For a complex Banach space $X$ and an open $U\subset\Bbb C$, is a "weakly holomorphic" $f:U\to X$ also strongly holomorphic? For any complex Banach space $X$ and any open $U\subset\Bbb C$, we say that a function $f:U\to X$ is holomorphic at a point $z_0\in U$ if for $|z-z_0|$ sufficiently small, we can express $f$ as $$f(z)=\sum_{k=0}^\infty a_k (z-z_0)^k$$ where $a_k\in X$ for all $k$, and where the sum converges absolutely. We say that $f:U\to X$ is (strongly) holomorphic if it is holomorphic at every point in $U$.
Meanwhile, we say that a function $f:U\to X$ is weakly holomorphic if for every $x'\in X'$, the function $$\langle x',f\rangle:U\to\Bbb C$$ is holomorphic. Now, it's easy to see that any holomorphic function is weakly holomorphic, but does the converse hold?
 A: For continuous functions on a segment with values in a Banach space their Riemann integral is defined, using Riemann sums, in a similar way to the classical case. Moreover, for every $g\colon I \to X$ continuous and $\phi \in X'$ we have
$$\phi( \int_I g) = \int_I \phi(g)$$
$\bf{Modification:}$ As @Arnaud: pointed out, from $f$ weakly continuous does not follow $f$ continuous.  So something must be done about the proof. If $f$ is assumed continuous, then we have the Cauchy formula weakly, so strongly.
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It's worth checking that for $f\colon \Omega \to X$, the function $f$ is continuous if and only if it is weakly- continuous, that is $\phi\circ f$ is continuous for every $\phi \in X'$ ( note that one can have sequences in $X$ that are weakly convergent, but not convergent in norm, so one needs and extra trick for this last statement)
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Now, it is enough to show that for a weakly holomorphic function we have the Cauchy integral formula:
$$f(a) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(z)}{z-a} dz$$
Now, this is true because it is true weakly. From here, one gets the power series expansion.
