Background Information:
Def – If $U$ is a bounded, open set, we say that $v\in C^2(U)\cap C(\overline{U})$ is subharmonic if
$$-\Delta v \leq 0 \ \ \ \text{in} \ U$$
Question:
Let $\phi:\mathbb{R}\to \mathbb{R}$ be smooth and convex (i.e., $\phi^{\prime \prime}(x) \geq 0 $). Suppose that $u$ is harmonic and define $v(x):= \phi(u(x))$. Prove that $v$ is subharmonic.
Attempted proof – We have
$$\Delta v(x) = \Delta\left[\phi(u(x))\right] = \phi'(u(x))\Delta u(x) + \phi^{\prime \prime}(u(x))|\nabla u|^2$$
Before I continue I really do not understand at all where $\phi^{\prime \prime}(u(x))|\nabla u|^2$ comes in by the rules of differentiation. Please explain this clearly as possible using necessary definitions of $\Delta$ if needed.
Best Answer
$$ \Delta v = \nabla \cdot (\nabla v), $$ and we have $$ \nabla v = \phi'(u) \, \nabla u $$ by the chain rule. We now use the product rule $$ \nabla \cdot (\psi \mathbf{F}) = \nabla \psi \cdot \mathbf{F} + \psi \, \nabla \cdot \mathbf{F} $$ where $\psi$ is a scalar function and $\mathbf{F}$ a vector, to obtain $$ \nabla \cdot (\nabla v) = \nabla \cdot ( \phi'(u) \, \nabla u) = (\phi''(u) \nabla u) \cdot \nabla u + \phi'(u) \, \nabla \cdot (\nabla u), $$ which simplifies to the expression you have.