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Problem of the Week #236 - Dec 06, 2016

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Show that if $f$ is an entire function such that $\int_{-\infty}^\infty \int_{-\infty}^\infty \lvert f(x + yi)\rvert^2\, dx\, dy < \infty$, then $f$ is identically zero.


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
This week's problem was solved correctly by Kokuhaku . You can read his solution below.



Note that $\displaystyle \int_{-\infty}^\infty \int_{-\infty}^\infty \lvert f(x + yi)\rvert^2\, dx\, dy = \int_0^\infty \int_0^{2\pi} \lvert f(re^{i\theta}) \rvert^2 r\, d\theta\,dr$ by using polar coordinates. Also, since $f$ is entire function we have $f(z)=\sum_{n=0}^\infty a_n z^n$, where $\displaystyle a_n = \frac{1}{2\pi i} \int_{\lvert z \rvert=r} \frac{f(z)}{z^{n+1}}dz = \frac{1}{2\pi} \int_0^{2\pi} \frac{f(r e^{i \theta})}{r^n} d\theta$ for every $r >0$ and $n \in \mathbb{N}_{\geqslant 0}$. We will prove that $a_n=0$ for $n \in \mathbb{N}_{\geqslant 0}$.

Using representation of $a_n$, we have $\displaystyle \lvert a_n \rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert f(re^{i \theta})\rvert}{r^n} d\theta$. Now, from Cauchy-Schwarz inequality for integrals we obtain $\displaystyle \lvert a_n \rvert ^2 \leqslant \frac{1}{2\pi r^{2n}} \int_0^{2\pi} \lvert f(re^{i \theta})\rvert^2 d\theta$, that is $\displaystyle \lvert a_n \rvert^2 r^{2n+1} \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert f(re^{i \theta})\rvert^2 r \,d\theta$. Now, integrating by $\displaystyle \int_0^\infty dr$ we find $\displaystyle \lvert a_n \rvert^2 \int_0^\infty r^{2n+1} \, dr \leqslant \frac{1}{2\pi} \int_0^\infty \int_0^{2\pi} \lvert f(re^{i \theta})\rvert^2 r \,d\theta \,dr$. Since RHS of last inequality is finite and $\displaystyle \int_0^\infty r^{2n+1} \, dr$ diverges, we must have $a_n=0$ for every $n \in \mathbb{N}_{\geqslant 0}$, from which we have $f(z)=0$ for all $z \in \mathbb{C}$.
 
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