# Strong convexity and the composition of convex functions

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.

• Also in the progress you mentioend, that is actually the definition of weakly convex. $f$ is $m$-strongly convex if $f-m\|\cdot\|^2/2$ is convex.
– Zim
Jan 11 at 22:46

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.
• Thank you for your reply. What if we take $g$ to be strongly convex then?
• @fq00 Great question! It might even be enough to have $g$ strictly increasing, but I'm not sure