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Let $B$ be a (complex) Banach space. A function $f : \mathbb{C} \to B$ is holomorphic if $\lim_{w \to z} \frac{f(w) - f(z)}{w - z}$ exists for all $z$, just as in the ordinary case where $B = \mathbb{C}$. Liouville's theorem for Banach spaces says that if $f$ is holomorphic and $|f|$ is bounded, then $f$ is constant.

The only way I know how to prove this uses the Hahn-Banach theorem: once we know that continuous linear functionals on $B$ separate points, we can apply the usual Liouville's theorem to $\lambda(f)$ for every such functional $\lambda : B \to \mathbb{C}$.

Can we avoid using Hahn-Banach? What if $B$ is in addition a Banach algebra?

Motivation: Liouville's theorem is useful in the elementary theory of Banach algebras, where it seems to me that we usually don't need the big theorems of Banach space theory (e.g. the closed graph theorem), and I would like to be able to develop this theory within ZF if possible. It would be very interesting if this were actually independent of ZF.

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    @Yemon: yes, I had Riemann or Bochner in mind and I'm quite certain that Hahn-Banach isn't needed anywhere important in Ryan's book linked to in my comment to Nate (a quick glance seems to indicate that it's only used in the proof of Pettis's measurability theorem).2012-06-12

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The usual argument (from Ahlfors) is to use the estimate $|f'(a)| \leq M/r$, where M is a bound for $|f|$ and $r$ is the radius of a large circle about $0$ containing $a$. This follows from Cauchy's integral formula. I believe there is no difficulty proving Cauchy's theorem and integral formula for Banach space valued functions using classical methods, since these just estimate absolute values (replace by norms) and then use completeness. First you need some integration, but the integral that is the limit of the integral on step maps suffices.

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    @Qiaochu: I just mentioned the Riemann integral as it is relatively simple to define, and I am very familiar with it. You could use the Bochner integral instead, which is a simple generalisation of the Lebesgue integral to Banach valued integrands. Whichever you are most familiar with really. However, if you are trying to completely avoid the Axiom of Choice, there are problems with using the Lebesgue and, by extension, the Bochner integral.2012-06-12