Let
- $d$ denote the Euclidean metric on $\mathbb R$ and $$\overline d(x,y):=\min(1,d(x,y))\;\;\;\text{for }x,y\in\mathbb R;$$
- $E$ be a normed $\mathbb R$-vector space;
- $\overline\rho$ denote the uniform metric on $\mathbb R^E$, i.e. $$\overline\rho(f,g):=\sup_{x\in E}\overline d(f(x),g(x))\;\;\;\text{for }f,g\in\mathbb R^E$$ and $\tau(\overline\rho)$ denote the topology on $\mathbb R^E$ generated by $\overline\rho$ (i.e. the uniform topology on $\mathbb R^E$);
- $\rho$ denote the sup metric on $B(E):=\{f\in\mathbb R^E:f(E)\text{ is bounded}\}$, i.e. $$\overline\rho(f,g):=\sup_{x\in E}d(f(x),g(x))\;\;\;\text{for }f,g\in B(E)$$ and $\tau(\rho)$ denote the topology on $B(E)$ generated by $\rho$;
- $\sigma_c(E',E)$ denote the topology of compact convergence on $E'$.
It's clear to me that $$\left.\tau(\overline\rho)\right|_{B(E)}=\tau(\rho),\tag1$$ i.e. the uniform topology on $B(E)$ is generated by $\rho$.
Moreover, I know that, in general, if $X$ is a compact topological space and $Y$ is a metric space the topology of compact convergence and the uniform topology on $Y^X$ coincide.
Let $B_{E'}$ denote the closed unit ball in $E'$ and $\delta>0$. By the Banach-Alaoglu theorem, $X:=\delta B_{E'}$ is $\sigma_c(E',E)$-compact. So, if $\tau:=\left.\sigma_c(E',E)\right|_X$, then $(X,\tau)$ is compact.
Can we conclude that $f:X\to\mathbb R$ is $\sigma_c(E',E)$-continuous if and only if $f:X\to\mathbb R$ is $\left\|\;\cdot\;\right\|_{E'}$-continuous?
