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Let $\Omega=B_1(0)$ and $u\in C(\Omega, R^n) \cap C^2(\Omega, R^n)$ be a vector valued map into the unit ball ( ie. $|u(x)|\le 1$ for all $x\in \Omega $, such that $$|\triangle u(x)| \le |\triangledown u(x)|^2$$ for all $x\in \Omega$

How can i show that $v:= |u^2| $ is a subharmonic and conclude that $sup_\Omega|u| \le sup_{\partial \Omega } |u| $ .

Thank you for your help .

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I guess that $\Omega$ is a subset of some $\mathbb{R}^m$, which makes $\nabla u(x)$ a matrix, so what do you mean by $|\nabla u(x)|^2$? –  Mercy Jun 16 '12 at 19:23
    
@Deltapsi I understood it to mean the same thing as for a vector: sum of squares of entries. (I.e., the square of the Frobenius / Hilbert-Schmidt norm.) This is what the proof required anyway. –  user31373 Jun 16 '12 at 19:57

1 Answer 1

up vote 1 down vote accepted

Write $u=(u_1,\dots,u_n)$. For each $k=1,\dots, n$ calculate $$\Delta (u_k^2)=\mathrm{div}\nabla (u_k^2) = \mathrm{div} (2u_k\nabla u_k)=2u_k\Delta u_k+2|\nabla u_k|^2$$ Hence, $\Delta |u|^2 = 2\sum_{k}u_k\Delta u_k+2|\nabla u|^2$. We like the second term, because it's nonnegative. What to do with the first? Use Cauchy-Schwarz, of course: $\left|\sum_{k}u_k\Delta u_k\right|\le |u||\Delta u|\le |\Delta u|\le |\nabla u|^2$. Thus $\Delta |u|^2\ge 0$.

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How do you prove that $\sum_ku_k\Delta u_k \ge 0$? –  Mercy Jun 16 '12 at 20:16
    
@Deltapsi I don't. –  user31373 Jun 16 '12 at 20:22
    
So why do you say that $\Delta|u|^2 \ge 0$? –  Mercy Jun 16 '12 at 20:45
    
@Deltapsi If $a$ and $b$ are real numbers such that $|a|\le b$, then $a+b\ge 0$. –  user31373 Jun 16 '12 at 20:59
    
I get it now, thanks! –  Mercy Jun 16 '12 at 21:17

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