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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 $U$?

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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 2 '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. –  Ilya Aug 2 '11 at 12:08
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 2 '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. –  Ilya Aug 2 '11 at 12:19
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 2 '11 at 12:48

3 Answers 3

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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.)

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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. –  Byron Schmuland Oct 4 '11 at 23:17

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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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 $C^\infty$.

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Do you mean locally the real part of a holomorphic function? –  Jesse Madnick Oct 4 '11 at 22:29
Without loss of generality $u$ is the real part of a holomorphic function since being $C^\infty$ is a local property. Does that make you feel better? –  Matt Oct 15 '11 at 23:10
Very much so, thanks. –  Jesse Madnick Oct 16 '11 at 0:39

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