I'm trying to prove: if a function $f: G\to V$, where $V$ is a complex Banach space and $G\subset \mathbb{C}$ is an open subset, is weakly analytic, then it is strong analytic. I found this answer and also read through Rudin's Thm 3.31. But the proof seems too complicated. I wonder whether the following direct estimate also works?
Let $z\in G$ and $\Delta_r(z):=\{w\in \mathbb{C}: 0\leq |w-z|<r\}$ such that $\overline{\Delta_{2r}(z)}\subset G$. Suppose $0< |h|<r$. On the boundary circle $\Gamma_{z,2r}:=\partial \Delta_{2r}(z)$, for every cts linear functional $x^*\in V^*$, we can apply $\mathbb{C}$-valued Cauchy Integral Formula, \begin{align*} x^*\left(\frac{f(z+h)-f(z)}{h}\right)&=\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)-(x^{*}\circ f)(z)}{(\lambda-z-h)(\lambda-z)}}d\lambda\\ &=\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)}{(\lambda-z-h)(\lambda-z)}}d\lambda-\frac{(x^{*}\circ f)(z)}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{d\lambda}{(\lambda-z-h)(\lambda-z)}}\\ &=\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)}{(\lambda-z-h)(\lambda-z)}}d\lambda-0\\ &=\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)}{(\lambda-z)^2}}d\lambda+\frac{h}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)}{(\lambda-z-h)(\lambda-z)^2}}d\lambda\\ &=x^*\left(\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda\right)+\frac{h}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{(x^{*}\circ f)(\lambda)}{(\lambda-z-h)(\lambda-z)^2}}d\lambda. \end{align*} So, if denoting $M_{f, \Gamma}:=\sup\|f(\Gamma)\|$, we have the following estimate $$\left|x^*\left(\frac{f(z+h)-f(z)}{h}- \frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda\right)\right|\leq |h| \|x^*\|\frac{M_{f, \Gamma}}{2r^2}$$
But now the upper bound depends on the functional norm $\|x^*\|$. Can we still conclude, using Hahn-Banach Thm, that $$\left\|\frac{f(z+h)-f(z)}{h}- \frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda\right\|_V\to 0, \text{ when } h\to 0?$$
Thanks in advance.
Update
It seems we can make that conclusion, by the canonical embedding to the bidual, $\Phi: V \to V^{**}$ and equality of norm $\|\Phi(x)\|_{V^{**}}=\|x\|_V$. With that, the above estimate is, for every $x^*\in V^*$, $$\left|\Phi\left(\frac{f(z+h)-f(z)}{h}- \frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda\right)(x^*)\right|\leq |h| \|x^*\|\frac{M_{f, \Gamma}}{2r^2}.$$ Hence, if letting $$\xi(z,r,h):=\frac{f(z+h)-f(z)}{h}- \frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda,$$ as $h\to 0$, $$\left\|\xi\right\|_V=\left\|\Phi(\xi)(x^*)\right\|_{V^{**}}=\sup_{x^*\in V^*} \frac{|\Phi(\xi)(x^*)|}{\|x^*\|}\leq |h|\frac{M_{f, \Gamma}}{2r^2}\to 0.$$ That is $$\lim_{h\to 0}\frac{f(z+h)-f(z)}{h}=\frac{1}{2\pi i}\int_{\Gamma_{z,2r}}{\frac{f(\lambda)}{(\lambda-z)^2}}d\lambda \in V.$$
Correct?