If $\tau$ is a vector topology on a normed space s.t. a closed ball is $\tau$-compact, is a continuous function on a closed ball even uniformly cont.?

Question

Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $r>0$, $X$ be a normed $\mathbb K$-vector space, $\tau$ be a vector topology on $X$ s.t. $\overline B_r^X(0):=\{x\in X:\left\|x\right\|_X\le r\}$ is $\tau$-compact and $h:E\to\mathbb C$ be $\tau$-continuous on $\overline B_r^X(0)$, i.e. the restriction $\left.h\right|_{\overline B_r^X(0)}$ of $h$ to $\overline B_r^X(0)$ is continuous with respect to $\left.\tau\right|_{\overline B_r^X(0)}=\{\Omega\cap\overline B_r^X(0):\Omega\in\tau\}$.

Are we able to show that $h$ is uniformly $\tau$-continuous on $\overline B_r^X(0)$?

Let $\varepsilon>0$. By assumption, for all $x\in\overline B_r^X(0)$, there is an open $\left.\tau\right|_{\overline B_r^X(0)}$-neighborhood $N_x$ of $x$ with $$\forall y\in N_x:|h(x)-h(y)|<\varepsilon\tag1.$$ Since $\overline B_r^X(0)$ is $\tau$-compact, there is a finite $F\subseteq\overline B_r^X(0)$ with $$\overline B_r^X(0)\subseteq\bigcup_{x\in F}N_x.\tag2$$ But at this point I'm stuck ...

Answer 1

It’s a general fact in any uniform space that a continuous function on a compact subset is uniformly continuous on that subset. Nothing specific to topological vector spaces.