Question About Harmonic Function Theory

Question

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 !

Answer 1

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).

Answer 2

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.

Answer 3

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.