Let $M$ be a Riemannian manifold and $S \subset M$ a compact submanifold of strictly lower dimension. Does every smooth function on $S$ extend to a harmonic function on a neighborhood of $S$?



Dirichlet-type condition on Riemannian manifold
This is not possible in general, for example because of constraints related to the analyticity of the extension.


For a concrete example, let $M = \mathbf{R}^2$, and $S \subset \mathbf{R}^2$ be a smooth, simple closed curve in the plane so that in the unit disc
\begin{equation}
S \cap D\_1 = \{ x\_2 = 0 \} \cap D\_1 = (-1,1) \times \{ 0 \}.
\end{equation}


On $S$ define a function $u\_0$ so that on the portion lying inside $D\_1$,
\begin{equation}
u\_0: x\_1 \in D\_1 \cap S \mapsto \mathrm{e}^{-1/x\_1^2}.
\end{equation}
This is not analytic, whereas a harmonic extension to a neighbourhood of $S$, say $u$ would be.


The existence of such a harmonic extension $u$ is therefore absurd. In conclusion: no matter how small $\delta > 0$ is chosen, there is no harmonic function $u: D\_\delta \to \mathbf{R}$ extending $u\_0$, let alone a function defined on a neighbourhood of $S$.


