Application of maximum princple for harmonic function

Question

Suppose u is a non constant harmonic function in $\mathbb{R}^n$, how can I show that the functions $F(r) = \sup_{z \in B(0,r)}|u(z)|$ and $G(r)= \int_{z \in B(0,1)}|u|^2(rz)dS_z$ are both strictly increasing functions? I know I have to apply the maximum principle but I don't know how to start. Any hint will be appreciated. For F I wanted to use the fact that the absolute value function is convex hence|u| will be subharmonic and then using the maximum principle for the subharmonic function the desired result will follow but I'm not too sure of that.

Answer 1

Note that $F$ is an increasing function for all functions $u$ regardless of whether they are harmonic. Let us now show that it is strictly increasing. This essentially follows from the following corollary of the strong maximum principle.

Corollary: Suppose that $\Omega$ is a domain in $\mathbb R^n$ and $u$ is harmonic in $\Omega$. If there exists $y \in \Omega$ such that $\vert u(y) \vert= \sup_\Omega \vert u \vert$ then $u$ is a constant in $\Omega$.

Proof: At $y$ the function $u$ either attains a maximum or a minimum. In either case the strong maximum principle implies $u$ is a constant in $\Omega$. $\square$

Suppose that $s<t$. Let $x \in \overline{B_s}$ be such that $\vert u(x) \vert =F(s)$. If $F(s)=F(t)$ then $\vert u \vert $ attains an interior maximum at $x$. Hence, the corollary implies $u$ is a constant in $B_t$. Since $u$ is harmonic in $\mathbb R^n$ and harmonic functions are analytic, if this occurs then $u$ is constant in $\mathbb R^n$, contradicting your assumption. Thus, $F(s)<F(t)$.

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For $G(r)$, I think you mean $$G(r) = \int_{\partial B_1} (u(rz))^2 \, d S_y $$ not$$G(r) = \int_{ B_1} (u(rz))^2 \, d S_y $$ (the second one doesn't make sense). Taking the derivative of $G$ w.r.t $r$ and making the change of coordinates $y=rz$, we have \begin{align*} G'(r) &= 2 \int_{\partial B_1} u(rz) Du(rz) \cdot z \, dS_z \\ &= 2r^{-n-1} \int_{\partial B_r} u(y) Du(y) \cdot \frac{y}r \, d S_y \\ &= 2r^{-n-1} \int_{\partial B_r}u(y) \frac{\partial u}{\partial \nu} \, d S_y \\ &= 2r^{-n-1}\int_{ B_r} \vert Du(y) \vert^2 \, dy \end{align*} were the last line follows from Green's identity and that $u$ is harmonic. Hence, $G'(r) \geqslant 0$ with equality if and only if $u$ is constant in $B_r$. As I mentioned earlier, if $u$ is constant in $B_r$ then $u$ is constant in $\mathbb R^n$. Hence, $G'(r)>0$ for all $r>0$, so $G$ is strictly increasing.