continuous extension of holomorphic function up to the boundary Denote $S=\{z \in \mathbb{C}|0 \leq \text{Re}(z) \leq 1\}$, let $X$ be a Banach space and let $f:S^\circ \rightarrow X$ be a holomorphic function. Under what assumptions does $f$ have a unique continuous extension $\tilde{f}:S \rightarrow X$? (Here $S^\circ$ denotes the set $\{z \in \mathbb{C}|0 < \text{Re}(z) < 1\}$).
The reason I ask this question:
Using the Weierstrass M-test I saw that a series of the form $\sum_{n=1}^{+\infty}f_n(z)$ converges absolutely and uniformly on $\boldsymbol{S^\circ}$ (where $f_n:S \rightarrow X$ are holomorphic in $S^\circ$ and continuous in $S$ for every $n \in \mathbb{N}$) so that I was able to define a function $f:S^{\circ} \rightarrow X$ by the formula $f(z)=\sum_{n=1}^{+\infty}f_n(z)$ (where the convergence happens with respect to the norm of $X$). I would like to extend $f$ continuously in $S$ (in a unique way). Is it possible?
 A: $f$ admits an extension to $S$ precisely when $f$ is uniformly continuous on bounded sets. This is for example the case if $f'$ is bounded on bounded sets.
(Proof: $\Rightarrow$: continuous functions on compact sets are uniformly continuous, $\Leftarrow$: uniformly continuous functions map Cauchy sequences to Cauchy sequences and $S^\circ$ is dense in $S$.)
In your particular setting $f$ admits an extension to $S$. Let $g_n=\sum_{m\leq n} f_m$, so that $g_n\to f$ uniformly on $S^\circ$. $g_n$ is a Cauchy sequence in $C(S,X)$ because $$\|g_n-g_m\|_{\infty,S^\circ}=\|g_n-g_m\|_{\infty,S}$$ as $S^\circ$ is dense in $S$. Thus (because of the completeness of $X$) $g_n$ converges in $C(S,X)$ to some function $g\in C(S,X)$. By uniqueness $g\rvert_{S^\circ}=f$, hence $g$ is the desired extension. (So basically: uniform convergence of continuous functions on dense sets implies uniform convergence on the whole set)
The extension is unique in any case because $S^\circ$ is dense in $S$ (and $X$ is Hausdorff).
