Given a nonnegative 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 !
Given a nonnegative 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 ! 


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 twodimensional 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. 


The function is superharmonic (due to the condition on Laplacian, note that subharmonic is wider spread but this is just a signchange) 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). 


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 $f(x) = \logx$, 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 $x^{2n}$ and cap it off above in $B_1$ by a paraboloid and smooth it out. 

