泛函分析2——Normed Linear Spaces

Preface

参考摘录于FUNCTIONAL ANALYSIS NOTES——Mr. Andrew Pinchuck

FUNCTIONAL ANALYSIS NOTES

2 Normed Linear Spaces

2.1 Definition

A norm on a linear space $X$ is a real-valued function $\Vert ?\Vert :X\to \mathbb{R}$ which satisfies the following properties:
For all $x,y\in X$ and $\lambda \in \mathbb{F}$
N1. $\Vert x\Vert \ge 0$;
N2. $\Vert x\Vert =0?x=0$
N3. $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$
N4. $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ (Triangle Inequality).
A normed linear space is a pair $(X,\Vert ?\Vert ),$ where $X$ is a linear space and $\Vert ?\Vert$ a norm on $X.$ The number $\Vert x\Vert$ is called the norm or length of $x$

Unless there is some danger of confusion, we shall identify the normed linear space $(X,\Vert ?\Vert )$ with the underlying linear space $X$.

Examples

[1] Let $X=\mathbb{F}$. For each $x\in X$, define $\Vert x\Vert =|x|$, then ($X,\Vert ?\Vert$)is a normed linear spaces

[2] Let $n$ be a natural number and $X={\mathbb{F}}^n$. For each $x=\left(x_1,x_2,\dots ,x_n\right)\in X,$ define
$$\begin{array}{l}\Vert x{\Vert }_p={\left({\sum }_{i=1}^n{\left|x_i\right|}^p\right)}^{\frac{1}{p}},\text{ for }1\le p<\infty ,\text{ and }\\ \Vert x{\Vert }_{\infty }={\max }_{1\le i\le n}\left|x_i\right|\end{array}$$
Then $\left(X,\Vert ?{\Vert }_p\right)$ and $\left(X,\Vert ?{\Vert }_{\infty }\right)$ are normed linear spaces.

[3] Let $X=\mathcal{B}[a,b]$ be the set of all bounded real-valued functions on $[a,b]$. For each $x\in X$, define
$$\Vert x{\Vert }_{\infty }=\underset{a\le t\le b}{\sup }|x(t)|$$
Then $\left(X,\Vert ?{\Vert }_{\infty }\right)$ is a normed linear space.

[4] Let X=C[a,b]={x:[a,b]→F∣xX=\mathcal{C}[a, b]=\{x:[a, b] \rightarrow \mathbb{F} \mid xX=C[a,b]={ x:[a,b]→F∣x is continuous }. For each $x\in X$, define
$$
\begin{aligned}

|x|_{\infty} &=\sup _{a \leq t \leq b}|x(t)| \

|x|{2} &=\left(\int{a}^{b}|x(t)|{2} d t\right)^{\frac{1}{2}}

\end{aligned}
$$
Then $\left(X,\Vert ?{\Vert }_{\infty }\right)$ and $\left(X,\Vert ?{\Vert }_2\right)$ are normed linear spaces.

[5] Let $X={?}_p,1\le p<\infty$. For each $x={\left(x_i\right)}_1^{\infty }\in X,$ define
$$\Vert x{\Vert }_p={\left(\sum _{i\in \mathbb{N}}{\left|x_i\right|}^p\right)}^{\frac{1}{p}}$$
Then $\left(X,\Vert ?{\Vert }_p\right)$ is a normed linear space.

[6] Let $X={?}_{\infty },c$ or $c_0.$ For each $x={\left(x_i\right)}_1^{\infty }\in X,$ define
$$\Vert x\Vert =\Vert x{\Vert }_{\infty }=\underset{i\in \mathbb{N}}{\sup }\left|x_i\right|$$
Then $X$ is a normed linear space.

[7] Let $X=\mathcal{L}\left({\mathbb{C}}^n\right)$ be the linear space of all $n\times n$ complex matrices. For $A\in \mathcal{L}\left({\mathbb{C}}^n\right),$ let $\tau (A)={\sum }_{i=1}^n(A{)}_{ii}$ be the trace of $A.$ For $A\in \mathcal{L}\left({\mathbb{C}}^n\right),$ define
$$\Vert A{\Vert }_2=\sqrt{\tau \left(A^{?}A\right)}=\sqrt{\sum _{i=1}^n\sum _{k=1}^n\overline{(A{)}_{ki}}(A{)}_{ki}}=\sqrt{\sum _{i=1}^n\sum _{k=1}^n{\left|(A{)}_{ki}\right|}^2}$$
where $A^{?}$ is the conjugate transpose of the matrix $A$.

Notation

Let $a$ be an element of a normed linear space $(X,\Vert ?\Vert )$ and $r>0.$

B(a,r)={x∈X∣∥x?a∥<r}B(a, r)=\{x \in X \mid\|x-a\|<r\}B(a,r)={ x∈X∣∥x?a∥<r} (Open ball with centre $a$ and radius $r)$

B[a,r]={x∈X∣∥x?a∥≤r}B[a, r]=\{x \in X \mid\|x-a\| \leq r\}B[a,r]={ x∈X∣∥x?a∥≤r} (Closed ball with centre $a$ and radius $r)$

S(a,r)={x∈X∣∥x?a∥=r}S(a, r)=\{x \in X \mid\|x-a\|=r\}S(a,r)={ x∈X∣∥x?a∥=r} (Sphere with centre $a$ and radius $r)$

Equivalent Norms

Definiton

Let $\Vert ?\Vert$ and $\Vert ?{\Vert }_0$ be two different norms defined on the same linear space $X.$ We say that $\Vert ?\Vert$ is equivalent to $\Vert ?{\Vert }_0$ if there are positive numbers $\alpha$ and $\beta$ such that
$$\alpha \Vert x\Vert \le \Vert x{\Vert }_0\le \beta \Vert x\Vert ,\text{ for all }x\in X$$

example

all norms on a finite-dimensional normed linear space are equivalent.

2.2 Open and Closed Sets

Definition

[1] A subset $S$ of a normed linear space $(X,\Vert ?\Vert )$ is open if for each $s\in S$ there is an $?>0$ such that $B(s,?)?S$

[2] A subset $F$ of a normed linear space $(X,\Vert ?\Vert )$ is closed if its complement $X\backslash F$ is open.

[3] Let $S$ be a subset of a normed linear space $(X,\Vert ?\Vert ).$ We define the closure of $S,$ denoted by $\bar{S},$ to be the intersection of all closed sets containing $S$

It is easy to show that $S$ is closed if and only if $S=\bar{S}$.

[4] metric on a set $X$ is a real-valued function $d:X\times X\to \mathbb{R}$ which satisfies the following properties: For all $x,y,z\in X$,
M1. $d(x,y)\ge 0$
M2. $d(x,y)=0?x=y$
M3. $d(x,y)=d(y,x)$
M4. $d(x,z)\le d(x,y)+d(y,z)$

Theorem

(a) If $(X,\Vert ?\Vert )$ is a normed linear space, then
$$d(x,y)=\Vert x?y\Vert$$
defines a metric on $X.$ Such a metric $d$ is said to be induced or generated by the norm $\Vert ?\Vert .$ Thus, every normed linear space is a metric space, and unless otherwise specified, we shall henceforth regard any normed linear space as a metric space with respect to the metric induced by its norm.
(b) the property of metric d If $d$ is a metric on a linear space $X$ satisfying the properties: For all $x,y,z\in X$ and for all $\lambda \in \mathbb{F}$,
(i) $d(x,y)=d(x+z,y+z)\text{ (Translation Invariance) }$

? (ii) $d(\lambda x,\lambda y)=|\lambda |d(x,y)$ (Absolute Homogeneity),
then
$$\Vert x\Vert =d(x,0)$$
defines a norm on $X$.

2.3 Quotient Norm and Quotient Map

[1] Let $M$ be a closed linear subspace of a normed linear space $X$ over $\mathbb{F}$. The quotient space $X/M$ is a normed linear space with respect to the norm(Quotient Norm)
$$\Vert [x]\Vert :=\underset{y\in [x]}{\inf }\Vert y\Vert =\underset{m\in M}{\inf }\Vert x+m\Vert =\underset{m\in M}{\inf }\Vert x?m\Vert =d(x,M),\text{ where }[x]\in X/M$$

${\inf }_{m\in M}\Vert x?m\Vert =d(x,M)$，这个我的理解是，x到m的范数最小，也就是x到M的距离，类比点到直线的距离。

[2] Let $M$ be a closed subspace of the normed linear space $X$. The mapping $Q_M$ from $X\to X/M$ defined by
$$Q_M(x)=x+M,x\in X$$
is called the quotient map (or natural embedding) of $X$ onto $X/M$.

2.4 Completeness of Normed Linear Spaces

Definition

Let ${\left(x_n\right)}_{n=1}^{\infty }$ be a sequence in a normed linear space $(X,\Vert ?\Vert )$.

converge of normed linear space sequence

(a) ${\left(x_n\right)}_{n=1}^{\infty }$ is said to converge to $x$ if given $?>0$ there exists a natural number $N=N(?)$ such that
$$\Vert x_n?x\Vert <?\text{ for all }n\ge N$$
Equivalently, ${\left(x_n\right)}_{n=1}^{\infty }$ converges to $x$ if
$$\underset{n\to \infty }{\lim }\Vert x_n?x\Vert =0$$
If this is the case, we shall write
$$x_n\to x\text{ or }\underset{n\to \infty }{\lim }{x}_n=x$$
Convergence in the norm is called norm convergence or strong convergence.

Cauchy sequence of normed linear space

(b) ${\left(x_n\right)}_{n=1}^{\infty }$ is called a Cauchy sequence if given $?>0$ there exists a natural number $N=N(?)$ such that
$$\Vert x_n?x_m\Vert <?\text{ for all }n,m\ge N$$
Equivalently, $\left(x_n\right)$ is Cauchy if
$$\underset{n,m\to \infty }{\lim }\Vert x_n?x_m\Vert =0$$

Cauchy sequence，在序列号趋于无穷大的时候，它的值就趋于稳定了。

Lemma

1. Let $C$ be a closed set in a normed linear space $(X,\Vert ?\Vert )$ over $\mathbb{F},$ and let $\left(x_n\right)$ be a sequence contained in $C$ such that ${\lim }_{n\to \infty }{x}_n=x\in X.$ Then $x\in C$

   赋范空间中的闭合子集中的一个序列，如果收敛，则极限值一定在这个闭合子集中。

2. Let $X$ be a normed linear space and $A$ a nonempty subset of $X.$
   $[1]|d(x,A)?d(y,A)|\le \Vert x?y\Vert$ for all $x,y\in X$
   $[2]|\Vert x\Vert ?\Vert y\Vert |\le \Vert x?y\Vert$ for all $x,y\in X$
   [3] If $x_n\to x,$ then $\Vert x_n\Vert \to \Vert x\Vert$
   [4] If $x_n\to x$ and $y_n\to y,$ then $x_n+y_n\to x+y$
   [5] If $x_n\to x$ and ${\alpha }_n\to \alpha ,$ then ${\alpha }_n{x}_n\to \alpha x$
   [6] The closure of a linear subspace in $X$ is again a linear subspace;
   [7] Every Cauchy sequence is bounded;
   [8] Every convergent sequence is a Cauchy sequence.

Proposition

Let $(X,\Vert ?\Vert )$ be a normed linear space over $\mathbb{F}$. A Cauchy sequence in $X$ which has a convergent subsequence is convergent.

就是说Cauchy sequence 如果有一个收敛的子序列，那么Cauchy sequence也是收敛的

Completeness

[1] A metric space $(X,d)$ is said to be complete if every Cauchy sequence in $X$ converges in $X$.

[2] A normed linear space that is complete with respect to the metric induced by the norm is called a Banach space.

就是说：Banach space是一个metric由norm给出的赋范空间

[3] Theorem Let $(X,\Vert ?\Vert )$ be a Banach space and let $M$ be a linear subspace of $X.$ Then $M$ is complete if and only if the $M$ is closed in $X$.

[4] The classical sequence space ${?}_p$ is complete.

2.5 Series in Normed Linear Spaces

赋范空间中的级数

Definition

[1] Let $\left(x_n\right)$ be a sequence in a normed linear space $(X,\Vert ?\Vert ).$ To this sequence we associate another sequence $\left(s_n\right)$ of partial sums, where $s_n={\sum }_{k=1}^n{x}_k$

[2] Definition Let $\left(x_n\right)$ be a sequence in a normed linear space $(X,\Vert ?\Vert ).$ If the sequence $\left(s_n\right)$ of partial sums converges to $s,$ then we say that the series ${\sum }_{k=1}^{\infty }{x}_k$ converges and that its sum is $s.$ In this case we write ${\sum }_{k=1}^{\infty }{x}_k=s$. The series ${\sum }_{k=1}^{\infty }{x}_k$ is said to be absolutely convergent if ${\sum }_{k=1}^{\infty }\Vert x_k\Vert <\infty .$

用部分和的形式定义赋范空间中的级数收敛

Theorem

[1] A normed linear space $(X,\Vert ?\Vert )$ is a Banach space if and only if every absolutely convergent series in $X$ is convergent.

[2] Let $M$ be a closed linear subspace of a Banach space $X.$ Then the quotient space $X/M$ is a Banach space when equipped with the quotient norm.

2.6 Bounded, Totally Bounded, and Compact Subsets of a Normed Linear Space

Definition

[1] A subset $A$ of a normed linear space $(X,\Vert ?\Vert )$ is bounded if $A?B[x,r]$ for some $x\in X$ and $r>0$
It is clear that $A$ is bounded if and only if there is a $C>0$ such that $\Vert a\Vert \le C$ for all $a\in A$.

[2] Let $A$ be a subset of a normed linear space $(X,\Vert ?\Vert )$ and $?>0.$ A subset $A_{?}?X$ is called an $?$ -net for $A$ if for each $x\in A$ there is an element $y\in A_{?}$ such that $\Vert x?y\Vert <?.$ Simply put, $A_{?}?X$ is an $?$ -net for $A$ if each element of $A$ is within an $?$ distance to some element of $A_{?}$

$A_{?}$表示这样一个集合，对于A中的每一个元素a你总能在$A_{?}$中找到对应的某个元素$a_{?}$，使得它俩的distance在$?$内。

A subset $A$ of a normed linear space $(X,\Vert ?\Vert )$ is totally bounded (or precompact) if for any $?>0$ there is a finite $?$ -net $F_{?}?X$ for $A$. That is, there is a finite set $F_{?}?X$ such that
$$A?\underset{x\in F_{?}}{?}B(x,?)$$

Proposition

[1] Every totally bounded subset of a normed linear space $(X,\Vert ?\Vert )$ is bounded.

[2] A subset $A$ of a normed linear space $(X,\Vert ?\Vert )$ is totally bounded if and only if for any $?>0$ there is a finite set $F_{?}?A$ such that
$$A?\underset{x\in F_{?}}{?}B(x,?)$$
[3] A normed linear space $(X,\Vert ?\Vert )$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

Theorem

[1] A subset $K$ of a normed linear space $(X,\Vert ?\Vert )$ is totally bounded if and only if every sequence in $K$ has a Cauchy subsequence.

[2] A subset of a normed linear space is sequentially compact if and only if it is totally bounded and complete .

Remark

It can be shown that on a metric space, compactness and sequential compactness are equivalent. Thus, it follows, that on a normed linear space, we can use these terms interchangeably.

Corollary

[1] A subset of a Banach space is sequentially compact if and only if it is totally bounded and closed

[2] A sequentially compact subset of a normed linear space is closed and bounded.

[3] A closed subset F of a sequentially compact normed linear space $（X;\Vert ?\Vert ）$is sequentially compact.