Prove that the Square of a Harmonic Function is Subharmonic

Question

Let $u$ be harmonic. Prove that for a ball contained in the domain of $u$ we have that: $$ u^2(x) \leq \frac{1}{|\partial B_r|}\int_{\partial B_r}u(y)^2 dS(y) $$ Is there any way to prove this without first proving that the laplacian of $u$ is positive? Can we do it directly from the mean value property?

Answer 1

The solution boils down to the fact that because $u$ is harmonic:

$$ u(x) = \frac{1}{|\partial B_r|}\int_{\partial B_r} u(y) dS(y) $$ Hence: $$ u(x)^2 = \left(\frac{1}{|\partial B_r|}\int_{\partial B_r} u(y) dS(y)\right)\left(\frac{1}{|\partial B_r|}\int_{\partial B_r} u(y) dS(y)\right) = \frac{1}{|\partial B_r|^2}\left(\int_{\partial B_r} u(y) dS(y)\right)^2$$

Applying the cauchy schwarz integral inequality to the functiosn $u$, $1$, we have that: $$ \left(\int_{\partial B_r} u(y) dS(y)\right)^2 \leq |\partial B_r| \int_{\partial B_r} u(y)^2 dS(y) $$ And this proves the result.

Answer 2

Hint: By the mean value property, you have equality if $2$ is replaced by $1.$