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I am learning PDE on myself as a beginner. It takes me like several hours to finally think out this proof. However, I feel something not right about my proof, especially choosing "$R$" part, it looks not sufficient at all.

My attempt:

$w(x)=u(x)-M\ge0$. Because $w(x)$ is bounded below by $0$, so we can find a point $x_0$ such that $w(x_0)=\min(w(x))\ge 0$. Suppose $w(x)$ is not bounded above, then for $N>10w(x_0)+10$, exists some $x_1$ such that $x_1 \ge N$. Consider the ball $B_R(x_0)$, where $R=\operatorname{dist}(x_1,x_0)$. Since $w(x)$ is harmonic, it satisfies the Mean Value equality. But $x_0$ is the minimum of $w(x)$ and $w(x)\ge 0$ and $N>10w(x_0)+10$, so it cannot satisfies the Mean Value equality on the ball $B_R(x_0)$. So $w(x)$ is bounded both below and above and harmonic on $R^n$. By Liouville's theorem $w(x)$ is constant, so $u(x)$ is constant.

Can anyone help me look at this? It is really important for me. Thanks so much!

  • 1
    $\begingroup$ I do not see why $x_0$ exists; that is, why should the function $w$ necessarily have a point $x_0$ that realizes the minimum over all of $\mathbb{R}^n$. $\endgroup$ – msteve Oct 7 '14 at 3:49
  • $\begingroup$ Agreeing with steve, I'll add that if you had $x_0$, the strong maximum principle does the job already. Do you know Harnack's inequality? This is the tool to use here. By the way, I edited the post to tone it down a bit. Mathematical discourse and shouting rarely go together well. $\endgroup$ – user147263 Oct 7 '14 at 3:55

Thm: Let $u \ge 0$ be harmonic, let $R > 0$, and assume $u \in C^1(\overline{B(x_0,R)})$. Then $$|\nabla u(x_0)| \le \frac{N}{R}u(x_0).$$

Proof: By assumption $u$ is harmonic, then $\Delta u = 0$. Differentiating both sides we get $0 = \frac{\partial \Delta u}{\partial x_i} = \Delta(\frac{\partial u}{\partial x_i})$, then $\frac{\partial u}{\partial x_i}$ is harmonic for $i = 1, \dots , N$. In particular the partial derivatives of $u$ satisfy the mean value property. This, together with the divergence theorem, gives the following computation $$\frac{\partial u}{\partial x_i}(x_0) = \frac{1}{\lambda(B(x_0,R))}\int_{B(x_0,R)}\frac{\partial u}{\partial x_i}(x)\, dx = \frac{1}{\lambda(B(x_0,R))}\int_{\partial B(x_0,R)}u(x)v_i\, dS(x)$$ Here $v_i(x)$ is the $i$-th component of the normal unit vector at $x$ pointing outward and $S$ is the Hausdorff measure. Take the absolute value in the previous equality and estimate as follows (recall also that $u$ is nonnegative by assumption!): $$\Big|\frac{\partial u}{\partial x_i}(x_0)\Big| \le \frac{1}{\lambda(B(x_0,R))}\int_{\partial B(x_0,R)}|u(x)|\, dS(x) = \frac{1}{\lambda(B(x_0,R))}\int_{\partial B(x_0,R)}u(x)\, dS(x).$$ One last application of the Mean Value Property shows $$\Big|\frac{\partial u}{\partial x_i}(x_0)\Big| \le \frac{N}{R\lambda(\partial B(x_0,R))}\int_{\partial B(x_0,R)}u(x)\, dS(x) = \frac{N}{R}u(x_0).$$ This concludes the proof of the theorem.

Finally, to get the desired result notice that if $u$ is harmonic in $\mathbb{R}^N$ we are allowed to send $R \to \infty$. This shows that every partial derivative equals zero, from which we deduce that $u$ is constant.


  • $\begingroup$ Thanks for your kind answer! I don't understand why $|\nabla u(x_0)| \le \frac{N}{R}\max_{\overline{B(x_0,R)}}{|u(x)|}$ goes to $|\nabla u(x_0)| \le \frac{N}{R}$$u(x_0)$? Can you explain it a bit for me? Thanks! $\endgroup$ – Tony Oct 7 '14 at 4:32
  • $\begingroup$ Yeah, I realized that I forgot to mention the crucial idea for proving this, which is: u harmonic implies that the derivatives are harmonic! I am sorry about that! I am editing my post with the details! $\endgroup$ – user67133 Oct 7 '14 at 5:04
  • $\begingroup$ I am done editing, hopefully everything is clear now! :D $\endgroup$ – user67133 Oct 7 '14 at 5:30
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    $\begingroup$ Thanks a lot! Your answer is really clear and helpful! Thanks for your time and effort!! $\endgroup$ – Tony Oct 7 '14 at 6:15

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