Showing uniqueness of holomorphic functions on Banach spaces using uniquess of scalar holomorphic functions? Theorem
Let $D ⊆ C$ be a connected open set, and let $z_n ∈ D$, $n ∈ \mathbb{N}$, be
a sequence which converges to a point $z_0 ∈ D$ such that $zn \neq z_0$ for all $n ∈ \mathbb{N}$. Further let $E$ be a Banach space, and let $f,g : D → E$ be two holomorphic functions such that $f(z_n) = g(z_n)$ for all $n ∈ N$. Then $f = g$ on $D$.
Proof
Let $E'$ be the dual of $E$. Then it follows from the theorem on uniqueness
of scalar holomorphic functions that, for all $Φ ∈ E'$, $Φ ◦ f = Φ ◦ g$. By the Hahn-Banach theorem this implies that $f = g$.
Questions


*

*I am aware of the theorem on uniqueness of scalar holomorphic functions, but I don't see how that gives us that $Φ ◦ f = Φ ◦ g$ for all $Φ ∈ E'$. Is it because $Φ ◦ f$ is scalar holomorphic function? In that case, I suppose the question is: is the composition of a holomorphic function ($f$) with a continuous function $(Φ ∈ E')$ a holomorphic function?

*I don't see how the Hahn-Banach theorem implies $f=g$. In general I do not have good intution on the Hahn-Banach theorem, so I would appreciate it greatly if anyone who answers this does so very explicitly.

 A: *

*The linearity of $\Phi$ is what allows you to conclude $\Phi\circ f$ holomorphic when $f$ is holomorphic $E$-valued function.  Indeed, you have estimate
$$
\begin{align*}
&\left\lVert\frac{(\Phi\circ f)(z)-(\Phi\circ f)(w)}{z-w}-\Phi(f'(w))\right\rVert\\
&=\left\lVert\Phi\left(\frac{f(z)-f(w)}{z-w}-f'(w)\right)\right\rVert\\
&\leq\lVert\Phi\rVert\cdot\underbrace{\left\lVert\frac{f(z)-f(w)}{z-w}-f'(w)\right\rVert}_{\to 0\text{ as }z\to w}
\end{align*}
$$
for $z,w\in D$.  Conversely if $\Phi\circ f$ is holomorphic for all $\Phi\in E'$ then $f$ is holomorphic, the proof is by Cauchy integral formula and uniform boundedness principle (skipped).
Now $\Phi\circ f$ and $\Phi\circ g$ are (scalar) holomorphic functions agreeing on $z_n\to z_0\in D$, so they are equal.

*For each fixed $z\in D$, $f(z)$ and $g(z)$ cannot be separated by any continuous functional $\Phi$.  So Hahn-Banach implies $f(z)=g(z)$ (since the "usual" Hahn-Banach can be stated as "there are enough continuous linear functionals to separate points").  Since this holds for all $z\in D$ we must have $f=g$.
