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This post Let $u$ harmonic. Then $\int_{\mathbb R^d}|u|^2<\infty \implies u=0$ answers this question in the case when $p=2$ using the Cauchy-Schwarz Inequality and an application of the Mean-Value Property. Since Cauchy-Schwarz only works for $p=2$, I was wondering if there is a generalization for $p \geq 2$. I tried it using Holder's Inequality, but I was not able to make that work.

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  • $\begingroup$ Holder would seem to give it directly for $1 < p < \infty.$ Where did you have a problem applying Holder to it? $\endgroup$ – Brian Moehring Nov 15 at 1:05
  • $\begingroup$ Here is my specific problem: When I try to apply Holder, I can't seem to get $|u|^p$ inside the integral by itself, so I can't apply the assumption $\int_{\mathbb{R}^n}|u|^p < \infty$. Since $||u||_1 \leq ||u||_p||u||_q$ by Holder, I always get two integrals --- $(\int_{\mathbb{R}^n} |u|^p)^{1/p}$ and $(\int_{\mathbb{R}^n} |u|^q)^{1/q}$ --- and I don't know how to work with the integral involving the conjugate exponent $q$. $\endgroup$ – phantom_dino Nov 15 at 1:21
  • $\begingroup$ The statement of Holder should be $\|fg\|_1 \leq \|f\|_p \|g\|_q.$ If you understand what was done in the Cauchy-Schwarz inequality in the post you linked, they set $$f = u, \qquad g = \mathbf{1}_{B(x,r)}$$ Do the same thing with Holder. $\endgroup$ – Brian Moehring Nov 15 at 1:26
  • $\begingroup$ Okay, I forgot to normalize $g=1_{B(x,r)}$, so $\frac{1}{|B(r,x)|^{1/q}}||1_{B(x,r)}||^{1/q}=1$ and the answer falls right out. Thanks for the help! $\endgroup$ – phantom_dino Nov 15 at 1:35
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Hint: Apply Hölder for $u$ and $1$ over $B(x,r)$.

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