What can you say about f if g is harmonic?

Question

Suppose that f : R → R is such that, whenever g : $R^n$→ R is harmonic, so is f(g(x)). What can you say about f?

This is my attempt , and I think f is a linear function.

Answer 1

Your proof is essentially correct*. It boils down to the chain rule for $\Delta g = \operatorname{div} \nabla g$, namely $$\nabla (f\circ g) = (f'\circ g)\nabla g$$ and $$\begin{split} \operatorname{div} ((f'\circ g)\nabla g) &= \nabla (f'\circ g)\cdot \nabla g+(f'\circ g)\operatorname{div} \nabla g \\ & = (f''\circ g) |\nabla g|^2 + (f'\circ g) \Delta g \end{split}$$ By the assumption, $(f''\circ g) |\nabla g|^2 \equiv 0$ for every harmonic function $g$.

(*) You should still consider the fact that the product may be zero due to the other factor, $|\nabla g|^2$, being zero. This concern can be disposed of in several ways, one of which is to choose a specific $g$ with nonvanishing gradient.

.. which, incidentally, suggests a simpler solution. The function $g(x)=x_1$ is harmonic, and $\Delta (f(x_1)) = f''(x_1)$.