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Given two functions, $f$ and $g$, both convex, and additionally $f$ is increasing. Using the fact that if a function is differentiable and convex on an interval, then such function is, on the interval, either increasing, or decreasing, or there is a number c such that $f$ is decreasing to the left of it and increasing to the right of it, prove that $(f\circ g)$ is convex.

What I know/tried:

I know that if I show that, given $x<y$, $f'(g(x))g'(x)<f'(g(y))g'(y)$ as it shows that the derivative is increasing, and consequently a convex function. Playing around with the possible configurations for $x$ and $y$ seems to be the strategy, but I only achieved, using the convexity of $g$ and $f$, and the increasing aspect of the function $f$, that the derivative of the composite function is positive if $x<y<c$ and $c<x<y$, which isn't enough.

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Differentiation is neither necessary nor useful here. For any $x, y$ and $0 \le t \le 1$, by definition of convex function $g(tx + (1-t)y) \le t g(x) + (1-t) g(y)$ so since $f$ is increasing, $$ f(g(tx + (1-t)y)) \le f(t g(x) + (1-t) g(y)) \le t f(g(x)) + (1-t) f(g(y))$$

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  • $\begingroup$ Yes, I agree that's easier. Doing that way was a previous problem, but the text specifically asks for a solution using it. $\endgroup$ – Fhoenix Nov 29 '17 at 13:09

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