# How do you show that an $L^p$ entire (holomorphic on the complex plane) function is $0$?

Just to clarify, I want to show that:

If $f$ is entire and $\int_{\mathbb{C}} |f|^p dxdy <\infty$, then $f=0$.

I think I can show that this is the case for $p=2$, but I'm not sure about other values of $p$...

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Right! Thank you, I didn't think about using Hölder's.

In short,

$$|f(0)| = \left|\int_{|z|=R} \frac{f(z)dz}{z} \right| < \left(\int_{|z|=R} |f(z)|^p |dz|\right)^{1/p} \left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q}$$

And $\left(\int_{|z|=R} |z|^{-q} |dz|\right)^{1/q} = R^{-1+1/q} (2\pi)^{1/q} = R^{-1/p} (2\pi)^{1/q}$

$$\int_0^\infty \left((2\pi)^{1/q} |f(0)| R^{1/p}\right)^p dR < || f ||_p^p <\infty$$

And the only way for this to fly is for $f(0)=0$, for $p\le 1$.

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$\newcommand\RR{\mathbb{R}}$ $\newcommand\NN{\mathbb{N}}$ $\newcommand\e{\varepsilon}$ $\newcommand\lbg{\lambda}$

If $f: \RR \to \RR$ is an analytic unbounded Lebesgue integrable function, then it must in particular satisfy this property (because it is continuous): $\forall \e,M > 0 \; \exists x_0 : |x| > |x_0| \Rightarrow \lbg(\{x \in \RR : |f(x)| > M\}) < \e,$ where $\lbg$ is the usual Lebesgue measure.

Then you just use that $f$ is analytic iff $\forall \text{ compact } K \subset \RR \; \exists C > 0 : x \in K, n \in \NN \Rightarrow \left\vert\frac{\partial^n f}{\partial x^n}(x)\right\vert \leq C^{n+1}n!$

But it is easy to see that this is impossible since $|f'|$ will have to be arbitrarily large - or in other words that you can go far enough out to the right on the x-axis that $|f|$ has to go from being larger than some big $M$ to smaller than some small $m$ on a very small interval.

This is missing quite a few details, but it should be pretty easy to fill out.

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Use Hölder's inequality and Cauchy's integral formula to show the function and its derivatives all vanishes at zero.

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+1, but what about 0<p<1? – Jonas Meyer Oct 8 '10 at 19:49