Smoothness and analyticity of harmonic functions

Remarks

Proof idea for part 1: by the mean value property, $u$ equals its convolution with a smooth radial mollifier away from the boundary of $U$. A convolution with a smooth function is smooth. Since $\Delta$ commutes with partial derivatives, $\Delta(\partial^\alpha u) = \partial^\alpha(\Delta u) = 0$. Proof idea for part 2: the gradient estimates give $\abs{\partial^\alpha u(a)} \le C^{k+1} k^k \sup_{B(a, r)} \abs{u} / r^{k}$ for $\abs{\alpha} = k$, with $C$ depending only on $n$. This growth makes the Taylor remainder tend to $0$ on a small ball. Example: $u(x, y) = e^x \cos y$ is harmonic on $\R^2$, and its Taylor series at $0$ converges to it everywhere. Harmonicity is needed and $C^2$ is not enough: $\abs{x}^3$ on $\R$ is twice continuously differentiable but not three times differentiable at $0$. In contrast, a merely continuous function with the mean value property is already harmonic, hence analytic.

Going deeper

For a student

This is the simplest case of elliptic regularity: solutions of elliptic equations with smooth or analytic coefficients are as regular as the coefficients allow. Analyticity implies a unique continuation property: a harmonic function on a connected open set that vanishes on a nonempty open subset vanishes everywhere. The heat equation shares smoothness but not analyticity in time, and the wave equation has neither, so this property marks the elliptic case.