# How to prove that f is convex?

This is from Bela Bollobas's book on functional analysis.

Given:

$$f: (a,b) \to (c,b) \text{ and } \phi: (c,b) \to \mathbb{R} \quad \phi^{-1}f, \phi \text{ are both convex }$$

To show:

$$f$$ is convex

What I've tried:

Looked at the definitions a lot of times, and the fact that $$\frac{f(x_2)-f(x_1)}{x_2 - x_1}$$ is non-decreasing in one variable when the other is held fixed. I also tried looking at the convex set. But I'm stuck. I also know that if $$\phi$$ is non-decreasing then we're done. I'm just looking for a hint.

This is not true. On $$(0,1)$$ let $$\phi (x)=\frac 1 x$$ and $$f(x)=\sqrt x$$. Then $$\phi^{-1}\circ f$$ and $$\phi$$ are convex but $$f$$ is not. If $$\phi$$ is also increasing then the result is true.