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

This entry was posted in math.CA and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s