Functional Analysis-Metric Space

Definition

A 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 $t=u^{q-1}$ implies $u=t^{p-1}$. Let $\alpha$ and $\beta$ be positive numbers, and be regarded as the width and height of a rectangle, the $$\alpha \beta \leq \int^{\alpha}_{0} t^{p-1} dt + \int^{\beta}_{0} u^{q-1} du=\frac{\alpha^p}{p} + \frac{\beta^q}{q}~~(Auxiliary Inequality)$$

For (B) let $(\tilde{\xi}_j)$ and $(\tilde{\eta}_j)$ with $\sum |\tilde{\xi}_j|^p=1~,~\sum |\tilde{\eta}_j|^q=1$. Then based on Auxiliary Inequality, we have $$|\tilde{\xi}_j| |\tilde{\eta}_j| \leq \frac{|\tilde{\xi}_j|^p}{p} + \frac{|\tilde{\eta}_j|^q}{q} ~~\Rightarrow~~\sum |\tilde{\xi}_j| |\tilde{\eta}_j| \leq \frac{1}{p} + \frac{1}{q}=1$$. We now take $\tilde{\xi}_j=\frac{\xi_j}{(\sum |\xi_k|^p)^{1/p}}$ and $\tilde{\eta}_j=\frac{\eta_j}{(\sum |\eta_m|^q)^{1/q}}$, then we have $$\sum^{\infty}_{j=1}|\xi_j \eta_j| \leq (\sum |\xi_k|^p)^{1/p} (\sum |\eta_k|^q)^{1/q}~~~~(\textbf{Holder~Inequality})$$

For (C), let $w_j=\xi_j+\eta_j$, then $|w_j|^p=|w_j||w_j|^{p-1}=|\xi_j+\eta_j||w_j|^{p-1} \leq (|\xi_j|+|\eta_j|)|w_j|^{p-1}$, then $$\sum |w_j|^p \leq \sum|\xi_j||w_j|^{p-1} + |\eta_j||w_j|^{p-1}$$, then based on the Holder Inequality, we have

$$\sum|\xi_j||w_j|^{p-1} \leq [\sum |\xi_k|^p]^{1/p} + [\sum(|w_m|^{p-1})^q]^{1/q} =  [\sum |\xi_k|^p]^{1/p}  [\sum |w_m|^p]^{1/q}$$

$$\sum |\eta_j||w_j|^{p-1}  \leq  [\sum |\eta_j|^p]^{1/p}  [\sum |(w_j)^{p-1}|^q]^{1/q} = [\sum |\eta_j|^p]^{1/p}  [\sum |w_j|^p]^{1/q}$$.

Therefore $$\sum |w_j|^p \leq \{ (\sum |\xi_j|^p)^{1/p} + (\sum |\eta_j|^p)^{1/p}  \} (\sum |w_j|^p)^{1/q}$$

That is $$(\sum |w_j|^p)^{1/p} = (\sum |\xi_j + \eta_j|^p)^{1/p}  \leq  (\sum |\xi_j|^p)^{1/p} + (\sum |\eta_j|^p)^{1/p} ~~~~ \textbf{Minkowski Inequality}$$

For (D), we use Minkowski inequality to verify $d(x,y) \leq d(x,z) + d(z,y)$, that is, $$d(x,y)=(\sum |\xi_j - \eta_j|^p)^{1/p} \leq (\sum [|\xi_j-\zeta_j|+|\zeta_j-\eta_j|]^p)^{1/p} \leq (\sum |\xi_j-\zeta_j|^p)^{1/p}  +   (\sum |\zeta_j-\eta_j|^p)^{1/p}=d(x,z)+d(z,y)$$

Open Set, Closed Set, Neighborhood

Ball, Sphere, Open Set, Closed Set, $r$-neighborhood, Neighborhood

Given a point $x_0 \in X$ and a real number $r>0$, we define

(1) Open Ball: $B(x_0;r)=\{x\in X | d(x,x_0)<r \}$, we also say $B(x_0;r)$ is the $\textbf{r}$-neighborhood of $x_0$

(2) Closed Ball:  $\tilde{B}(x_0;r)=\{x\in X| d(x,x_0)\leq r \}$  

(3) Sphere:  $B(x_0;r)={x\in X | d(x,x_0)=r}$  

$x_0$ is called center, and $r$ is called radius. If each point of a subset $M$ of a metric space $X$ has a ball, then we say $M$ is an open set. If $M$'s complement is open, we say $M$ is a closed set. If a subset $M$ of $X$ contains an $r$-neighborhood of $x_0$, we say $M$ is the neighborhood of $x_0$ and $x_0$ is an interior point of $M$. The interior of $M$ is the set of all interior points of $M$. 

Topological Space

The collection of all open subsets of $X$, $\mathscr{T}$, has the following properties:

(T1) $\emptyset \in \mathscr{T}~,~X \in \mathscr{T}$

(T2) The union of any members of $\mathscr{T}$ is a member of $\mathscr{T}$

(T3) The intersection of finitely many members of $\mathscr{T}$ is a member of $\mathscr{T}$

We define $(X,\mathscr{T})$ the topological space. A metric space is a topological space. Every metric space is a topological space, but not every topological space is a metric space. I would actually prefer to say every metric space induces a topological space on the same underlying set. A metric space has a notion of distance, while a topological space only has a notion of closeness. If we have a notion of distance then we can say when things are close to each other. However, distance is not necessary to determine when things are close to each other.

Continnous Mapping

 Let $X=(X,d)~,~Y=(Y,\tilde{d})$ be 2 metric spaces. A mapping from $X\rightarrow Y$ is said to be continuous at a point $x_0 \in X$ if for any arbitrarily small $\varepsilon > 0$, there always exists a $\delta > 0$ such that (as shown in the following figure) $$\tilde{d}(Tx,Tx_0)<\varepsilon~,~for~all~x~satisfying~d(x,x_0)<\delta$$. T is said to be continuous if it is continuous at every point of $X$.

Theorem: A mapping $T: X \rightarrow Y$ is continuous if and only if the inverse image of any open subset of Y is an open subset of X.  

accumulation point, closure

Let $M$ be a subset of a metric space X. Then a point $x_0$ of $X$ (which may or may not be a point of $M$) is called an accumulation point of M (or limit point of M) if every neighborhood of $x_0$ contains at least one point $y\in M$ distinct from $x_0$. The set consisting of the points of $M$ and the accumulation points of $M$ is called the closure of M and is denoted by $\bar{M}$. It is the smallest closed set containing M. Before we go on, we mention another unusual property of balls in a metric space. Whereas in $\textbf{R}^3$ the closure $\bar{B}(x_0;r)$ of an open ball $B(x_0;r)$ is the closed ball $\tilde{B}(x_0;r)$, this may not hold in a general metric space. 

 Dense Set, Separable Space

A subset $M$ of a metric space $X$ is said to be dense in X if $\bar{M}=X$. $X$ is said to be separable if it has a countable subset which is dense in $X$.

 

转载于:https://www.cnblogs.com/chaseblack/p/6661225.html