If $E$ is a normed vector space and $\tau$ is a vector topology on $E$ s.t. $f$ is $\tau$-continuous on every closed ball, is $f$ $\tau$-continuous? Let $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$, $E$ be a normed $\mathbb K$-vector space, $\tau$ be a vector topology on $E$ such that $\overline B_1(0):=\{x\in E:\left\|x\right\|_E\le1\}$ is $\tau$-compact and $f:E\to\mathbb C\setminus\{0\}$ be $\tau$-continuous and uniformly $\left\|\;\cdot\;\right\|_E$-continuous.
Since $f$ is uniformly $\left\|\;\cdot\right\|_E$-continuous, there is a $\delta>0$ with $$|f(x)-f(y)|<\frac\varepsilon2\tag1$$ for all $x,y\in E$ with $\left\|x-y\right\|_E<\delta$. Let $r>0$, $k\in\mathbb N$, $0=t_0<\cdots<t_k=1$ with $\max_{1\le i\le k}|t_{i-1}-t_i|<\frac\delta r$. Then $$|f(t_{i-1}x)-f(t_ix)|<\frac\varepsilon2\tag2$$ for all $1\le i\le k$ and $x\in\overline B_r(0)$ and $$\frac{f(t_ix)}{f(t_{i-1}x)}\in B_{\frac12}(1)\subseteq\mathbb C\setminus(-\infty,0]\tag3$$ for all $x\in\overline B_r(0)$.

By $(3)$, $$g_r(x):=\sum_{i=1}^k\ln\frac{f(t_ix)}{f(t_{i-1}x)}\;\;\;\text{for }x\in\overline B_r(0)$$ is well-defined and $\ln$ is continuous on $\mathbb C\setminus(-\infty,0]$. Note that $$g(x):=g_r(x)\;\;\;\text{for }x\in E\text{ and }r>0\text{ with }x\in\overline B_r(0)$$ is a well-defined function on $E$.
Question: Are we able to conclude that $g$ is $\tau$-continuous and $\left\|\;\cdot\;\right\|_E$-continuous?

It should be obvious that for any $r>0$, the function $g_r$ is continuous with respect to the subspace topologies induced by $\tau$ and $\left\|\;\cdot\;\right\|_E$ on $\overline B_r(0)$. But is this enough to conclude?
 A: Let $H$ be an infinite dimensional Hilbert space and let $D\subseteq H$ be a proper dense subspace.   Consider the locally convex
topology $\sigma (H, D)$ on $H$ given by the family of semi-norms
$\{p_\eta \}_{\eta \in D}$, where
$$
  p_\eta (\xi ) = |\langle \xi , \eta \rangle |, \quad (\xi \in H).
  $$
It is easy to see that $\sigma (H, D)$ is a Hausdorff topology, coarser than the weak topology  (notice that the
weak topology is what you get by choosing $D=H$).
Consequently the identity map
$$
  \text {id} : \big (\bar B_1(0), \sigma (H, H)\big ) \to \big (\bar B_1(0), \sigma (H, D)\big )
  $$
is continuous and, since its  domain  is known to be  compact, while its
co-domain is clearly Hausdorff, we conclude that it is a homeomorphism, which means that $\sigma (H,D)$ coincides with
$\sigma (H,H)$ on $\bar B_1(0)$.
In particular any norm-continuous linear functional is continuous on $\big (\bar B_1(0), \sigma (H, D)\big )$, but it is well known
that the continuous functionals on $H$ relative to $\sigma (H, D)$ are only those of the form
$$
  \xi \mapsto \langle \xi ,\eta \rangle ,
  $$
with $\eta \in D$.
In conclusion, if $\zeta $ is any vector in $H\setminus D$, then the linear functional
$$
  \xi \mapsto \langle \xi ,\zeta \rangle
  $$
is continuous relative to $\sigma (H,D)$ on $\bar B_1(0)$, and hence also on any bounded set, but it is not continuous on the whole
of $H$ with this topology.
