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?

Solution 1:

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 $f''(y/2) < 0$.

Solution 2:

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 $f\in C^1.$