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?
 A: 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.
