In my functional analysis class we defined a map $x:\Omega \to F,$ where $\Omega\subset \mathbb{C}$ is open and $F$ is a complex Banach space, to be differentiable in $z_0\in \Omega$ if the limit \begin{equation}\lim_{z\to z_0}\frac{1}{z-z_0}\left( x(z)-x(z_0)\right) \end{equation} exists in $F$ and to be holomorphic in $z_0$ if there exists a nbhd of $z_0$ in which $x$ is differentiable in every point.

Then the following remark was made: $x$ is holomorphic in $z_0$ if and only if $x$ has a power series expansion \begin{equation}x(z)=\sum_{k=0}^\infty a_k (z-z_0)^k, \end{equation} where $a_k \in F,$ near $z_0.$

She mentioned that this is proven exactly as in standard complex analysis. In my complex analysis class however we proved the fact that every holomorphic function has a power series axpansion via the cauchy integral formula \begin{equation}f(z)=\frac{1}{2\pi i}\int_{\partial B_{z_0}(R)}\frac{f(w)}{w-z} dw. \end{equation} I, however, can't see how this could be generelised to Banach spaces. When I asked my professor how one could do that, she replied that it is possible to prove equivalence without the cauchy integral theorem. In her complex analysis script, however, she does it using the cauchy integral formula.

So my question is how to prove this equivalence in general Banach spaces? Is there some kind of cauchy formula too?

Thanks in advance


The Cauchy integral theorem generalizes to your setting. For the integral to make sense you can either use Bochner integration theory or convince yourself that the definition of line integrals that you know from complex analysis works equally well in this setting, if you ask your Riemann sums to converge in the norm of the Banach space.

Using any approach you find that $\langle \int_\Gamma g(w) dw \vert \phi\rangle = \int_\Gamma \langle g(w)\vert \phi\rangle dw $ for all $\phi \in F’$ where $\langle \cdot \vert \cdot \rangle$ is the pairing of $F$ with its dual.

Moreover it follows from your definition of complex differentiability that for a holomorphic function $f:\Omega \rightarrow F$ also $z \mapsto \langle f(z) \vert \phi \rangle$ is holomorphic (in the usual sense) for any $\phi \in E’$. (The converse is also true.)

Combining these two ideas you get from the usual Cauchy integral formula that

$$ \left \langle \int_\Gamma \frac{f(w)}{w-z} \frac{dw}{2\pi i} - f(z) \vert \phi \right \rangle = \int_\Gamma \frac{\langle f(w) \vert \phi \rangle}{w-z} \frac{dw}{2\pi i} - \langle f(z) \vert \phi \rangle = 0 $$ This is true for all $\phi\in E’$ and thus formula $$ \int_\Gamma \frac{f(w)}{w-z} \frac{dw}{2\pi i} = f(z) $$ follows from Hahn-Banach.

Also, here seems to be a proof of the power series expansion without integration theory.


Yes, we have Cauchy's integral formula too for $F$-valued holomorphic functions. Rudin treats the basics of the theory in chapter 3 of his Functional Analysis. He uses the Pettis integral, which is quite convenient in this area, but the formula also holds if one uses any other sensible integral for continuous $F$-valued functions, like the Bochner integral.

In Rudin's treatment, a function $x \colon \Omega \to F$ for which the limit of the difference quotients exists is called strongly holomorphic, and a function $y \colon \Omega \to F$ such that $\lambda \circ y \colon \Omega \to \mathbb{C}$ is holomorphic for every $\lambda \in F^{\ast}$ is called weakly holomorphic. It is then easy to see that every strongly holomorphic function is weakly holomorphic, and that the Cauchy integral formula holds for weakly holomorphic functions. Using the Pettis integral that is immediate, since

$$\lambda\bigl(y(z_0)\bigr) = \frac{1}{2\pi i} \int_{\gamma} \frac{\lambda\bigl(y(z)\bigr)}{z-z_0}\,dz$$

then by definition of the integral means

$$\frac{1}{2\pi i}\int_{\gamma} \frac{y(z)}{z-z_0}\,dz.$$

For the Bochner integral (or various other integrals for vector space-valued functions) one needs to prove that integration commutes with application of continuous linear functionals (or more generally continuous linear maps to some normed space). That's not too difficult either.

Then, from the integral formula, one obtains the power series expansion in essentially the same way as for $\mathbb{C}$-valued holomorphic functions. And the power series representation shows that every weakly holomorphic function is in fact strongly holomorphic.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.