# convexity of $f(x)+f(g(x))$ [closed]

suppose $$f$$ is a convex function. i.e.,

for all $$0\leq\lambda\leq 1$$: $$$$f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$$$$ Consider the following function: $$$$h(x)=f(x)+f(g(x))$$$$ what can we say about convexity of $$h$$ w.r.t $$x$$?

more specifically, under what condition $$h$$ is a convex function as well? i.e., $$$$h(\lambda x+(1-\lambda y))\leq \lambda h(x)+(1-\lambda)h(y)$$$$

note that there is no restriction on the convexity of $$g$$.

• What kind of conditions are you thinking of? – A simple case would be $f(x) = x$, so hat $h(x) = x + g(x)$ which is convex iff $g$ is convex. I doubt that anything can be said without restrictions on $g$. Dec 10 '20 at 9:56
• like for instance, if I allow g to be restricted, can we find all the family of functions like g that leads to the convexity of h? Dec 10 '20 at 10:00
• What have you tried? Dec 10 '20 at 10:18
• Also, can we make restrictions on $f$, or only on $g$? Dec 10 '20 at 10:24
• @supinf It's a general question, so if you have some idea to restrict f which results in the convexity of h, I would appreciate it if you could share it. Dec 10 '20 at 11:05

If $$g: \Bbb R \to \Bbb R$$ is a linear function then $$h(x) = f(x) + f(g(x))$$ is convex for any convex function $$f: \Bbb R \to \Bbb R$$.
Conversely, linear functions are the only functions $$g$$ with that property:
• Choosing the convex function $$f(x) = x$$ implies that $$h(x) = x + g(x)$$ is convex, so $$g$$ must be convex.
• Choosing the convex function $$f(x) = -x$$ implies that $$h(x) = -x - g(x)$$ is convex, so $$g$$ must be concave.