My CV泛函分析作业12.4
22. Prove that in $l^1$ space, strong convergence equals to weak convergence.
proof:
Obviously, weak convergence can be inferred from weak convergence.
So we just need to prove the opposite direction. In fact we only need to prove \(x_n \stackrel{\omega}{\to}x_0\Rightarrow x\stackrel{||\cdot||}{\to}x_0.\) If exist sequences ${x_n}^\infty_{n=1}$, it is weakly converging bunt not strongly converging. Might as well set $||x_n||_1=1$.
We can always choose a sequence $x_{N_1}$, s.t. exist $k_1$, \(x_{N_1}=\{x_{N_1}^{(1)},x_{N_1}^{(2)},\dots,x_{N_1}^{(3)}\},\\ \sum^{k_1}_{i=1}\ |x_{N_1}^{(i)}|<\frac 1 4.\) And exist a sequence $x_{N_2}$, s.t. exist $k_2$ \(\sum^{k_2}_{i=k_1+1}\ |x_{N_2}^{(i)}|>\frac 2 3,\\ \sum^{k_1}_{i=1}\ |x_{N_2}^{(i)}|<\frac 1 4.\) By parity of reasoning, we can take $x_{N_1},x_{N_2},x_{N_3},\dots$ \(f(x)=\sum_{i=1}^\infty\ x_iy_i.\) And for $x_{N_j}$, $y_i=sgn(x_{N_j}^{(i)})$. From the difinition of every sequence \(f(x_{N_j})\geq\sum_{i=k_j}^{k_{j+1}}\ x_{N_j}^{(i)}\cdot sgn(x_{N_j}^{(i)})-\sum_{i=1}^{k_j-1}\ |x_{N_j}^{(i)}|-\sum_{i=k_{j+1}+1}^\infty\ |x_{N_j}^{(i)}|\geq\frac 2 3-(1-\frac 2 3)=\frac 1 3.\) But this is conflict to $x_n\stackrel{\omega}{\longrightarrow}x_0$. Q.E.D.
24. Find a sequence that weakly but not stongly converges in $L^p\ [a,b]\quad(1<p<\infty)$.
proof:
We sepecifically choose the sequence from $L^2[0,1]$. We denote the sequence as ${x_n}_{i=1}^\infty$, $x_n(t)=sin\ n\pi t$.
Then according to Riemann-Lebesgue theorem, we can see the continuous linear function in duel space \(f(x)=\int_0^1sin\ n\pi t\cdot y(t)\ dt\to0\quad(n\to\infty,\ y(t)\in L^q[0,1]).\) That is to say $\forall$ continuous linear function in X’, there must be \(f(x_n)\to0\Rightarrow x_n\stackrel{\omega}{\longrightarrow}x_0=0.\) But by the norm of $L^2[0,1]$, $||x_n||=\frac1 2$.
25. Set ${x_n}{n=1}^\infty\in C[a,b] $. Prove that $x_n\stackrel{\omega}{\rightarrow}x\Rightarrow\lim{n\to\infty}x_n(t)=x(t)$.
We define $f(x)=x(t)$. Obviously it is linear. \(|f(x)|=|x(t)|\leq b \|x\|.\) So it is well defined. \(f(x_n)\to f(x)\Leftrightarrow x_n(t)\to x(t).\) Q.E.D.
26. Set $X,Y$ are both Banach space, $T\in L(X,Y)$.
Prove if $x_n\stackrel{\omega}{\longrightarrow}x$, then $Tx_n\stackrel{\omega}{\longrightarrow}Tx$.
proof:
\[x'(x_n)\to x'(x)\Rightarrow T'y(x_n)\to T'y(x)\Rightarrow y(Tx_n)\to y(Tx)\Rightarrow Tx_n\stackrel{\omega}{\longrightarrow}Tx.\]27. Set $M$ is subspace of normed linear space $X$ , ${x_n}_{i=1}^\infty \subset M$ , $x_n\stackrel{\omega}{\longrightarrow}x_0$, then $x_0\in M$.
proof:
According to ==Hahn-Banach Theorem==, if $x_0\notin M$, then exist a $x’$, \(x'(x_n)=0,\quad x'(x_0)=1.\) But by the definition of weak convergence, we know \(x'(x_n)\to x'(x_0)=0.\) This is conflict to $x’(x_0)=1$. Q.E.D.
28. Set $X,Y$ are both Banach space, also $X\neq {0}$. Prove if $L(X,Y)$ is a Banach space, then $Y$ is Banach space.
proof:
Take a Cauchy sequence ${y_n}$, then for all $T\in L(X,Y)$
According to Hahn-Banach Theorem, $\exist\ f,\ |f|=1$ \(|f(x)|=\|x\|.\) Take $x_0\in X$. Define $F_n(x)=f(x)y_n$. Obviously, it is linear by f. \(|F_n(x)|=|f(x)y_n|\leq\|f\|\|y_n\|\|x\|.\) So it is well defined. \(\|(F_n-F_m)\|=\sup_{\|x\|=1}|f(x)(y_n-y_m)|\leq\|f\|\|y_n-y_m\|.\) Now we take $y$, s.t. \(y=\frac{F(x_0)}{\|x_0\|}.\)
\[\|y_n-y\|=|f(\frac{x_0}{\|x_0\|})y_n-\frac{F(x_0)}{\|x_0\|}|\leq\|(F_n-F)\|\|\frac{x_0}{\|x_0\|}\|=\|F_n-F\|.\]30. Set $X,Y$ are both Banach space, $T\in L(X,Y)$. If $R(T)=Y$, then exist a comstant number $M$, s.t. for all $y\in Y$
\[y=Tx,\quad \|x\|\leq M\|y\|.\]proof:
We try to take a bijective operator by take quotient space $X/N(T)$ \(\bar T[x]=Tx',\quad x'\in [x].\) By this definition, it is linear and bounded. \(\|\bar T\|=\inf_{x\in X/N(T)}\frac{\|\bar Tx\|}{\|x\|}=\inf_{x\in X/N(T)}\frac {\|y\|}{\|x\|}=\frac 1 M.\)