Banach-Space-Valued Analytic Functions 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. 
 A: 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.
[...]
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$$\left|\left|\frac{f(h)-f(0)}{h}-f'(0)\right|\right|= \left|\left|\frac{f(h)}{h}\right|\right|=||f(1)||\quad(h>0).$$
