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


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

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.

  • 1
    $\begingroup$ Ahh thank you. I guess I found an errata. $\endgroup$ Nov 2 '20 at 9:50

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