Can a convex/concave function be transformed to a concave/convex function in a generic way?

Question

My question is that let’s say I have a concave increasing function, can I transform it into a new convex increasing function? I know you can take log or exp of the function till it changes one to another, but I mean in a more generic and certain way.

Answer 1

Here is one idea, surely not what you intended. Negation flips about the horizontal, changing concave to convex. Reflecting in a vertical line $x=c$ then turns convex decreasing to convex increasing. Shifting vertically with constants retains these properties.

---

         
          ^{$3+x^2$ is concave increasing, $10-x^2$ is convex decreasing. $10-(x-4)^2$ reflects around the vertical $x=4$.}

---

Answer 2

If we assume that $f$ is strictly increasing and concave it follows that it has an inverse $f^{-1}$ that is strictly increasing.

• $f^{-1}$ is convex because for all $x,y$ in the range of $f$ we have $u,v$ such that $x=f(u),y=f(v)$ and $$ f^{-1}\Big(\frac{x+y}{2}\Big)=f^{-1}\Big(\frac{f(u)+f(v)}{2}\Big)\,. $$ Note that $$ f^{-1}\Big(\frac{x+y}{2}\Big)\le \frac{f^{-1}(x)+f^{-1}(y)}{2}\,, $$ which is the convexity of $f^{-1}\,,$ is equivalent to $$ \frac{x+y}{2}\le f\Big(\frac{f^{-1}(x)+f^{-1}(y)}{2}\Big)\,, $$ that is, to $$ \frac{f(u)+f(v)}{2}\le f\Big(\frac{u+v}{2}\Big)\,, $$ which is the concavity of $f$.