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Let , $f:\Bbb C \to \Bbb C$ be an entire function and $\displaystyle \int_{\Bbb R^2}|f(x+iy)|\,dx\,dy<\infty$. Then prove that $f(z)=0$ for all $z\in \Bbb C$.

Put $x=r\cos \theta$ and $y=r\sin \theta$. Then , integration becomes $\displaystyle \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi}|f(re^{i\theta})|r\,d\theta\,dr<\infty$. But from here I'm unable to conclude nothing.! How I proceed? Any hint please.

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    Louisville's theorem? – 2017-01-23
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    How you apply Liouville's theorem ? – 2017-01-23
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    @SteveD What is "Louisville's Theorem?" There is a [Liouville's Theorem](https://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)) to which you are likely referring. – 2017-01-23
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    It's not immediately clear that $f$ is necessarily bounded -- in real calculus, there are unbounded functions that belong to $L^p$ for all $p$ finite but not in $L^\infty$ – 2017-01-23
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    Also, I believe you have the order of integration incorrect ($r$ does not range between $0$ and $2\pi$) – 2017-01-23
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    @Nitin I don't mean that the limit of $r$. I just cleared it. – 2017-01-23
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    @nitin: this function is smooth. – 2017-01-23
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    @drmv: yes, clearly I meant that. Phone autocorrected it. – 2017-01-23
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    @SteveD what does smoothness have to do with anything? OP was integrating $\theta$ between 0 and $\infty$ and $r$ between $0$ and $2\pi$. Doesn't matter, it's been fixed. – 2017-01-23

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Hint: Use the estimate from Cauchy's integral formula $$ \lvert f^{(n)}(0) \rvert \le \frac{n!}{2 \pi r^n} \int_0^{2\pi} \lvert f(re^{i\theta}) \rvert \, d\theta $$ to show that $$ \int_0^{\infty}\int_0^{2\pi}|f(re^{i\theta})|r\,d\theta \, dr $$ cannot be finite unless $f^{(n)}(0) = 0$ for all $n \ge 0$.

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    I'm not getting..please extend something. – 2017-02-04
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    All too easy! (+1) – 2017-02-09
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    @S717717: $\int_0^R\int_0^{2\pi}|f(re^{i\theta})|r\,d\theta \, dr \ge \int_0^R \frac{2\pi}{n!}\lvert f^{(n)}(0) \rvert r^{n+1} \, dr = ...$ – 2017-02-09