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Given a non-negative function $u $ defined on $\mathbb{R}^2 $ , and satisfies : $ \Delta u \leq 0 $ .

How can I prove that $u$ must be constant? Is there an easy way to do it ?

Thanks !

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This is wrong as this is satisfied by u(x,y) = x. –  Matthias Ludewig Jul 7 '12 at 8:55
@Kofi: your example is not non-negative. (Except perhaps if non-negative function should in this context have a particular meaning unknown to me. In that case, sorry for the noise.) –  quid Jul 7 '12 at 9:03
@Kofi: Your example is indeed wrong, since it is not non-negative... –  jason mfash Jul 7 '12 at 9:42

3 Answers 3

The function is superharmonic (due to the condition on Laplacian, note that subharmonic is wider spread but this is just a sign-change) and bounded below thus it is constant, by some analog of Liouville's theorem.

Some more details:

Things like this can, as commented by Mateusz Wasilewski, be found in certain complex analysis textbooks. (Though as said subharmonic and bounded above is I think a more common formulation, but this is just a sign change.) Here are some lecture notes that contain an essentially selfcontained exposition; see Theorem 8 and the remark following it (note that the definition of subharmonic is different and things are for the complex plane, but this is fine, compare the page linked above).

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Thanks a lot!!!! –  jason mfash Jul 7 '12 at 14:01
@jason mfash: you are welcome! –  quid Jul 8 '12 at 1:01

I don't know whether this counts (probably it doesn't) as an easy solution, but you can use Ito's lemma to conclude that $u(W_{t})$ (where $W_{t}$ denotes a two-dimensional Wiener process) is a nonnegative supermartingale, hence it converges almost surely. However, it is known that we can approach every point on a plane by a Wiener process, so $u$ must be constant because the limit is unique.

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@Wasilewki: It's probably a possible solution, but unfortunately I have no idea what Ito's lemma, Wiener process nor supermartingale are... Thanks anyway ! ( I was thinking about some PDE method or something) –  jason mfash Jul 7 '12 at 10:21
Apparently it is not true in higher dimensions, so I think that one must play with exact integral formulae for harmonic functions (maybe use elementary complex analysis). As I am far from an expert on that field (and in fact, on any branch of mathematics), I do not guarantee that I can provide an answer along these lines. By the way, my last name is Wasilewski. –  Mateusz Wasilewski Jul 7 '12 at 11:50

This is actually pretty cool. Superharmonic functions bounded below in $\mathbb{R}^2$ are constant, while there are nonconstant superharmonic functions bounded below in $\mathbb{R}^n$ for $n \geq 3$. Here is a proof that doesn't use complex analysis, and only uses that the fundamental solution in $\mathbb{R}^2$ ($\log(|x|)$) is unbounded from above and below, and the maximum principle.

Slide $u$ so that its minimum on $\partial B_1$ is $0$. Take the fundamental solution $f(x) = -\log|x|$, which is $0$ on $\partial B_1$. Since $u$ is bounded below and log is unbounded, $\epsilon f(x) < u(x)$ for $|x|$ sufficiently large (depending on $\epsilon$). By the maximum principle, $u(x) \geq \epsilon f(x)$ in $\mathbb{R}^2 - B_1$ for all $\epsilon$. Taking $\epsilon$ to $0$, we see that $u \geq 0$ outside $B_1$. But then, we see that $u$ takes its minimum in $\bar{B_1}$, and by the mean value inequality any superharmonic function with an interior minimum must be constant!

This result is false in higher dimensions. For a counterexample, just take the fundamental solution $|x|^{2-n}$ and cap it off above in $B_1$ by a paraboloid and smooth it out.

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What do you mean by "smooth it out" ? I understand that taking $\min(1,|x|^{2-n})$ resp. $\min(p(x),|x|^{2-n})$ with $p$ a suitable paraboloid can give a superharmonic function in the viscosity sense that is continuous resp. $\mathcal C^1$, but I don't see a way of having such a function that is $\mathcal C^2$. –  Laurent Oct 31 '13 at 0:48
Smooth $u$ with a positive mollifier. One sees via integration by parts ($u$ is $C^{1,1}$ on a smooth surface and smooth otherwise) that the mollification is also superharmonic. –  Connor Mooney Nov 30 '13 at 5:06

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