Smoothness of harmonic functions 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 $U$?
 A: 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.)
A: 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 $C^\infty$.
A: 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.
