# Is a relatively weakly compact subset necessarily norm bounded (by the Banach-Steinhaus theorem)?

Let $$E_i$$ be a normed vector space (EDIT: Assume $$E_2$$ is complete), $$\langle\;\cdot\;,\;\cdot\;\rangle$$ be a duality pairing between $$E_1$$ and $$E_2$$ with $$\left|\langle x_1,x_2\rangle\right|\le\left\|x_1\right\|_{E_1}\left\|x_2\right\|_{E_2}\tag1,$$ $$\sigma(E_2,E_1)$$ denote the topology on $$E_2$$ generated by the seminorms $$p_{x_1}(x_2):=\left|\langle x_1,x_2\rangle\right|\;\;\;\text{for }(x_1,x_2)\in E_1\times E_2$$ and $$\mathcal F\subseteq E_2$$ be relatively $$\sigma(E_2,E_1)$$-compact.

Assume$$^1$$ $$\left\|x_2\right\|_{E_2}=\sup_{\left\|x_1\right\|_{E_1}\le1}p_{x_1}(x_2)\;\;\;\text{for all }x_2\in E_2\tag2.$$

Are we able to conclude that $$\mathcal F$$ is norm bounded, i.e. $$\sup_{x_2\in\mathcal F}\left\|x_2\right\|_{E_2}<\infty$$?

By $$(2)$$ and the Banach-Steinhaus theorem, it would be sufficient to show that $$\sup_{x_2\in\mathcal F}p_{x_1}(x_2)<\infty\tag3.$$ Can we show this?

$$^1$$ Do we actually explicitly need to assume this or is $$(2)$$ guaranteed to hold in the present setting?

• Perhaps look up the hypotheses for Banach-Steinhaus. Dec 21 '20 at 9:39
• @GEdgar What do you mean? The Banach-Steinhaus theorem is: If $X$ is a Banach space, $Y$ a normed vector space and $\mathcal F$ a family of bounded linear operators from $X$ to $Y$ with $\sup_{A\in\mathcal F}\left\|Ax\right\|<\infty$ for all $x\in X$, then $\sup_{A\in\mathcal F}\left\|A\right\|<\infty$. Dec 21 '20 at 9:47
• OK. Note that your question is for "normed vector space" and not "Banach space". Dec 21 '20 at 9:51
• @GEdgar Yeah, you're right. We need to assume that $E_2$ is complete. Dec 21 '20 at 9:55
• @0xbadf00d Hello! $E_1$ needs to be complete, not $E_2$. Dec 21 '20 at 11:15

If you assume that $$\tag{1} ||x_2||_{E_2}=\sup_{||x_1||_{E_1}\leq 1}p_{x_1}(x_2)\ \text{ for all } x_2\in E_2$$ and $$\mathcal{F}\subseteqq E_2$$ is an $$\sigma(E_2,E_1)-$$compact subset of $$E_2$$ then indeed $$\mathcal{F}$$ is bounded with respect to the $$||.||_2$$ norm. This follows from the uniform boundedness theorem applied to the space $$E_1$$ ( here we need to assume that $$E_1$$ is a Banach space instead of only being a normed space ). Indeed, consider the linear functionals $$f_{x_2}:E_1\to \mathbb{R}$$ where $$f_{x_2}(x_1)=\langle x_1,x_2 \rangle$$ for every $$x_2 \in \mathcal{F}$$. Then, by the assumption $$\tag{2}|\langle x_1,x_2\rangle|\leq ||x_1||_{E_1}\cdot ||x_2||_{E_2}$$ $$f_{x_2}$$'s are continuous. Now, since $$\mathcal{F}$$ is $$\sigma(E_2,E_1)-$$compact it follows that for fixed $$x_1\in E_1$$ the set $$\{f_{x_2}(x_1):\, x_2\in \mathcal{F}\}$$ is a bounded subset of $$\mathbb{R}$$. Hence, by the uniform boundedness theorem it follows that $$\sup_{x_2\in \mathcal{F}}||f_{x_2}||<\infty$$ But, by $$(1)$$, $$||f_{x_2}||=\sup_{||x_1||\leq 1}p_{x_1}(x_2)=||x_2||$$. Hence, $$\sup_{x_2\in \mathcal{F}}||x_2||<\infty$$.
Now if you dont have that $$(1)$$ holds necessarily then the conclusion it is not true in general. An example is the following: Consider $$E_1=E_2=\ell_2$$ and for every $$x,y\in \ell_2$$ the duality pairing $$\langle\;\cdot\;,\;\cdot\;\rangle_{\ell_2'} :\ell_2 \times \ell_2\to \mathbb{R}$$ given by $$\langle x,y\rangle_{\ell_2'}=\sum_{n=1}^{\infty}\frac{1}{n}x_n y_n$$ $$(2)$$ follows by the Cauchy-Schwarz inequality. Now, let $$\mathcal{F}=\{ne_n:\, n\in \mathbb{N}\}\cup \{0\}$$ where $$e_n$$ is the standard basis of $$\ell_2$$. Obviously, $$\mathcal{F}$$ is not bounded in $$\ell_2$$. Now for every $$x\in \ell_2$$ let $$f_x:\ell_2\to \mathbb{R}$$ given by $$f_x(y)=\sum_{n=1}^{\infty}\frac{1}{n}x_ny_n$$ Then, the topology $$\sigma(E_2,E_1)$$ is the coarsest topology for which the functionals $$f_x$$'s are continuous. We claim that $$\mathcal{F}$$ is compact with respect to $$\sigma(E_2,E_1)$$. First observe that $$\tag{3} f_x(\mathcal{F})=\{x_n:\, n\in \mathbb{N}\}\cup\{0\}$$ for every $$x=(x_n)_{n=1}^{\infty}\in \ell_2$$. Since, $$x_n\to 0$$ it follows that $$f_x(\mathcal{F})$$ is a compact subset of $$\mathbb{R}$$. Let $$(W_i)_{i\in I}$$ be an $$\sigma(E_2,E_1)$$-open cover of $$\mathcal{F}$$. Let $$i_0$$ such that $$0\in W_{i_0}$$. There is an $$\epsilon>0$$ and a finite set $$\{x_1,...,x_m\}$$ such that $$0\in W'\subseteq W_{i_0}$$ where $$W'=\{y\in E_2:\, |f_{x_i}(y)|<\epsilon,\,\ i=1,...,m\}$$ By the description of $$(3)$$, if we fix an $$N$$ such that $$|x_i(n)|<\epsilon$$ for every $$i=1,...,m$$ and $$n\geq N$$ we obtain the inclusion $$\{ne_n:\ n\geq N\}\cup\{0\}\subseteq W'$$ Since, the finite set $$\{ne_n:\, 1\leq n\leq N\}$$ can be covered by finitely many $$W_i$$'s it follows that $$\mathcal{F}$$ can be covered by finitely many $$W_i$$'s. Hence, $$\mathcal{F}$$ is an $$\sigma(E_2,E_2)-$$compact subset of $$E_2$$ which is unbounded in the $$||.||_{\ell_2}$$ norm.
• Thank you for your answer. Please note that I'm only assuming that $\mathcal F$ is relatively $\sigma(E_2,E_1)$-compact. Does the claim still hold? And aren't we able to show that any duality pairing which satisfies $(1)$ in the question also satisfies $(2)$ in the question (which is $(1)$ in your post)? Dec 24 '20 at 8:09
• @0xbadf00d If $\mathcal{F}$ is relatively compact then you apply the same argument to the set $\overline{\mathcal{F}}$ and you conclude that $\overline{\mathcal{F}}$ is bounded. Hence, $\mathcal{F}$ must be bounded as a subset of $\overline{\mathcal{F}}$. For a general duality pairing we cant conclude that your $(2)$ holds even if satisfies $(1)$. In the example that i provided, the duality pairing $\langle\;\cdot\;,\;\cdot\;\rangle_{\ell_2'}$ cannot satisfy your $(2)$ condition since the set $\mathcal{F}=\{ne_n:\, n\in \mathbb{N}\}\cup \{0\}$ is compact but not norm bounded in $\ell_2$. Dec 24 '20 at 9:03
• What am I missing? Let $x_2\in E_2$ and $c:=\sup_{\|x_1\|_{E_1}\le1}p_{x_2}(x_1)$. By $(1)$ (in my question), $c\le\left\|x_2\right\|_{E_2}$. Now assume $x_2\ne 0$. Then, since $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing, there is a $x_1\in E_1$ with $\|x_1\|_{E_1}=1$ and $\langle x,y\rangle\ne0$. Thus, $\|x_2\|_{E_2}\le c$. Dec 24 '20 at 9:10
• @0xbadf00d Can you elaborate more on how from $||x_1||_{E_1}=1$ and $\langle x, y\rangle \neq 0$ we obtain $||x_2||_{E_2}\leq c$? Dec 24 '20 at 9:15
• I was a little bit too hasty. I was trying to mimic the prove of $$\left\|x\right\|_E=\sup_{\|\varphi\|_{E'}\le1}|\varphi(x)|\;\;\;\text{for all }x\in E$$ for every normed space $E$, which is an immediate consequence of the Hahn-Banach theorem and in turn yields that $$\langle x,\varphi\rangle:=\varphi(x)\;\;\;\text{for }(x,\varphi)\in E\times E'$$ is a duality pairing between $E$ and $E'$. Dec 24 '20 at 10:14