My question concerns strong convexity and the composition of functions. Let $f:\Omega \subseteq \mathbb{R}^n \rightarrow \mathbb{R}$ be a continuous differentiable function. Recall that $f$ is convex if its epigraph $\{ (\mathbf{x}, \mu) \in \Omega \times \mathbb{R} \colon \mu \geq f(\mathbf{x}) \}$ is a convex set. Furthermore, $f$ is strongly convex if it holds $$( \nabla_{\mathbf{x}} f - \nabla_{\mathbf{y}}f )^T (\mathbf{x} - \mathbf{y})\geq m \| \mathbf{x} - \mathbf{y} \|_2^2$$ for a constant $m > 0$. Note that strong convexity is a strictly stronger definition than convexity.
It is well-known that if $f$ is convex and $g$ is convex non-decreasing over an univariate domain, then the function $g \circ f$ is also convex. Does this property extends to strong convexity? Specifically, if $f$ is a strongly convex function as described above and $g$ is convex non-decreasing, is the function $g \circ f$ also strongly convex?
Progress I made on this question: It can be easily proven that $f$ is strong convex iff. the function $f(\boldsymbol{x}) - \frac{m}{2} \| \boldsymbol{x} \|_2^2$ is convex. Hence, to answer this question it is sufficient to prove that $(g \circ f)(\boldsymbol{x}) - \frac{\beta}{2} \| \boldsymbol{x} \|_2^2$ is a convex function for some parameter $\beta > 0$, given that $f$ is strongly convex and $g$ is convex non-decreasing.
No; here is a counterexample:
Let $f=\|\cdot\|^2$ which is strongly convex. However, if we let $g$ be the zero function, then $g\circ f$ is also the zero function, which is convex, but not strongly convex.