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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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    I don't have time to think through the details (especially without going beyond ZF) but I would try to mimic [Nelson's argument for Liouville's theorem](http://dx.doi.org/10.1090/S0002-9939-1961-0259149-4). This should work at least if $B$ is a Banach algebra.2012-06-12
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    Where does the usual argument using Cauchy's estimates fail?2012-06-12
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    Slightly off-topic but I have some old course notes where I think I managed to prove non-emptiness of spectrum without using vector-valued integrals, specifically to avoid Hahn-Banach. If I forget to check back on MSE after a few days, feel free to drop me an email and I will take a closer look.2012-06-12
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    @t.b.: Nelson's argument is indeed nice, but in this context you'd need to integrate $B$-valued functions, and in my experience most such arguments bring in linear functionals.2012-06-12
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    @t.b. How are you defining averages over spheres/circles? I seem to remember it was this point which led me to try and bypass vector-valued integrals, I guess one could probably use Riemann sums as the integrands are well-behaved, but off the top of my head I'm not sure this avoids Hahn-Banach2012-06-12
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    @Nate: As I said, I haven't checked the details. I don't think functionals enter the *definition* the Riemann or the Bochner integral on such a simple domain as a circle or a disk and prove its basic properties, see e.g. [Ryan](http://books.google.com/books?id=7xRlVTVSNpQC&pg=PA25). Approximate $f$ uniformly by simple or step functions which you know how to integrate and show that the integral doesn't depend on the approximating sequence. There might be some arguments that require a bit more than mere ZF (countable choice, maybe), however I don't see where functionals would enter the picture.2012-06-12
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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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    Full details seem to be in Dieudonne's _Foundations of Modern Analysis_. He works with Banach spaces throughout.2012-06-12
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    Right. Somehow I had convinced myself that this wouldn't work without more effort. I guess I just need to learn an approach to Banach space-valued integrals that avoids Hahn-Banach...2012-06-12
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    @Qiaochu: Just use the standard Riemann sum method for the integrals. Nothing advanced about Banach spaces is required beyond the definition.2012-06-12
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    Everything that you would hope to hold for Banach spaces follows without any major modifications over the standard proofs. Integration, differentiation, every complex differentiable function is infinitely differentiable and expands as a power series, Cauchy's integral formula, etc.2012-06-12
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    @George: hmm. I guess Riemann sums ought to be fine for integrals of continuous Banach-valued functions from $S^1$, which I think is all that is needed here... (I admit that I am only familiar with the Darboux integral and the Lebesgue integral and am actually not completely sure how to prove things about the _Riemann_ integral, but it shouldn't be too bad, I guess.)2012-06-12
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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