Suitable composition of concave and convex functions is convex? Let $f:[0,1]\to[0,1]$ be a strictly increasing continuous concave function with $f(0)=0$ and $f(1)=1$.
Let $g$ be the inverse of $f$.
Then $g$ is strictly increasing and convex.
It seems that the function $h(x)=f(\frac12g(x))$ is always convex.
Is this true?
If yes, why?
 A: If $g(x) = y$, I get
$$ h''(x) = \dfrac{f''(y/2) f'(y) - 2 f'(y/2) f''(y)}{4 f'(y)^3} $$
so for $h$ to be convex requires
$$ \dfrac{f''(y/2)}{f'(y/2)} \ge 2 \dfrac{f''(y)}{f'(y)}$$
For a counterexample, take $f$ that is strictly concave on $[0,1/2]$ but linear
on $[1/2, 1]$, so that if $1/2 \le y < 1$ we have $f''(y) = 0$ but $f''(y/2) < 0$.
A: To show that $g$ is increasing,assume to the contrary that there exists $x,y\in I$ such that $x<y$ but $g(x)>g(y)$. Since $f$ is increasing and $g(x),g(y)\in[0,1]$ then $f(g(x))>f(g(y))\Rightarrow x>y$ which is a contradiction.
Now lets prove the second part. Let $x,y\in I$ and put $t=g(x),\:s=g(y)$. By concavity of $f$ we have $$f(\lambda t+(1-\lambda)s)\geq\lambda f(t)+(1-\lambda)f(s)\Rightarrow g\Big(f(\lambda t+(1-\lambda)s)\Big)\geq g\big(\lambda f(t)+(1-\lambda)f(s)\big) \text{(remember g was increasing )}$$
Thus $\lambda t+ (1-\lambda)s\geq g(\lambda x+(1-\lambda)y))$ that is $\lambda g(x)+(1-\lambda )g(y)\geq g(\lambda x+(1-\lambda)y))$. Done! Im not assuming that $f\in C^1.$
