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In the book on PDEs I'm reading there is a section on harmonic functions. To prove that these functions are in the class $C^\infty$ the author use standard mollifiers which I am not comfortable with. If there another proof of the $C^\infty(U)$ for the functions $u$ such that $\Delta u = 0$ on $U$?

SBF
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    But then you should take this as a motivation to learn about mollifiers! It's not that hard and extremely useful. – t.b. Aug 02 '11 at 12:06
  • @Theo: thanks for advise ) only if it is extremely useful. Are they used then to introduce the integration on manifolds? I remember there $C^\infty$ functions with compact supports. – SBF Aug 02 '11 at 12:08
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    Well, we're getting off-topic here: for integration on manifolds you use partitions of unity. However, concerning mollifiers: I told you that they are extremely useful and I mean it. In fact, I can't imagine that you'll get very far in your book on PDEs without learning about them at some point. – t.b. Aug 02 '11 at 12:14
  • @Theo: I thought that for the partitions of unity one uses $C^\infty_c$ function, doesn't it? Anyway, I will follow your advise - but still I am interested in the mollifier-free proof of smoothness for the harmonic functions. – SBF Aug 02 '11 at 12:19
  • I would second Theo B's comments... perhaps even to say that the technique of "mollifiers" is of greater general interest and utility than the specific smoothness of harmonic functions, which is just an example! :) – paul garrett Aug 02 '11 at 12:29
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    The other possible approach is via the Poisson kernel and uniqueness theorem/maximum principle (take a small disk and note that the harmonic function is the Poisson integral of its boundary values). – fedja Aug 02 '11 at 12:48
  • Mollifiers are cool. I try to use them as often as possible. Learn about them! – JT_NL Aug 02 '11 at 13:09
  • Yes, partitions of unity can be constructed with smooth functions of compact support. However, you don't convolve (convolution only makes sense in a context where you have some group structure around like on $\mathbb{R}^n$). And I didn't intend my comment to mean that it's not interesting to know whether there are proofs of smoothness of harmonic functions that avoid them. fedja's approach sounds promising. – t.b. Aug 02 '11 at 13:29

3 Answers3

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Suppose $u$ is harmonic in $U$ (that is, $u\in C^2(U)$ and $\Delta u = 0$ in $U$). Let $x$ be a point of $U$ and $B= B(x,r)$ the open ball centered at $x$ with radius $r>0$ so small that $\overline B\subset U$. Then $$ u(y) = \int_S P(y,z)\,\sigma(dz),\qquad y\in B, $$ where $S=S(x,r)$ is the boundary of $B$, $\sigma$ is the surface area measure on $S$, and $P(y,z)$ is the Poisson kernel for $B$: $$ P(y,z) = {r^2 - |y|^2\over rc_d|y-z|^2}, $$ $c_d$ being the surface area of the unit sphere in $R^d$. As the Poisson kernel is manifestly smooth in $y\in B$, the smoothness of $u$ follows from the above and standard theorems for differentiatng under an integral. The Poisson integral representation shown above can be proved using the Green/Stokes theorem. (See, for example, the first chapter of Doob's book on potential theory, or Helms' book on the same subject, or "Green, Brown, and Probability" by K.L. Chung.)

John Dawkins
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    The right hand side of the first equation does not depend on the function $u$. I guess the integral is missing a factor of $u(z)$ or so. –  Oct 04 '11 at 23:17
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In two dimensions you can do it like this: If $u$ is a harmonic function, then $u$ is the real part of a holomorphic function, which is differentiable infinitely many times. Therefore $u$ is also $C^\infty$.

Beni Bogosel
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In an answer I posted last month, I showed that the mean-value property is sufficient to show that harmonic functions are $C^\infty$ on the interior of their domains. I don't know if this makes you feel any more comfortable, but it might be worth a look.

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robjohn
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