Analytic Banach Space Valued Functions

I know in my last post I said that I would be focusing on the Prime Number Theorem, but I want to make a brief digression on a topic related to an earlier post, namely holomorphic functions taking values in a Banach space. A discussion about Stein interpolation in my analysis class today is the impetus behind this digression.

Let ${(X,\left\|\cdot\right\|_{X})}$ be a Banach space, and ${\Omega \subset \mathbb{C}}$ be a domain. Recall that a function ${f: \Omega \rightarrow X}$ is said to be differentiable at ${z}$ if there exists ${f'(z) \in X}$ such that

$\displaystyle \lim_{h \rightarrow 0}\left\|\frac{f(z+h) - f(z)}{h} - f'(z)\right\|_{X} = 0$

In this post, I want to prove an equivalent characterization of holomorphicity that relates such a function ${f}$ to complex-differentiable functions as studied in a first course in complex analysis.

Theorem 1 A function ${f: \Omega \rightarrow X}$ is holomorphic if and only if for every bounded linear function ${l \in X^{*}}$, the function ${l(f): \Omega \rightarrow \mathbb{C}}$ defined by ${l(f)(z) := l(f(z))}$ is holomorphic.

Proof: We showed in the post on the generalization of Liouville’s theorem that the function ${l(f)(z)}$ is holomorphic with derivative ${l(f'(z))}$ if ${f}$ is holomorphic, which is the ${\Rightarrow}$ direction. We now prove ${\Leftarrow}$. Let ${z_{0} \in \Omega}$ and let ${\Gamma}$ be a postively oriented circle centered at ${z_{0}}$ with radius ${r}$. For each ${l \in X^{*}}$, the function ${F_{l}(z) = l(f(z))}$ is an analytic function of ${z}$. For ${z,z'}$ in the interior of ${\Gamma}$, we have by the Cauchy integral formula applied to ${F_{l}(z), F_{l}(z_{0})}$, and ${F_{l}(z')}$ that

$\displaystyle l\left(\frac{f(z) - f(z_{0})}{z-z_{0}} - \frac{f(z') - f(z_{0})}{z'-z_{0}}\right) = \frac{1}{2\pi i}\int_{\Gamma}\left[\frac{1}{z-z_{0}}\left(\frac{1}{\zeta - z} - \frac{1}{\zeta - z_{0}}\right) - \frac{1}{z'-z_{0}}\left(\frac{1}{\zeta - z'} - \frac{1}{\zeta - z_{0}}\right)\right]l(f(\zeta))d\zeta$

$\displaystyle = \frac{1}{2\pi i}\int_{\Gamma}\left[\frac{1}{(\zeta - z)(\zeta - z_{0})} - \frac{1}{(\zeta - z')(\zeta - z_{0})}\right]l(f(\zeta))d\zeta$

$\displaystyle = \frac{1}{2\pi i}\int_{\Gamma}\frac{z-z'}{(\zeta - z)(\zeta - z')(\zeta - z_{0})}l(f(\zeta))d\zeta$

Under the canonical injection, we can identify ${f(\zeta) \in X^{**}}$. Since ${F_{l}(\zeta)}$ is continuous in ${\zeta}$ and ${\Gamma}$ is compact, there exists ${C_{l} > 0}$ such that ${\left|l(f(\zeta))\right| \leq C_{l}}$ for all ${\zeta \in \Gamma}$. Thus, the family of linear operators ${\mathcal{F} = \left\{f(\zeta) : \zeta \in \Gamma\right\} \subset X^{**}}$ is pointwise bounded. By the principle of uniform boundedness, we have

$\displaystyle \sup_{\zeta \in \Gamma}\left\|f(\zeta)\right\|_{X^{**}} \leq C < \infty \Rightarrow \sup_{\zeta \in \Gamma} \left|l(f(\zeta))\right| \leq C\left\| l \right\|_{X^{*}}$

Hence if ${\left|z-z_{0}\right|, \left|z'-z_{0}\right| \leq \frac{r}{2}}$,

$\displaystyle \left|l\left(\frac{f(z) - f(z_{0})}{z-z_{0}} - \frac{f(z') - f(z_{0})}{z'-z_{0}}\right)\right| \leq \frac{1}{2\pi}\frac{\left|z-z'\right|}{\frac{r}{2}\frac{r}{2}r}C\left\|l\right\|_{X^{*}}2\pi r = \frac{4C}{r^{2}}\left|z-z'\right|\left\|l\right\|_{X^{*}}$

Taking the supremum over all ${l \in X^{*}, \left\|l\right\|_{X^{*}} = 1}$, we conclude that

$\displaystyle \left\|\frac{f(z) - f(z_{0})}{z-z_{0}} - \frac{f(z') - f(z_{0})}{z'-z_{0}}\right\|_{X} \leq \frac{4C}{r^{2}}\left|z-z'\right|,$

which converges to ${0}$ as ${z,z' \rightarrow z_{0}}$. This shows that the difference quotients are Cauchy. Since ${X}$ is complete, ${\frac{f(z) - f(z_{0})}{z-z_{0}} \rightarrow f'(z_{0}) \in X}$ as ${z \rightarrow z_{0}}$. $\Box$