Weak convergence and weak_star convergence

Weak Convergence

Let $X$ be a normed vector space, a sequence $\{X_n{\}}_{n=1}^{\infty }\subset X$ is weakly convergent to $x\in X$ if $\forall l\in X^{\ast }$, ${\lim }_{n\to \infty }l(x_n)=l(x)$

Proposition

1. Strong convergence implies weak converge.
2. Weak converges does not implies strong convergence. For example consider $X=l^2(\mathbb{R})$, then for $f_n(x)=1_{[n,n+1]}(x)$, $f_n$ converges weakly to 0, but not converge strongly to 0.
3. If $\{x_n\}$ is norm convergent (strong convergent) then ${\sup }_n\Vert x_n{\Vert }_X<\infty$ and similarly if $\{x_n\}$ is weakly convergent then $\forall l\in X^{\ast }$, ${\sup }_n|l(x_n)|<\infty$, also called weakly bounded

Weakly Bounded and Bounded Set

Suppose $X$ is a normed vector space, then $E\subset X$ is weakly bounded if ${\sup }_{x\in E}|f(x)|<\infty$ for all $f\in X^{\ast }$. $E$ is bounded if ${\sup }_{x\in E}\Vert x{\Vert }_X<\infty$.

Theorem

Suppose $X$ is a normed vector space, then $\{x_n\}$ is weakly bounded iff $\{x_n\}$ is bounded.

Proof We’ll just show harder direction. Let $E\subset X$ be a weakly bounded set, for every $f\in X^{\ast }$, we have ${\sup }_{x\in E}|f(x)|<\infty$. Consider the canonical embedding $T:X\to X^{\ast \ast }$ defined by $Tx(f)=f(x)$ for all $f\in X^{\ast }$. For any fixed $f$, ${\sup }_{x\in E}|(Tx)(f)|={\sup }_{x\in E}|f(x)|<\infty$ and $X^{\ast }$ is a Banach space, with $|T_x(f)|=|f(x)|\le \Vert f\Vert \Vert x\Vert$ the operators are continuous, the Uniform Boundedness Principle applies. Therefore, ${\sup }_{x\in E}\Vert Tx{\Vert }_{X^{\ast \ast }}<\infty$ Since $X^{\ast }$ is a normed vector space, the space $X^{\ast \ast }$ is a Banach space. The elements $Tx$ for $x\in E$ form a subset of $X^{\ast \ast }$. We examine the norms of these functionals: $\Vert Tx{\Vert }_{X^{\ast \ast }}={\sup }_{f\in X^{\ast },\Vert f{\Vert }_{X^{\ast }}=1}|(Tx)(f)|={\sup }_{f\in X^{\ast },\Vert f{\Vert }_{X^{\ast }}=1}|f(x)|$ By the Hahn Banach Theorem, for every $x\in X$, there exists some $f_x\in X^{\ast }$ such that $\Vert f_x{\Vert }_{X^{\ast }}=1$ and $f_x(x)=\Vert x{\Vert }_X$. Thus, $\Vert Tx{\Vert }_{X^{\ast \ast }}=\Vert x{\Vert }_X\text{for all }x\in X$ Now, consider the family of bounded linear functionals $\{Tx{\}}_{x\in E}$ on the Banach space $X^{\ast }$. Since $\Vert Tx{\Vert }_{X^{\ast \ast }}=\Vert x{\Vert }_X$, we conclude that: ${\sup }_{x\in E}\Vert x{\Vert }_X<\infty$ Hence, $E$ is bounded.

${\text{Weak}}^{\ast }$ convergence

Let $X$ be a normed space, if $X=Y^{\ast }$ for some normed vector space $Y$, then we say $\{x_n\}{\text{weak}}^{\ast }$ converges to x, if $\forall y\in Y$, ${\lim }_{n\to \infty }(x_n-x)(y)=0$

Proposition

1. Weak convergence implies ${\text{weak}}^{\ast }$ convergence. As for $X=Y^{\ast }$, we always have $Y\hookrightarrow Y^{\ast \ast }$, so we test against more vectors when we test weak convergence.
2. In finite dimensional spaces, converges in norm $⟺$ weak convergence $⟺{\text{weak}}^{\ast }$ convergence.
3. If $X$ is reflexive(i.e. $X^{\ast \ast }=X$), then weak convergence $⟺{\text{weak}}^{\ast }$ convergence.

Convergence for sequences of operators between normed vector spaces

Let $X,Y$ be normed vector spaces. Let $T_n:X\to Y$ be a sequence of bounded linear operators. Let also $T:X\to Y$ be a bounded linear operator. Then

1. $T_n$ converge in norm$⟹$ $T_n$ converge strongly to $T$, that is ${\lim }_{n\to \infty }\Vert T_n-T{\Vert }_{\mathcal{B}(X,Y)}=0⟹{\lim }_{n\to \infty }\Vert T_{n}f-Tf{\Vert }_Y=0$, since $\Vert T_{n}f-Tf{\Vert }_Y\le \Vert T_n-T{\Vert }_{\mathcal{B}(X,Y)}\Vert f{\Vert }_X$.
2. $T_n$ converge strongly to T$$\implies$$T_n converge weakly to $T$, that is ${\lim }_{n\to \infty }(l(T_{n}f)-l(Tf))=0⟹{\lim }_{n\to \infty }\Vert T_{n}f-Tf{\Vert }_Y=0$, as $|l(T_{n}f)-l(Tf)|=|l(T_{n}f-Tf)|\le \Vert l{\Vert }_{Y^{\ast }}\Vert T_{n}f-Tf{\Vert }_Y$.
3. $T_n$ converge weakly to $T$ $/⟹$ $T_n$ converge strongly to $T$. Example:$T_n:{\ell }^2\to {\ell }^2\text{ shift operator }{T}_n(x_1,x_2,\dots )=(\underset{n\text{ terms}}{\underbrace{0,0,\dots ,0}},x_1,x_2,\dots )$ Then $T_n$ converges to $0$ weakly: $\forall x\in {\ell }^2,\forall y\in {\ell }^2,|⟨T_{n}x,y⟩|=\left|{\sum }_{j=n+1}^{\infty }{x}_{j-n}{y}_j\right|\le {\left({\sum }_{j=n+1}^{\infty }|y_j{|}^2\right)}^{1/2}\Vert x{\Vert }_{{\ell }^2}\to 0\text{as }n\to \infty$ $\text{but }{T}_n\text{ does not converge strongly to }0\text{:}$ Let $x=(1,0,0,\dots )\in {\ell }^2$. Then $\Vert T_{n}x{\Vert }_{{\ell }^2}^2=\Vert x{\Vert }_{{\ell }^2}^2\nrightarrow 0.{\lim }_{n\to \infty }\Vert T_{n}x-0{\Vert }_{{\ell }^2}\ne 0$.
4. $T_n$ converge strongly to T$$\not\implies $T_n$ converge in norm. Example $T_n:{\ell }^2\to {\ell }^2\text{ forgetting operator }{T}_{n}x=(\underset{n\text{ terms}}{\underbrace{0,0,\dots ,0}},x_{n+1},x_{n+2},\dots )\text{if }x=(x_1,x_2,\dots )$ $T_n$ converges strongly to $0$: $\forall x\in {\ell }^2,\Vert T_{n}x-0{\Vert }_{{\ell }^2}^2={\sum }_{j=n+1}^{\infty }|x_j{|}^2\to 0\text{as }n\to \infty$ $\text{But }{T}_n\text{ does not converge to }0\text{ in norm: }\Vert T_n{\Vert }_{\mathcal{B}({\ell }^2,{\ell }^2)}^2=1\text{ since }\Vert T_n{e}_{n+1}{\Vert }_{{\ell }^2}=\Vert e_{n+1}{\Vert }_{{\ell }^2}=1.$