Question About Harmonic Function Theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:44:28Z http://mathoverflow.net/feeds/question/101564 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101564/question-about-harmonic-function-theory Question About Harmonic Function Theory jason mfash 2012-07-07T08:42:46Z 2012-08-06T20:52:57Z <p>Given a non-negative function $u$ defined on $\mathbb{R}^2$ , and satisfies : $\Delta u \leq 0$ . </p> <p>How can I prove that $u$ must be constant? Is there an easy way to do it ? </p> <p>Thanks ! </p> http://mathoverflow.net/questions/101564/question-about-harmonic-function-theory/101568#101568 Answer by Mateusz Wasilewski for Question About Harmonic Function Theory Mateusz Wasilewski 2012-07-07T09:57:59Z 2012-07-07T09:57:59Z <p>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.</p> http://mathoverflow.net/questions/101564/question-about-harmonic-function-theory/101572#101572 Answer by quid for Question About Harmonic Function Theory quid 2012-07-07T12:31:37Z 2012-07-07T12:31:37Z <p>The function is <a href="http://en.wikipedia.org/wiki/Superharmonic_function" rel="nofollow">superharmonic</a> (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. </p> <p>Some more details:</p> <p>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 <a href="http://www.dm.unibo.it/~arcozzi/subharmonic.pdf" rel="nofollow">some lecture notes</a> 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).</p> http://mathoverflow.net/questions/101564/question-about-harmonic-function-theory/104146#104146 Answer by Connor Mooney for Question About Harmonic Function Theory Connor Mooney 2012-08-06T20:52:57Z 2012-08-06T20:52:57Z <p>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.</p> <p>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, $\epsilon f(x) &lt; u(x)$ 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!</p> <p>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.</p>