Does $\phi\circ g$ convex imply $\phi\circ f\circ g$ convex?

Question

Assume that $\phi$ is an increasing function and $g$ is a decreasing function with $\phi\circ g$ convex. If $f$ is an increasing convex function does this imply $\phi\circ f\circ g$ is a convex function?

I came to this from an applied problem where we can use the fact that $f\circ \phi\circ g$ is convex and I'd also like to say something about what happens if we did $f$ and $\phi$ in the opposite order.

Answer 1

Counterexample: $$\phi(x)=\begin{cases} x+1,\quad & x\le 0 \\ e^x,\quad &x\ge 0\end{cases}$$ $$g(x)=\begin{cases} \ln(-x),\quad & x\le -1 \\ -x-1,\quad &x\ge -1\end{cases}$$ Observe that $\phi(g(x))=-x$, which is convex.

Let $f(x)=x-1$. Then $$f(g(x))=\begin{cases} \ln(-x)-1,\quad & x\le -1 \\ -x-2,\quad &x\ge -1\end{cases}$$ which changes sign at $x=-e$. Consequently, $$\phi(f(g(x)))=\begin{cases} -x/e, \quad & x\le -e \\ \ln(-x), \quad & -e\le x\le -1 \\ -x-1, \quad & x\ge -1 \end{cases}$$ Convexity fails on $[-e,-1]$.