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In complex analysis, Liouville's theorem is that every bounded entire function is constant. To prove it, Cauchy intergral formula is used$$f(z) = \frac{1}{2\pi i}\int_C\frac{f(s)}{s-z}ds$$ where C is circle with radius of R, and $|f'(z)|\le \frac{M}{2\pi R}$ by using bounded condition($|f(z)|\le M)$ and Cauchy's differentiation formula. Then, $f'(z)$ goes to $0 $ as R goes to infinity. So, f(z) is constant. The same method can be used in proving Liouville's theorem for harmonic function?

In Partial differential equation, Liouville's theorem is that $u(\mathbf x)$ is constant if $u(\mathbf x)$ is harmonic function in $\Bbb R^N (N=2,3) $ and $u(\mathbf x)$ is bounded by $M>0$.

By mean value principle, $$u(\mathbf x) = \frac{1}{|\partial B_R(\mathbf x)|}\int_{\partial B_R(\mathbf x)}u(s)ds$$ Then, how can I differentiate $u(\mathbf x)$?

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  • $\begingroup$ If you're just looking for a proof of Liouville's theorem for harmonic functions, then just use the fact that over a simply-connected domain, every harmonic function is equal to the real part of a holomorphic function. $\endgroup$ – Ashvin Swaminathan May 26 '18 at 17:50
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    $\begingroup$ @AshvinSwaminathan, your idea only works for $\dim = 2$. $\endgroup$ – Martín-Blas Pérez Pinilla May 26 '18 at 17:53
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Yes the same method can be used. A Cauchy-type estimate on the derivative of a harmonic function holds, and the proof is very much like that for the holomorphic case.

Let $B$ denote the open unit ball in $\mathbb R ^n.$ Claim: There is a constant $C$ such that

$$|\nabla u (0)| \le C\sup_{\partial B} |u|$$

for any function harmonic on a neighborhood of $\overline B.$

Proof: This comes from differentiating through the integral sign in the Poisson integral representation of $u$ on the unit sphere. Doing this will give an appropriate bound on $|\partial u/\partial x_k (0)|,$ $k=1,\dots ,n.$ This proves the claim.

Now suppose $u$ is harmonic on $\mathbb R ^n$ and $|u|\le M$ everywhere. For any $r>0,$ we can set $u_r(x) = u(rx).$ Note each $u_r$ is harmonic on $\mathbb R^n$ and $|u_r|\le M.$ Using the above, we get

$$|\nabla u_r (0)| \le C\sup_{\partial B} |u_r|\le C\cdot M.$$

But $\nabla u_r (0) = r\nabla u (0).$ This shows $|\nabla u (0)| \le CM/r.$ Now let $r\to \infty$ to see $|\nabla u (0)| =0.$ Finally, apply this to any translate of $u$ to see $|\nabla u | =0$ everywhere. This implies $u$ is constant.

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Quote (the full paper!) of A Proof of Liouville's Theorem:

Consider a bounded harmonic function on Euclidean space. Since it is harmonic, its value at any point is its average over any sphere, and hence over any ball, with the point as center. Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since the function is bounded, the averages of it over the two balls are arbitrarily close, and so the function assumes the same value at any two points. Thus a bounded harmonic function on Euclidean space is a constant.
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