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August 26th, 2014, 01:04 PM  #1 
Member Joined: Mar 2013 Posts: 71 Thanks: 4  prove that $gof:I \rightarrow \mathbb R$ is also convex
Good afternoon! I'm in trouble to prove the folowing problem: "Let $f:I \rightarrow \mathbb R$ and $g:J \rightarrow \mathbb R$ be convex mappings with $f(I) \subset J$, $g$ monotonic, non decreasing. a) prove that $gof:I \rightarrow \mathbb R$ is also convex. b) give another demonstration of (a), using the fact that $f$ and $g$ are twice differentiable. c) give an exemple to show that if $g$ is not non decreasing, then the result doesn't necessarily hold My solution: a) Since $f$ is convex, then for $x,y \in I$, $0 \le t \le 1, 0 \le r \le 1, t+r=1, f(tx+ry)\le tf(x)+rf(y)$. Now I use the fact that $g$ is monotonic nondecreasing, i.e, $x<y \Rightarrow g(x) \le g(y)$ to write: $gof(tx+ry) \le g(tf(x)+rf(y))$. But since $g$ is convex, then $gof(tx+ry) \le tgof(x)+rgof(y)$. It proves that $gof$ is convex (doesn't it?). b) If I manage to prove that $[gof]''(x)\ge 0$, then I'm done. $[gof]'(x)=g'(f(x)).f'(x)$ and it follows that $[gof]''(x)=g'(f(x)).f''(x)+f'(x)[g''(f(x)).f'(x)]=g'(f(x)).f''(x)+g''(f(x)).[f'(x)]^2$. I cannot conclude that $[gof]''(x) \ge 0$, because $g'(f(x))$ may change signal. Please, give me a hint! c) Please, help me to find a counterexemple. Thanks. 

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