This is an exercise in Conway's 《A course in Functional Analysis》.
X is a complex Banach space. $G$ is an open set in the complex plane and $f:G\rightarrow X$ is a function such that for each $x^{*}$ in $X^{*}$,dual space of $X$, $x^*\circ f:G\rightarrow \Bbb{C}$ is analytic. Show $f$ is analytic.
Any help would be appreciated.
Here is the proof of this problem in Rudin's Functional Analysis. But I have a puzzle about it and I write it as a comment that follows.
Assume $0\in G$. We shall prove that $f$ is strongly continuous at 0,i.e. continuous in the norm-topology. Define $$\Delta_r=\{z\in \Bbb{C}:|z|\le r\}$$ Then $\Delta_{2r}\subset G$ for some $r\gt0$. Let $\Gamma$ be the positively oriented boundedary of $\Delta_{2r}$. Fix $x^{*}\in X^{*}$. Since $x^{*}(f)$ is holomorphic,$$\frac{x^{*}(f)(z)-x^{*}(f)(0)}{z}=\frac{1}{2\pi i}\int_{\Gamma}{\frac{x^{*}(f)(\lambda)}{(\lambda-z)\lambda}}d\lambda,~~~0\lt|z|\lt2r$$ Let $M(x^{*})$ be the maximum of $|x^{*}(f)|$ on $\Delta_{2r}$. If $0\lt|z|\le r$, it follows that $$|z^{-1}x^{*}[f(z)-f(0)]|\le r^{-1}M(x^{*})$$ The set of all quotients $$\left\{ \frac{f(z)-f(0)}{z}:0\lt|z|\le r \right\}$$ is therefore weakly bounded in $X$. Since original norm of x in $X$ is equal to its norm in $X^{**}$ and using Principle of uniform boundedness,we get this set is bounded in X.
It follows easily that $$f(z)\rightarrow f(0)\text{ in norm, as }z\rightarrow 0.$$i.e. $f$ is continuous on $G$.