0
$\begingroup$

2.5: Let u be harmonic and nonnegative, show that u is constant. (Hint use the previous exercise).

The previous exercise was posted in another question, stated the following.

2.4: Let $u:B(0,R)\subset \mathbb{R^d}\rightarrow\mathbb{R}$ be harmonic and nonnegative. Prove the following version of the Harnack inequality: $$\dfrac{R^{d-2}(R-|x|)}{(R+|x|)^{d-1}}u(0)\leq u(x)\leq \dfrac{R^{d-2}(R+|x|)}{(R-|x|)^{d-1}}u(0)$$

I used the poisson's integral for the ball that states the following.

$$u(x)=\dfrac{R^2-|x|}{n\alpha(n)R}\int_{\partial B(0,R)}\dfrac{g(y)}{|x-y|^n}dS(y)$$ and the fact that for $y\in \partial B(0,R)$, $|y|=R$ and $|x|-|y|\leq |x-y|\leq |x|+|y|$ that is $|x|-R\leq |x-y|\leq |x|+R$.

Any hints on how to apply 2.4 to 2.5? thanks in advance.

$\endgroup$
1
1
$\begingroup$

Just let $R\to\infty$. Both sides of the inequality converge to $u(0)$.

$\endgroup$

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.