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I need to solve this problem: Let $u$ be a harmonic function inside the open disk $K$ centered at the origin with radius $a$. We are also given that $\int_K u^2(x,y)dxdy=M<\infty.$ Show that $|u(x,y)|\leq \frac{1}{a-\sqrt{x^2+y^2}}\left(\frac{M}{\pi}\right)^{1/2}$ for any $(x,y)\in K$.

I tried to use the mean value property (average over area)for harmonic functions using a smaller ball of radius $r=a-\sqrt{x^2+y^2}$ centered at $(x,y)$ which would lie entirely inside $K$. That would lead me straight to the answer IF $u^2$ were harmonic, something that doesn't happen, so my reasoning is wrong... Haven't gotten much further yet. Any ideas?

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  • $\begingroup$ The Cauchy-Schwarz inequality? $\endgroup$
    – user103402
    Nov 10 '13 at 1:08
  • $\begingroup$ Ok, you say I have to use something like $|\int_K fg dx dy|\leq \sqrt{\int_K f^2 dx dy} \sqrt{\int_K g^2 dx dy}$. So setting f=1 and g=u will help $|\int_K u dx dy|\leq \sqrt{\int_K dx dy} \sqrt{\int_K u^2 dx dy}$... I got it! thanks $\endgroup$
    – arestes
    Nov 10 '13 at 1:34
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Simplified a bit (with Jensen's inequality instead of Cauchy-Schwarz): let $D$ be the disk of radius $r$ centered at $(x,y)$. The average value of $u$ on $D$ is $u(x,y)$. By Jensen's inequality, the average value of $u^2$ on $D$ is at least $u(x,y)^2$. On the other hand, since the total integral of $u^2$ is $M$, the average of $u^2$ over $D$ is at most $M/(\pi r^2)$. Hence $u(x,y)^2\le M/(\pi r^2)$, which was to be proved.

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