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Let $u:\mathbb R^d\longrightarrow \mathbb R$ a harmonic function. Show that,

1) $\int_{\mathbb R^d}|u|^2<\infty \implies u=0$

2) $\int_{\mathbb R^d}|\nabla u|^2<\infty \implies u$ constant.

My attempts

Let consider the ball $B(x,r)$. Then, $$|u(x)|=\left|\frac{1}{|B(x,r)|}\int_{B(x,r)}u(y) d y\right|\leq \frac{1}{|B(x,r)|}\int_{B(x,r)}|u|\underset{CS}{\leq}\frac{1}{\sqrt{|B(x,r)|}}\sqrt{\int_{B(x,r)}|u|^2}\leq \frac{C}{\sqrt{|B(x,r)|}}.$$ Letting $r\to \infty $, we get $|u(x)|=0$ and thus $u=0$.

Is it correct ? (In fact, $u\in L^1(\mathbb R^d)$ would be enough, no ?)

2) We have that $|\nabla u|$ is harmonic, and thus the claim follow for the previous exercise.

Is it correct ?

MSE
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  • I think I do recall that for harmonic $u$ the norm of the gradient is subharmonic, but I don't recall that it is harmonic. Are you sure? The reasoning in 1) looks ok, but, for domains of unbounded volume, $L^1$ is not a subset of $L^2 $ nor is $L^2 $ subset of $L^1$, so "$u \in L^1$ would be enough" may be true but is not to the point. – Thomas Dec 29 '16 at 17:29
  • @Thomas: Thank you for your very constructive answer. I'm not sure $|\nabla u|$ harmonic (I'm maybe wrong). I'll think about the fact that it's sub-harmonic, I'll probably conclude with this. – MSE Dec 29 '16 at 18:18
  • Don't worry about the harmonicity of the gradient (it isn't by the way). The condition implies each partial derivative of $u,$ which is harmonic, satisfies the condition in 1). – zhw. Dec 29 '16 at 20:46

1 Answers1

2

Your solution of (1) is correct. And you are right that $u\in L^1$ would lead to the same conclusion, but as Thomas noted, neither $L^1$ nor $L^2$ contains the other.

For (2), consider individual derivatives $u_{x_i}$ instead of the norm of the gradient. Since $\Delta u_{x_i} = (\Delta u)_{x_i}=0$, part 1) applies to $u_{x_i}$. Hence all the partial derivatives are zero.