with \(r_1<1\), otherwise it already satisfies.

Since \(F_2\) is also a nowhere dense set, there exists another smaller ball \(\overline{S}(x_2,r_2)\subset \overline{S}(x_1,r_1)\), such that \(\overline{S}(x_2,r_2)\cap F_2=\varnothing\). We could also let \(r_2<\frac{1}{2}\).

Continue this routine, we have a sequence of nested closed ball \(\{\overline{S}(x_k,r_k)\}\) with \(r_k\frac{1}{2^k}\), which satisfies \(\overline{S}(x_k,r_k)\cap F_k=\varnothing\).

By Theorem of Nested Closed Balls, since \(X\) is complete, there exists a unique point \(x_0\in \bigcap_{k=1}^\infty \overline{S}(x_k,r_k)=X\), but \(x_0\not\in \bigcup_{k=1}^\infty F_k=X\), which contradicts!

\(\square\)

Compact Sets

Here we use nested closed ball theorem to give the version of Balzano-Weierstrass Theorem. However, bounded is not sufficient for a sequence to have a limit point in general metric space, specially in infinite-dimensional metric space.

Definition of Bounded set

Assume \(X\) is a metric space, and \(A\subset X\). \(A\) is said to be a bounded set, if there exists a open ball \(K=S(x,r)\) (or closed ball \(K=\overline{S}(x,r)\)) such that \(A\subset K\).

Definition for relative compact set & compact set

Assume \(X\) is a metric space, \(A\subset X\). If \(\forall \{x_n\}\subset A\), \(\exists \{n_k\}_{k\geq 1}\) and \(x\in X\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\), then we call \(A\) a relative compact set. If the limit point \(x\in A\), then \(A\) is called compact set.

If \(X\) itself is a compact set, then \(X\) is called compact metric space.

Properties for relative compact set & compact set

(i) Any set with finite number elements is compact set.

(ii) Assume \(A\subset X\) is a relative compact set, then \(\forall B\subset A\), \(B\) is also a relative compact set.

(iii) Assume \(A\subset X\) is a compact set, then \(\forall \text{ closed set }B\subset A\), \(B\) is also a compact set.

(ii) By definition. \(\forall \{x_n\}_{n\geq 1}\subset B\subset A\), so \(\exists \{n_k\}_{k\geq 1}\) and \(x\in X\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\), so \(B\) is a relative compact set.

(iii) \(\forall \{x_n\}_{n\geq 1}\subset B\subset A\), \(\exists \{n_k\}_{k\geq 1}\) and \(x\in A\), such that \(\lim\limits_{k\rightarrow \infty}x_{n_k}=x\). Since \(B\) is a closed set, the limit point should be in \(B\), i.e. \(x\in B\). So \(B\) is a compact set.

\(\square\)

Relative compact set

Now we give a description for relative compact set.

Inspired by finding a limit point in \(\mathbb{R}^n\), we find the core is to formulate a finite-number of "net" to cover the bounded set, and one of its piece has infinite number of points. So here comes the abstract of the specific idea.

Definition for \(\varepsilon\)-net

Assume \(X\) is a metric space with metric \(\rho\), \(A,B\subset X\). If for a given number \(\varepsilon>0\), \(\forall x\in A\), \(\exists y\in B\), such that \(\rho(x,y)<\varepsilon\), then we call \(B\) is an \(\varepsilon\)-net of \(A\).

Note we do not ask \(B\cap A\neq \varnothing\).

Definition for Totally Bounded Set

Assume \(X\) is a metric space, \(A\subset X\). If \(\forall \varepsilon>0\), \(\exists \varepsilon\)-net with finite number of elements for \(A\), then we call \(A\) is a Totally Bounded Set.

Properties for totally bounded set

(i) Any set with finite number of elements is a totally bounded set.

(ii) If \(A\subset X\) is a totally bounded set, then \(\forall B\subset A\), \(B\) is also a totally bounded set.

(iii) If \(A\subset X\) is a totally bounded set, then \(\forall\varepsilon>0\), \(\exists B\subset A\), s.t \(B\) is a \(\varepsilon\)-net of \(A\).

(i) Choose the set itself as an \(\varepsilon\)-net, which has finite number of elements.

(ii) \(\forall \varepsilon>0\), \(\exists C\) with finite number of elements, such that \(B\subset A\subset \bigcup\limits_{x\in C}S(x,\varepsilon)\), so \(C\) is also an \(\varepsilon\)-net of \(B\).

(iii) \(\forall \varepsilon>0\), \(\exists C=\{y_n\}_{1\leq n\leq N}\) is an \(\frac{\varepsilon}{2}\)-net of \(A\). That is, \(\forall x\in A\), \(\exists y\in C\), such that \(\rho(x,y)<\frac{\varepsilon}{2}\). By assumption, \(\exists \{n_k\}_{1\leq k\leq K} (K\leq N)\), such that \(S(y_{n_k},\frac{\varepsilon}{2})\cap A\neq \varnothing\), then choose \(z_k\in S(y_{n_k},\frac{\varepsilon}{2})\cap A\). Then define \(B=\bigcup_{k=1}^K \{z_k\}\), which satisfies

\[ \forall x\in A, \exists z_k\in B, \rho(x,z_k)<\varepsilon. \]

So \(B\) is a \(\varepsilon\)-net of \(A\).

Theorem for totally bounded set

Totally bounded set is bounded and separable.

(i) easy to see it is bounded. Choose one element of an \(\varepsilon\)-net as center of ball, estimate the sufficient radious of the ball.

(ii) Leverage the \(\frac{1}{n}\)-net to formulate a dense denumerable subsequence.

Assume \(B_n=\{y_{n_k}\}_{1\leq k\leq K_n}\) is a \(\frac{1}{n}\)-net for totally bounded set \(A\). Then define

\[ B=\bigcup_{n=1}^\infty B_n \]

which is denumerable. Now we show that \(B\) is dense in \(A\). Actually, \(\forall x\in A\), \(\forall \varepsilon>0\), \(\exists n\) such that \(\frac{1}{n}<\varepsilon\), so \(\exists y_{n_k}\in B_n\subset B\), such that \(\rho(x, y_{n_k})<\frac{1}{n}<\varepsilon\).

Or \(\forall x\in A\), \(\exists x_n\in B_n\), such that \(\rho(x, y_{n_k})<\frac{1}{n}\), \(\{x_n\}\rightarrow x\), which means \(B\) is dense in \(A\).

\(\square\)

.

Hausdorff's Theorem for relative compact description

Assume \(X\) is a metric space with metric \(\rho\), \(A\subset X\) is a relative compact set, then \(A\) is totally bounded set.

On the other hand, If \(X\) is a complete metric space, if \(A\subset X\) is a totally bounded set, then \(A\) is also a relative compact set.

The core relationship between relative compact set and totally bounded set lies in every sequence has a Cauchy subsequence.

  • We first prove that \(A\subset X\) is a totally bounded set, iff every sequence in \(A\) has a Cauchy subsequence.

  • Necessary.

If \(A\) is totally bounded set, then \(\exists\) finite-number \(\varepsilon\)-net for \(A\). Choose an arbitrary sequence \(x_n\subset A\).

For \(1\)-net \(B_1=\{y_{j_1}\}_{1\leq j_1\leq J_1}\), we have \(\bigcup\limits_{j\in B_1}B(y_{j_1}, 1) \supset \{x_n\}\) , so \(\exists y_1\in B_1\) such that \(B(y_1,1)\cap \{x_n\}\) has infinitely many number. Choose \(x_{n_1}\in B(y_1,1)\cap \{x_n\}\), which satisfies \(\rho(x_{n_1}, y_1)<1\).

For \(\frac{1}{2}\)-net, \(B_2=\{y_{j_2}\}_{1\leq j_2\leq J_2}\), we have \(\bigcup\limits_{j_2\in B_2}B(y_{j_2}, \frac{1}{2}) \supset \{x_n\}\) and \(\bigcup\limits_{j_2\in B_2}B(y_{j_2},\frac{1}{2}) \cap B(y_1,1) \cap \{x_n\}\) has infinitely many number. So \(\exists y_2\in B_2\) such that \(B(y_2,\frac{1}{2})\cap B(y_1,1)\) has infinitely many number of \(\{x_n\}\). Choose \(n_2>n_1\), \(x_{n_2}\in B(y_2,\frac{1}{2})\cap \{x_n\}\), which satisfies \(\rho(x_{n_2}, y_2)<\frac{1}{2}\).

Continue this routine, we have for each \(k\),

\[ \bigcap_{i=1}^k B(y_i, \frac{1}{i}) \]

has infinitely many number of \(\{x_n\}\). Choose \(n_k>n_{k-1}\), \(x_{n_k}\in \bigcap\limits_{i=1}^k B(y_i, \frac{1}{i})\). So a subsequence \(\{x_{n_k}\}_{k\geq 1}\) is chosen, which satisfies \(\forall n>k\)

\[ \rho(x_{n_k}, x_{n})\leq \rho(x_{n_k}, y_k)+\rho(y_k,x_{n})\leq \frac{1}{k}+\frac{1}{k}=\frac{2}{k}. \]

which is a Cauchy sequence.

Actually, iterately choose point from \(\{x_n\}\) to formulate \(\frac{1}{k}\)-net, which is also a totally bounded set.

  • Sufficient.

Choose a point \(x_1\in A\), if \(O(x_1,1)\supset A\), then \(A\) is a totally bounded set. Otherwise, \(A-O(x_1,1)\neq \varnothing\), choose \(x_2\in A-O(x_1,1)\)，if \(O(x_1,1)\cup O(x_2，1) \supset A\), then \(A\) is a totally bounded set. Otherwise, \(A-O(x_1,1)-O(x_2,1)\neq \varnothing\). Continue this routine, if

\[ A-\bigcup_{k=1}^n O(x_k,1)\neq \varnothing, \]

we choose a point \(x_{n+1}\) to check whether

\[ \bigcup_{k=1}^{n+1}O(x_k,1)\supset A. \]

If we could not end in finite steps, we have a sequence \(\{x_n\}\), which satisfies \(\rho(x_i, x_j)\geq 1\) for \(i\neq j\). However by assumption, this sequence must have a Cauchy subsequence, which contradicts!

In this proof we could see that if for all sequence of \(A\), it has a convergent subsequence, then it must be totally bounded set.

\(\square\)

  • Necessary.

If \(A\) is a relative compact set, then any sequence \(\{x_n\}\) must have a convergent subsequence, \(\{x_{n_k}\}\), which is also a Cauchy sequence, so by lemma \(A\) is totally bounded set.

  • Sufficient.

If \(A\) is a totally bounded set, then for any sequence \(\{x_n\}\), it must have a Cauchy subsequence \(\{x_{n_k}\}\). Since \(X\) is complete, \(\{x_{n_k}\}\) must converge to a point \(x\in X\), which means \(A\) is relative compact set.

\(\square\)

The following theorem is useful in proof. Because it is hard to find an \(\varepsilon\)-net which proved to have finite number, but easy to show it is relative compact.

Theorem for relative compact set

Assume \(X\) is a complete metric space, \(A\subset X\) is a relative compact set, iff \(A\) has a relative compact \(\varepsilon\)-net.

  • Necessary.

If \(A\) is a relative compact set, \(A\) itself is a relative compact \(\varepsilon\)-net. Or we could do as follows.

We know \(A\) is also totally bounded, so there exists a finite-number \(\frac{\varepsilon}{2}\)-net \(B\) for \(A\). Since \(B\) could be covered by a \(\varepsilon\)-net of itself, so \(B\) is totally bounded, since \(X\) is complete, \(B\) is relative compact.

  • Sufficient.

If \(A\) has a relative compact \(\frac{\varepsilon}{2}\)-net \(B\), then \(B\) is totally bounded, so \(B\) could be covered by finite-number \(\frac{\varepsilon}{2}\)-net \(B'\subset B\), then \(B'\) is a finite-number \(\varepsilon\)-net for \(A\), so \(A\) is totally bounded, since \(X\) is complete, \(A\) is relative compact.

\(\square\)

Compact set description

Compact set is complete

Assume \(X\) is a metric space, if \(U\subset X\) is a compact set, then \(U\) is complete.

By definition. For all Cauchy's sequence \(\{x_n\}\in U\), it has a subsequence \(\{x_{n_k}\}\) converging to a point \(x\in U\), then since

\[ |x_n-x|\leq |x_n-x_{n_k}|+|x_{n_k}-x| \]

so \(x_n\) converges to \(x\in U\), so \(U\) is complete.

Compact nested set

Assume \(\{K_n\}\subset X\) is a non-empty compact set sequence, which satisfies

\[ K_1\supset K_2\supset \cdots\supset K_n\supset \cdots \]

then \(\bigcap_{n=1}^\infty K_n\neq \varnothing\).

Choose a point \(x_n\in K_n\) and we have a sequence \(\{x_n\}\subset K_1\). Since \(K_1\) is a compact set, \(\exists \{x_{n_k}\}\rightarrow x\in K_1\).

For any given \(N\geq 1\), choose \(n_k\geq N\), then by nested property we have \(x_{n_k}\in K_N\). Since \(K_N\) is closed (readers could prove this), we have \(x\in K_N\). So

\[ x\in \bigcap_{n=1}^\infty K_n. \]

Compact set \(K\), is closed, because for a limit point \(x\), there exists a sequence \(x_n\subset K\) converges to it. By compact set, it has a subsequence \(x_{n_k}\) converges to \(y\in K\). By

\[ |x-y|\leq |x-x_{n_k}|+|x_{n_k}-y| \]

we have \(x=y\). So the limit point is in \(K\).

In \(\mathbb{R}\), bounded set is described by Balzano-Weierstrass Theorem, and closed interval is described by Heine-Borel finite-covering Theorem. Corresponding to closed interval, compact set could be described by Open-covering Theorem.

Definition of Open Covering

Assume \(X\) is a metric space, \(A\subset X\). \(\{G_c\}_{c\in J}\) is a family of open sets in \(X\). If

\[ A\subset \bigcup_{c\in J}G_c \]

then \(\{G_c\}_{c\in J}\) is called an open covering of \(A\). If \(J\) has finitely namy number, then \(\{G_c\}_{c\in J}\) is called a finitely open covering of \(A\).

Gross's Open Covering Theorem

Assume \(X\) is a metric space, \(A\subset X\). \(A\) is a comapct set, iff for any open covering \(\{G_c\}_{c\in J}\) of \(A\), we could choose a finitely open covering from it.

  • Necessary condition.

Assume \(\{G_c\}_{c\in J}\) is an open covering of \(A\). If

\[ \begin{align} \forall \varepsilon>0, \exists x\in A, \forall c\in J, s.t. B_\varepsilon(x)\not\subset G_c,\label{opposite} \end{align} \]

then let \(\varepsilon_n=\frac{1}{2^n}\), then we have a sequence \(\{x_n\}\in A\), such that

\[ B_{\frac{1}{2^n}}(x_n)\not\subset G_c. \]

Since \(A\) is a compact set, \(\exists \{x_{n_k}\}\) converging to \(x_0\in A\). Also because \(A\subset \bigcup_{c\in J}G_c\), there exists \(c_0\in J\), such that \(x_0\in G_{c_0}\). Choose \(N\) such that

\[ \frac{1}{2^N}<d(x_0,\partial G_{c_0}), \]

then \(B_{\frac{1}{2^n}}(x_n)\subset G_{c_0}\) whenever \(n>N\), which contradicts the assumption \(\ref{opposite}\)! So actually we have its opposite

\[ \exists \varepsilon_0>0,\forall x\in A, \exists c\in J, s.t. B_{\varepsilon_0}(x)\subset G_{c} \]

Still because \(A\) is compact, it is also relative compact, or to be specific, totally bounded. So from \(\{B_{\varepsilon}(x)\}_{x\in A}\), we have a finitely many number \(\{B_{\varepsilon}(x_l)\}_{1\leq l\leq L}\) which covers \(A\). So for each \(l\), we choose a \(G_{c_l}\) such that \(B_{\varepsilon}(x_l)\subset G_{c_l}\), then

\[ A\subset \bigcup_{l=1}^L G_{c_l}. \]
  • Sufficient condition.

We still use contradiction. For any \(\{x_n\}\subset A\), if \(A\) is not compact, then every point of \(A\) is not a limit point of \(\{x_n\}\), i.e.

\[ \forall y\in A, \exists \delta_y, n_y>0, s.t. x_n\not\in B_{\delta_y}(y)\quad \text{ whenever } n\geq n_y \]

from above we formulate a huge open covering of \(A\): \(\bigcup_{y\in A}B_{\delta_y}(y)\). By theorem condition, we could find finitely namy number \(L\) such that

\[ A\subset \bigcup_{l=1}^L B_{\delta_{y_l}}(y_l). \]

Let \(N=\max\{n_{y_1}, \cdots, n_{y_L}\}\), then

\[ x_n \not\in \bigcup_{l=1}^L B_{\delta_{y_l}}(y_l),\quad\text{ whenever } n\geq N. \]

which contradicts the assumption! The core is, we could choose a finite number of open covering to expel \(x_n\) whenever \(n>N\).

Here any open covering allow the open set to be strange, usually converges to a point, so we could not choose a finite-number of the sets to cover if it is only totally bounded.

Continuous mapping on compact set

Image of continuous mapping

Assume \(X\), \(Y\) are metric space, with metric \(\rho\), \(\rho'\), respectively. \(A\subset X\) is a compact set, \(T\) is a continuous mapping from \(X\) to \(Y\), then the image of \(T\), denoted as \(T(A)\) is also a compact set in \(Y\).

Moreover, the above condition could give us stronger condition, which is, continuous mapping on compact set is also uniformly continuous. That is, \(T\) is a uniformly continuous on \(A\).

\(\forall \{y_n\}\subset Y\), there exsits \(\{x_n\}\) such that \(Tx_n=y_n\). Since \(A\) is a compact set, \(\{x_n\}\) has a subsequence \(\{x_{n_k}\}\) converging to a point \(x\in X\), then by continuity of \(T\), \(\forall \varepsilon>0\), \(\exists \delta>0\), \(\forall x_{n_k}\in B_\delta(x)\),

\[ \rho'(y_{n_k}, Tx)=\rho'(Tx_{n_k}, Tx)<\varepsilon. \]

So \(Tx\) is a converging point of \(\{y_{n_k}\}\), and \(Tx\in T(A)\).

We could also use open sets in image, by continuous function, its preimage open sets have finite-number covering \(A\), so image itself has finite-number covering \(T(A)\).

  • For uniformly continuity. We use contradiction. \(\exists \varepsilon_0>0\), \(\forall \delta_n=\frac{1}{n}>0\), \(\exists x_n, y_n\in A\), such that
\[ \rho'(Tx_n, Ty_n)\geq \varepsilon_0,\text{ whenever } \rho(x_n,y_n)<\frac{1}{n}. \]

since \(A\) is a compact set, then \(\{x_n\}\) has a subsequence \(\{x_{n_k}\}\) converging to \(x\in A\). Actually, \(y_{n_k}\) and \(x\) could also be close enough, by the following inequation

\[ \rho(y_{n_k}, x_0)\leq \rho(y_{n_k}, x_{n_k})+\rho(x_{n_k},x)\rightarrow 0 (k\rightarrow \infty). \]

However by continuous mapping

\[ \rho'(Tx_{n_k}, Ty_{n_k})\leq \rho'(Tx_{n_k}, Tx)+\rho'(Tx, Ty_{n_k})\rightarrow 0 (k\rightarrow \infty), \]

which contradicts!

We could also use finite-open covering to prove. That is, choose a open covering \(\bigcup_{x\in A}B(x,\varepsilon_0)\supset A\), then there exsits finite-number \({x_n}_{1\leq n\leq N}\) such that

\[ \bigcup_{n=1}^N B(x_n, \varepsilon_0)\supset A. \]

\(\forall x,x'\in A\), let \(\rho(x,x')<\varepsilon_0\), then they must lie in same open ball \(B(x_{0}, \varepsilon_0)\). By continuous mapping, \(\forall \varepsilon>0\), \(\exists \delta>0\), choose \(\varepsilon_0<\delta\), we have

\[ \rho'(Tx, Tx_{0})<\frac{\varepsilon}{2},\quad \rho'(Tx', Tx_{0})<\frac{\varepsilon}{2}, \]

so

\[ \rho'(Tx,Tx')<\rho'(Tx, Tx_{0})+\rho'(Tx', Tx_{0})<\varepsilon. \]

Corollary: Continuous functional

Assume \(X\) is a metric space, and \(A\subset X\) is a compact set. \(f\) is a continuous functional on \(A\), then \(f\) is bounded, and could arrive at its infimum and supremum.

\(f\) maps a compact set \(A\) into \(T(A)\subset \mathbb{R}\), which is closed and bounded.

Contraction Mapping Method

We have already give an introduction of contraction mapping in proving the existence and uniqueness of solution to an ODE. Now we give a more detailed discussion.

Theorem for contraction mapping

Assume \(X\) is a complete metric space, with metric \(\rho\). \(T\) is a mapping from \(X\) to \(X\), and satisfies \(\forall x,y\in X\)

\[ \rho(Tx,Ty)\leq \theta \rho(x,y), \]

where \(\theta\in [0,1)\) is constant. Then \(T\) has a unique fixed point in \(X\).

Actually, the above definition guarantees that starting from an arbitrary point, we could construct a Cauchy sequence. Choose \(x_0\in X\),

\[ x_1=Tx_0,x_2=Tx_1,\quad x_n=Tx_{n-1}, \]

easy to show that \(\{x_n\}\) is a Cauchy sequence. Let \(k=\rho(x_1,x_0)\), so \(\rho(x_2,x_1)\leq \theta\rho(x_1,x_0)=k\theta\), by induction we have

\[ \rho(x_n, x_{n-1})\leq k\theta^{n-1}. \]

So \(\forall \varepsilon>0\), choose \(n\) such that \(k\frac{\theta^{n-1}}{1-\theta}<\varepsilon\), then

\[ \begin{align*} \rho(x_n,x_{n+k})&\leq \rho(x_n, x_{n+1})+\cdots, \rho(x_{n+k-1}, x_{n+k})\\ &\leq (1+\theta+\cdots+\theta^{k-1})k\theta^{n-1}\\ &=\frac{1-\theta^{k}}{1-\theta}k\theta^{n-1}\\ &\leq \frac{1}{1-\theta}k\theta^{n-1}<\varepsilon. \end{align*} \]

Then by \(X\) is complete, \(\{x_n\}\) has a convergent point \(x\in X\). Now we show that it is a fixed point.

\[ x=\lim_{n\rightarrow \infty}x_{n+1}= \lim_{n\rightarrow \infty}T^n x_0=T\lim_{n\rightarrow \infty}T^n x_0=Tx. \]

it is unique, since for two fixed points \(x,y\), we have

\[ \rho(x,y)=\rho(Tx,Ty)\leq \theta\rho(x,y) \]

which means \(\rho(x,y)=0\), so \(x=y\).

Example. Prove the following intergal equation has a unique solution when \(\lambda\) is sufficiently small.

\[ x(t)=f(t)+\lambda \int_a^b K(t,s)x(s)ds \]

where \(f(t)\in L^2[a,b]\) and \(K(t,s)\) is kernel function defined on \([a,b]\times [a,b]\), which satisfies

\[ \int_a^b\int_a^b |K(t,s)|^2ds dt<\infty. \]
  • Show that under metric
\[ \rho(x,y)=\left(\int_a^b|x(t)-y(t)|^2dt\right)^{\frac{1}{2}} \]

the mapping \(T\)

\[ (Tx)(t)=f(t)+\lambda \int_a^b K(t,s)x(s)ds \]

is a mapping from \(X\) to \(X\), or to be more specific, \(Tx\in L^2[a,b]\). Notice we only have to show that the latter in the integral equation is L2-integrable. Actually

\[ \begin{align*} \int_a^b dt\left|\int_a^b K(t,s)x(s)ds\right|^2&\leq \int_a^b dt \left[\int_a^b |K(t,s)|^2 ds \int_a^b |x(s)|^2 ds\right]\\ &=\int_a^b |x(s)|^2 ds \int_a^b dt \int_a^b |K(t,s)|^2 ds<\infty. \end{align*} \]

Choose \(\lambda\) such that \(|\lambda|\left[\int_a^b dt\int_a^b |K(t,s)|^2 ds\right]^\frac{1}{2}<1\), then

\[ \begin{align*} \rho(Tx,Ty)&=\left(\int_a^b |\lambda|\left|\int_a^b K(t,s)[x(s)-y(s)]ds\right|^2 dt\right)^\frac{1}{2}\\ &\leq \left(\int_a^b |\lambda| \left[\int_a^b |K(t,s)|^2 ds\right] \left[\int_a^b |x(s)-y(s)|^2 ds\right] dt\right)^\frac{1}{2}\\ &= |\lambda| \left[\int_a^b |x(s)-y(s)|^2 ds\right]^\frac{1}{2} \left(\int_a^b \left[\int_a^b |K(t,s)|^2 ds\right] dt\right)^\frac{1}{2}\\ &\leq |\lambda|\left[\int_a^b dt\int_a^b |K(t,s)|^2 ds\right]^\frac{1}{2}\rho(x,y) \end{align*} \]

means \(T\) is a contraction mapping.

Extended Contraction mapping theorem

Assume \(X\) is a complete metric space, with metric \(\rho\). \(T\) is a mapping from \(X\) to \(X\). If two constants \(\exists n_0>0\), \(\theta\in [0,1)\), such that

\[ \rho(T^{n_0}x,T^{n_0}y)\leq \theta \rho(x,y),\quad \forall x,y\in X, \]

then \(T\) has a unique fixed point in \(X\).

If we view \(T^{n_0}\) as a mapping \(T'\), then apply Theorem for contraction mapping, we have a unique fixed point \(x\in X\), such that \(x=T'x\). We show that this is also the unique fixed point of \(T\).

Since \(Tx=T T'x=T T^{n_0}x=T^{x_0+1}x=T' (Tx)\), so \(Tx\) is a fixed point of \(T'\). By uniqueness of fixed point, we have \(x=Tx\).

If \(T\) has another fixed point \(y\) of \(T\), then \(Ty=y\). So by iterating over \(n_0\), \(T^{n_0}y=T^{n_0-1}Ty=T^{n_0-1}y=\cdots=y\), we have \(y\) is a fixed point of \(T^{n_0}\), by uniqueness of fixed point for \(T^{n_0}\), we have \(x=y\).