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Functional Analysis-Metric Space

时间:2017-04-03 09:37:40      阅读:200      评论:0      收藏:0      [点我收藏+]

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Definition

metric space is a pair $(X,d)$ where $X$ is a set and $d$ is a metric(or distance function) $d$ on $X$, that is, a function defined on $X\times X$ such that for all $x,y,z\in X$ we have 4 axioms of a metric:

(1) $d$ is real-valued, finite and nonnegative.

(2) $d(x,y)=0$ if and only if $x=y$

(3) $d(x,y)=d(y,x)$  (Symmetry)

(4) $d(x,y)\leqq d(x,z)+d(z,y)$  (Triangle Inequality)

$X$ is called underlying set of $(X,d)$. If $x,y$ are fixed, we call $d(x,y)$ the distance from $x$ to $y$. A subspace of $(Y,\tilde{d}=d\mid_{ Y\times Y})$ of $(X,d)$ is obtained if we take a subset $Y\subset X$ and restrict $d$ to $Y\times Y$. $\tilde{d}$ is called the metric induced on $Y$ by $d$. Instead of $(X,d)$ we may simply write $X$ if there is no danger of confusion.

Examples

(1)Real line $\textbf{R}$: $(\textbf{R}, d=|\xi-\eta|)$

(2)Euclidean plane: $(\textbf{R}^2, d=\sqrt{(\xi_1-\eta_1)^2+(\xi_2-\eta_2)^2})$

(3)Euclidean plane: $(\textbf{R}^2, d=|\xi_1-\eta_1|+|\xi_2-\eta_2|)$

(4)Unitary space $\textbf{C}^n$: $(\textbf{C}^n, d=\sqrt{|\xi_1-\eta_1|^2+...+|\xi_n-\eta_n|^2})$ where $\textbf{C}^n$ is n dimensional unitary space. Its point $\xi=(\xi_1,...,\xi_n)$ or $\eta=(\eta_1,...,\eta_n)$ is a n dimensional complex number.

(5)Sequence space $l^{\infty}$:  In this space each point is a bounded sequence of complex number. The bounded means that any dimension $\xi_j,~(j=1,2,...)$ of any point $x=(\xi_1,\xi_2,...)$(or briefly $x=(\xi_j)$) in $l^{\infty}$ has a bound $|\xi_j|\leqq c_x$ where $c_x$ depends on point $x$ but doesn‘t depend on $j$. For any 2 points $x$ and $y=(\eta_1,\eta_2,...)$, the metric is defined by $d(x,y)=\mathop{sup}\limits_{j\in N}|\xi_j-\eta_j|,~N\in(1,2,...)$.

(6)Space $l^p$( $p\geq 1$ ): In this space each point is a sequence $x=(\xi_j)=(\xi_1,\xi_2,...),~\xi_j$ can be complex number(complex space) or real number(real space) which must ensure $|\xi_1|^p+|\xi_2|^p+...$ converges, i.e, $\sum^{\infty}_{j=1}|\xi_j|^p \le \infty$. The metric is $d(x,y)=(\sum_{j=1}^{\infty} |\xi_j-\eta_j|^p)^{1/p}$. If $p=2$, we have Hilbert (sequence) space $l^2$.

(7)Sequence space $s$: In this space each point is a bounded or unbounded sequence of complex number, with the metric defined by $d(x,y)=\sum^{\infty}_{j=1}\frac{1}{2^j}\frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}$.

(8)Function space $C[a,b]$($C$ suggests "continuous"): In this space each point is a function of a real independent variable(自变量) $t$ which is defined and continuous on a given closed interval $J=[a,b]$. The metric is $d(x,y)=\mathop{max} \limits_{t\in J} |x(t)-y(t)|$.

(9)Bounded function space $B(A)$($B$ suggests "bounded"): In this space each point $x\in B(A)$ is a function bounded on a given set $A$, with the metric $d(x,y)=\mathop{sup} \limits_{t\in A} |x(t)-y(t)|$. If $A=[a,b]\subset R$, then we write $B([a,b])$ for $B(A)$.

(10)Discrete metric space: $d(x,x)=0,~~d(x,y)=1~x\neq y$

Verify $l^p$ Space

To demonstrate that a space is a metric space, we must verify the above 4 axioms are all true. Actually, only the last one, i.e, the triangluar inequality, is difficult to verify. Now let‘s take some time to apply the triangular inequality to the example (6),i.e, the $l^p$ space. So our goal is to verify $d(x,y)\leq d(x,z)+d(z,y)$. In the context of $l^p$ space, $d(x,y)=(\sum_{j=1}^{\infty} |\xi_j-\eta_j|^p)^{1/p}$, therefore the goal is to verify $$(\sum_{j=1}^{\infty} |\xi_j-\eta_j|^p)^{1/p} \leq (\sum_{j=1}^{\infty} |\xi_j-\zeta_j|^p)^{1/p} + (\sum_{j=1}^{\infty} |\zeta_j-\eta_j|^p)^{1/p}$$. Before verifying this, let me first illustrate

(A) an auxiliary inequality

(B) the Holder inequality from (A) 

(C) the Minkowski inequality from (B)

(D) the triangle inequality (C)

For (A) let $p>1$ and define $q$ by $\frac{1}{p}+\frac{1}{q}=1$ where $p,~q$ are called conjugate exponents. Another form is $1/(p-1)=q-1$, so that 

Functional Analysis-Metric Space

标签:fusion   tin   one   max   analysis   nal   closed   ber   init   

原文地址:http://www.cnblogs.com/chaseblack/p/6661225.html

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