Convex functions of subharmonic functions

Remarks

Proof idea: a convex function on $\R$ is continuous, and an increasing continuous function preserves upper semicontinuity. For the sub-mean inequality on a sphere with normalised measure $\sigma$, monotonicity and Jensen's inequality give $$\varphi(u(a)) \le \varphi\Big(\int u \, d\sigma\Big) \le \int \varphi(u) \, d\sigma.$$ In part 2 the first step is an equality, by the mean value property of $h$, so monotonicity of $\varphi$ is not needed. If $\int u \, d\sigma = -\infty$, then $u(a) = -\infty$ and $\varphi(u(a)) = \inf \varphi$, which is below every value of $\varphi \circ u$.

Examples: if $u$ is subharmonic, so are $e^{u}$ and $\max(u, 0)$; if $h$ is harmonic, so are $\abs{h}$ and $h^2$. Monotonicity is needed in part 1: $u(x) = \abs{x}^2$ is subharmonic and $\varphi(t) = -t$ is convex, but $\varphi \circ u = -\abs{x}^2$ is not subharmonic.

Going deeper

For a student

This is the multi-dimensional analogue of the fact that an increasing convex function of a convex function is convex. Combined with the subharmonicity of $\log\abs{f}$ for holomorphic $f$, it shows that $\abs{f}^p$ is subharmonic for all $p > 0$, which is the starting point of the theory of Hardy spaces. It also shows that $\log u$ need not be subharmonic when $u$ is, since $\log$ is concave; functions with $\log u$ subharmonic form the smaller class of log-subharmonic functions.