Is a holomorphic function on $\mathbb{C}$ that's $L^2$ integrable necessarily bounded?

Question

I'm working on a homework problem and if I can prove this claim then I am finished. Intuitively the answer should be yes, but I can't think about how I would attempt to rigorously prove this. I have an entire function $f:\mathbb{C}\rightarrow\mathbb{C}$ and I know that $|f|^2$ is integrable over $\mathbb{C}$. I want to say this implies that $f$ is bounded and hence the $0$ function by Liouville's theorem.

The idea is that if $f$ isn't bounded, then its modulus has to approach infinity as $|z|\rightarrow\infty$ for some particular direction. But this "direction" should be on a set that doesn't have measure $0$ and hence the integral of $|f|^2$ wouldn't be finite, which would contradict my claim.

Liouville theorem only implies that $f$ is constant not $0$. — C. Dubussy, Mar 01 '16 at 20:45
@C.Dubussy: But there are not a lot of functions which are constant and square-integrable. — Roland, Mar 01 '16 at 20:46

score 4 · Accepted Answer · answered Mar 01 '16 at 20:47

4

Hint: Write $f(z)=\sum_{n=0}^{\infty}a_nz^n.$ Compute the integral of $|f|^2$ over $D(0,R),$ using polar coordinates and Parseval's theorem.

answered Mar 01 '16 at 20:47

zhw.

105,693

Great, thanks!! – Jake Mar 01 '16 at 21:15

Is a holomorphic function on $\mathbb{C}$ that's $L^2$ integrable necessarily bounded?

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