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Is it true that a positive harmonic function on $\mathbb{R}^n$ must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness.

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Since it is not clear whether the Wikipedia proof uses boundedness or not, please allow me to give a detailed proof that only uses nonnegativity.

Let $u$ be a nonnegative harmonic function in $\mathbb{R}^n$, and let $x,y\in\mathbb{R}^n$. Denote by $B_R(y)$ the ball of radius $R>0$ centred at $y$, and similarly by $B_r(x)$ the ball of radius $r>0$ centred at $x$. We consider such balls with $R>0$ large and $r=R-|x-y|$, that is, $B_r(x)\subset B_R(y)$ and the boundary of $B_r(x)$ is tangent to the boundary of $B_R(y)$. Then we have $$ u(x) = \frac1{|B_r|}\int_{B_r(x)}u \leq \frac1{|B_r|}\int_{B_R(y)}u = \frac{|B_R|}{|B_r|}u(y), $$ where we have used the mean value property in the first and last equalities, and the nonnegativity of $u$ in the second inequality. Now we send $R\to\infty$ and get $$ u(x) \leq u(y). $$ As $x$ and $y$ were arbitrary, we conclude that $u$ is constant.

Note that the Harnack inequality is lurking behind this proof. Also, it was crucial to consider balls with different radii in order to make use of nonnegativity. The Wikipedia proof uses balls of equal radii.

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  • $\begingroup$ Sorry, I'm confused, so we have that $|B_{R}| \geq |B_{r}|$, but how do we get $u(y) \geq u(x)$ from the inequality $u(x) \leq \frac{|B_{R}|}{|B_{r}|}u(y)$? $\endgroup$
    – user135520
    Commented Feb 26, 2018 at 21:41
  • $\begingroup$ Oh, I see, maybe one can think of $|B_{R}|$ and $|B_{r}|$ being the same size as $R \to \infty$. $\endgroup$
    – user135520
    Commented Feb 26, 2018 at 21:43
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    $\begingroup$ @user135520 Note that $r$ is tied to $R$, so when $R\to\infty$, so does $r$. $\endgroup$
    – timur
    Commented Feb 28, 2018 at 1:03
  • $\begingroup$ In the right side of the inequality, why do we have $\frac{1}{|B_r|}$, as opposed to $\frac{1}{|B_R|}$. Also, how do we know that the first inequality holds? Are we using that $r<R$? $\endgroup$
    – Chubwagon
    Commented Apr 19, 2018 at 16:06
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    $\begingroup$ @Chubwagon: The middle inequality is due to monotonicity of the integral, as the integrand is nonnegative. $\endgroup$
    – timur
    Commented Apr 20, 2018 at 2:05
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Taking $n=2$ just to simplify:

Note that, given $y\in\mathbb{R}^2$, let $r=||y||$. Then, $\forall n\in\mathbb{N}$: $$B(y,n)\subset B(0,2n)$$ Thus $$\int_{B(y,n)}u(z)dz\leq\int_{B(0,n+r)}u(z)dz$$ $$u(y) = \frac{1}{\pi n^2}\int_{B(y,n)}u(z)dz\leq\frac{(n+r)^2}{n^2}\frac{1}{\pi(n+r)^2}\int_{B(0,n+r)}u(z)dz=\left(\frac{n+r}{n}\right)^2u(0)$$ That is $$u(y)\leq\left(\frac{n+r}{n}\right)^2u(0),\qquad\forall n \in\mathbb{N}$$ Takin $lim_n\rightarrow \infty$, we have that $u(y)\leq u(0)$. But since $y\in\mathbb{R}^2$ was arbitrary, we conclude that $u(z)\leq u(0),\forall z\in\mathbb{R}^2$.
But then, given $R>0$, we have that $\max_{z\in\overline{B(0,R)}}u(z) \leq u(0)$, thus, $\max_{z\in\overline{B(0,R)}}u(z) = u(0)$. Since $0$ is an interior point, by the strong maximum principle, $u$ is constant equal to $u(0)$ in the ball. Since $R$ was arbitrary, it follows that $u(z) = u(0),\forall z\in\mathbb{R}^2$.

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One can use Louisville's Theorem for harmonic functions to prove this. Note that if $u$ is a harmonic function then so is $e^{-u}$ and if $u$ is bounded below, then $e^{-u}$ is bounded and thus constant. This then forces $u$ to be constant. This shows the slightly stronger fact that if $u$ is a harmonic function bounded below, then $u$ is constant.

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  • $\begingroup$ Why is $e^-u$ harmonic? If I take the Laplacian of $e^{-u}$ I get $- e^{-u} \triangle u + e^{-u} |\nabla u|^2 $. $\endgroup$ Commented Feb 11, 2020 at 2:40
  • $\begingroup$ You're right. I don't know what I was thinking. See math.stackexchange.com/questions/2441878/… for a proof that the only time $e^f$ is harmonic for $f$ harmonic is when $f$ is constant. $\endgroup$ Commented Feb 12, 2020 at 16:57

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