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 $U$?

Solutions Collecting From Web of "Smoothness of harmonic functions"

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.)

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 $C^\infty$.