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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    $\begingroup$ 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. This should work at least if $B$ is a Banach algebra. $\endgroup$
    – t.b.
    Commented Jun 12, 2012 at 4:02
  • $\begingroup$ Where does the usual argument using Cauchy's estimates fail? $\endgroup$
    – Jose27
    Commented Jun 12, 2012 at 4:55
  • $\begingroup$ 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. $\endgroup$
    – user16299
    Commented Jun 12, 2012 at 5:07
  • $\begingroup$ @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. $\endgroup$ Commented Jun 12, 2012 at 5:10
  • $\begingroup$ @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-Banach $\endgroup$
    – user16299
    Commented Jun 12, 2012 at 5:10

1 Answer 1


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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    $\begingroup$ Full details seem to be in Dieudonne's Foundations of Modern Analysis. He works with Banach spaces throughout. $\endgroup$ Commented Jun 12, 2012 at 6:24
  • $\begingroup$ 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... $\endgroup$ Commented Jun 12, 2012 at 6:38
  • $\begingroup$ @Qiaochu: Just use the standard Riemann sum method for the integrals. Nothing advanced about Banach spaces is required beyond the definition. $\endgroup$ Commented Jun 12, 2012 at 7:34
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    $\begingroup$ 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. $\endgroup$ Commented Jun 12, 2012 at 7:38
  • $\begingroup$ @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.) $\endgroup$ Commented Jun 12, 2012 at 17:08

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