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 $U$?
Smoothness of harmonic functions
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0Yes, 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. – 2011-08-02
3 Answers
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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0The 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. – 2011-10-04
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.
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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0Very much so, thanks. – 2011-10-16