1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.