TITLE: Maximizing the value of $\int_0^1 f(x)f^{-1}(x)\ \mathrm dx$
QUESTION [37 upvotes]: I am trying to find the maximum size of the integral $\int_0^1 f(x)f^{-1}(x)\ \mathrm dx$ for differentiable, increasing $f$ with $f(0)=0$ and $f(1)=1$. I made up this exercise for myself and thought it would be easy, but I can't do it.
I feel the answer should be $\frac 1 3$ intuitively, which comes from $f(x)=x$. So far I've tried integration by parts but then I don't know what to do.
Edit: here is the integration by parts I tried, though I think it doesn't lead anywhere:
$$\int^1_0 f(x)f^{-1}(x)\ \mathrm dx=\int_0^1f^{-1}(x)\ \mathrm dx-\int_0^1f'(x)\left(\int_0^x f^{-1}(t)\ \mathrm dt\right)\ \mathrm dx\text.$$
I thought this could help because $f'(x)>0$ since $f$ is increasing and the other factor in this integral is also positive by default.

REPLY [5 votes]: We can make rigorous the argument given by Flounderer using the comment given by Varun Vejalla. Also we can prove that the identity function is indeed a maximizer computing the second variation which turns out to be strongly concave.
Using the change of variable $x=f(t)$ we have
$$
\int_0^1 f(x) f^{-1}(x)\,dx=\int_0^1 f(f(t))f'(t) t \,dt.
$$
We compute the Euler-Lagrange equation: let $h(t)=f(t)+\delta g(t)$ where $g\in S:=\{G \in C^1[0,1] : g(0)=g(1)=0\}$ and $\delta$ real number.
We
$$
u[g](\delta)=\int_0^1 t (f(t)+\delta g(t))(f(f(t)+\delta g(t))+\delta g(f(t)+\delta g(t)))\,dt
$$
Now the Euler-Lagrange equations is
$$0=\frac{d}{d\delta}u[g](0)=\int_0^1 t \{  g'(t) f(f(t))+ f'(t) f'(f(t)) g(t)+f'(t) g(f(t)) \}\,dt=\int_0^1 t [f(f(t))g(t)]' \,dt+ \int_0^1 t f'(t) g(f(t)) \,dt.$$
Integrate by parts the first integral and substitute $f(t)=x$ to obtain
$$
-\int_0^1 f(f(t)) g(t)+\int_0^1 g(x) f^{-1}(x) dx=\int_0^1 g(t) \lvert f^{-1}(t)-f(f(t))\rvert\, dt \quad \forall g\in S.
$$
An application of the fundamental lemma gives
$$
f^{-1}(t)=f(f(t)),\quad \text{that is} \quad t=f(f(f(t))).
$$
By symmetry we can suppose $t\leq f(t)$ and since $f$ increases we obtain $t\leq f(t)\leq f(f(t))\leq f(f(f(t))=t$. Hence $f(t)=t$ is the unique critical point for the functional.
We can prove that $f(t)=t$ is indeed a maximum since $u''[g](0) \leq - \alpha \int_0^1 g^2(t)\,dt$ for some $\alpha>0$.
If we compute the second variation in $\delta=0$ in Flounderer's argument we obtain $-2 \int_0^1 g^2(t)\, dt$, namely the strongly concave condition with $\alpha =2$. We can make the argument rigorous using the approach above