I need some help with the following:
If $f$ and $g$ are two harmonic functions in $R^2$, i.e., $\Delta f = 0$ and $\Delta g = 0$. Can we say that the scalar function $\nabla f \cdot \nabla g$ is harmonic too, i.e., $\Delta (\nabla f \cdot \nabla g) = 0$? I have the feeling that some properties of the mixed partial derivatives are needed. Thanks in advanse.
Hint:: Let $f(x,y)=g(x,y)=xy$.
We have the easy-to-prove identity $$ \Delta(fg) = g\Delta f + 2\nabla f \cdot \nabla g + f\Delta g, $$ and applying another Laplacian to this gives $$ \Delta\Delta(fg) = \Delta(g\Delta f + 2\nabla f \cdot \nabla g + f\Delta g ) = 2\Delta(\nabla f \cdot \nabla g ). $$ Therefore for harmonic $f$ and $g$, $\nabla f \cdot \nabla g$ is harmonic if and only if $fg$ is biharmonic.