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

for all $0\leq\lambda\leq 1$: \begin{equation} f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y) \end{equation} Consider the following function: \begin{equation} h(x)=f(x)+f(g(x)) \end{equation} 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., \begin{equation} h(\lambda x+(1-\lambda y))\leq \lambda h(x)+(1-\lambda)h(y) \end{equation}

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

  • $\begingroup$ 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 $g$. $\endgroup$
    – Martin R
    Dec 10 '20 at 9:56
  • $\begingroup$ 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? $\endgroup$
    – Jason
    Dec 10 '20 at 10:00
  • $\begingroup$ What have you tried? $\endgroup$
    – supinf
    Dec 10 '20 at 10:18
  • $\begingroup$ Also, can we make restrictions on $f$, or only on $g$? $\endgroup$
    – supinf
    Dec 10 '20 at 10:24
  • $\begingroup$ @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. $\endgroup$
    – Jason
    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.

A function which is both convex and concave is necessarily linear.


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