Please explain Evans 's PDE Liouville 's Theorem

Question

Here is the proof :

d. Liouville's Theorem. We assert now that there are no nontrivial bounded harmonic functions on all of $\Bbb R^n$.

THEOREM 8 (Liouville's Theorem). Suppose $u:\Bbb R^n\to\Bbb R$ is harmonic and bounded. Then $u$ is constant.

Proof. Fix $x_0\in\Bbb R^n$, $r\gt0$, and apply Theorem 7 on $B(x_0,r)$: $$\eqalign{ |Du(x_0)|&\leq\dfrac{\sqrt nC_1}{r^{n+1}}\lVert u\rVert_{L^1(B(x_0,r))}\\&\leq\dfrac{\sqrt nC_1\alpha(n)}{r}\lVert u\rVert_{L^\infty(\Bbb R^n)}\to0,}$$ as $r\to\infty$. Thus $Du\equiv0$, and so $u$ is constant.

I don't know why he can change L1 with the ball $B(x_0,r)$ to L infinity with Rn

Answer 1

If $u \in L^1(\Omega) \cap L^\infty(\Omega)$, then

$$\| u \|_1 = \int_\Omega |u| dx \leq \int_\Omega \| u \|_\infty dx = |\Omega| \| u \|_\infty$$

where $|\Omega|$ is the volume of $\Omega$. Now compute the volume of the ball.

Answer 2

He hasn't just replaced it by the $\infty$ norm, but $\alpha(n) r^n$ (the volume of the unit ball) times the infinity norm.

Answer 3

Remark that \begin{align} \| u \|_{L^1(B(x_0,r))} &= \int_{B(x_0,r)} |u(x)| dx \\ &\leq \left( \sup_{x \in B(x_0,r)}|u(x)| \right) \int_{B(x_0,r)} dx\\ & = \| u \|_{L^{\infty}(B(x_0,r))} \textrm{vol}(B(x_0,r)) \\ &\leq \| u \|_{L^{\infty}(\mathbb{R^n})} \alpha(n)r^n. \end{align}