A harmonic global Sobolev distribution vanishes

Statement

If $h\in H^s(\mathbb R^n)$ for some real $s$ and $\Delta h=0$ in distributions, then $h=0$ . This uses the whole-space inhomogeneous Sobolev space, including its condition at zero frequency.

Fourier proof

Fourier transformation gives $|\xi|^2\widehat h=0$ . On any open set disjoint from zero one can divide a test function by $|\xi|^2$ , proving that $\widehat h$ is supported at $\{0\}$ . But $\widehat h$ is represented by a locally square-integrable function. Such a function supported on the measure-zero set $\{0\}$ vanishes almost everywhere. Fourier inversion gives $h=0$ .

Why the hypothesis matters

Nonzero constants are harmonic tempered distributions, yet belong to no global inhomogeneous $H^s$ . One cannot replace the stated Sobolev membership by temperateness or by local Sobolev membership.