This is Chapter VII, $\S$3, exercise 4, from Conway's book: A Course in Functional Analysis:
Let $X$ be a Banach space and $G\subset \mathbb{C}$ an open subset. We say that $f: G \to X$ is analytic if the limit $$\lim_{h \to 0} \frac{ f(z+h)-f(z)}{h}$$ exist in $X$ for all $z \in G$.
Prove that if $f: G \to X$ is a function such that for each $x^* \in X^*$, the function $x^*\circ f : G \to \mathbb{C}$ is analytic (in the usual way), then $f$ is analytic.
Since weak convergence does not imply the strong one, I feel that I am missing something to prove this. Any thoughts? Thanks in advance.
We give the proof the OP wants, also an example showing that the result fails for functions on the line.
The standard proof via the vector-valued Cauchy Integral Formula seems like the "right" proof, because that vector-valued CIF is going to be fundamentally important soon anyway, when we get to Banach algebras and operator theory and so on. But, if we want a proof that we can do before we get to vector-valued integrals, here it is. Doesn't use Uniforrm Boundedness either; uses nothing but Hahn-Banach on the Banach space side.
Say $\overline{D(z,r)}\subset G$. Say $0<\delta<r/2$ and $|s|,|t|\in(0,\delta)$. If $x^*\in X^*$ with $||x^*||\le1$ then CIF applied to $x^*\circ f$ shows that $$\left|x^*\left(\frac{f(z+s)-f(z)}{s}-\frac{f(z+t)-f(z)}{t}\right)\right|\le c\delta\sup_{|w-z|=r}|x^*(f(w))| \le c\delta\sup_{|w-z|=r}||f(w)||.$$The right side is independent of $x^*$; we sneak in a $\sup_{||x^*||\le1}$ on the left side and we obtain $$\lim_{s,t\to0}\left|\left|\frac{f(z+s)-f(z)}{s}-\frac{f(z+t)-f(z)}{t}\right|\right|=0.$$
Free Bonus An example of $f:\Bbb R\to X$ such that $x^*\circ f$ is $C^1$ for every $x^*\in X^*$ although $f$ is not differentiable in norm, showing that there has to be something "complex" about the proof of the result above:
Define $f:\Bbb R\to c_0$ by $$f(t)=\left(e^{it},\frac12 e^{2it}, \frac13e^{3it},\dots\right).$$If $x^*\in X^*=\ell^1$ then dominated convergence shows that $x^*\circ f$ is $C^1$. On the other hand, it's clear that $$f'(t)=\left(ie^{it},ie^{2it},\dots\right)$$is the derivative of $f$ in some sense (for example in the sense of weak* convergence in $c_0^{**}$); if $f$ were differentiable in norm the derivative would be the $f'$ above, but $f'(t)\notin c_0$.
I suspect that there is no such example with $X$ reflexive.
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No wait, there's a simple example in a Hilbert space. (In the example above you may note that $||f'(t)||_{c_o^{**}}$ is constant. This was why I couldn't find the Hilbert space example. As soon as I realized that, in a Hilbert space, if $t\mapsto f'(t)$ is weakly continuous but not norm continuous then $||f'(t))||$ cannot be continuous the example fell right out.)
Define $f:\Bbb R\to L^2(\Bbb R)$ by setting $f(t)=0$ for $t\le0$, while for $t>0$ $$f(t)(x)=\begin{cases}0,&(x\le 0), \\(t^{1/2}-x^{1/2})^+,&(x>0).\end{cases}$$We leave the details to the reader, since this post is long enough already. Two hints: The weak derivative comes out to $$f'(t)=\begin{cases}0,&(t\le0),\\ \frac{t^{-1/2}}{2}\chi_{(0,t)},&(t>0),\end{cases}$$and in verifying that $f$ is not differentiable in norm you can avoid an argument that requires the numbers to come out just right by noting that$$\left|\left|\frac{f(h)-f(0)}{h}-f'(0)\right|\right|= \left|\left|\frac{f(h)}{h}\right|\right|=||f(1)||\quad(h>0).$$