# Find all differentiable functions $f : [0 ; \infty) \to \Bbb{R}$ such that $\int_{0}^{1}(f'(x))^2 dx \ge f(x + f(x))-f(x)$

I came across this problem in a book, and the solution isn't printed.It goes like this:

Find all differentiable functions $f : [0 ; \infty) \to \Bbb{R}, f(0)=0,$ such that:

1) $f'$ is strictly positive and increasing

2) $\int_{0}^{1}(f'(x))^2 dx \ge f(x+f(x))-f(x)$, for all x in the domain

So far I haven't found a function that satisfies both conditions. Since $f'$ is strictly positive, it means that $f$ is strictly increasing, therefore it has only positive values, since $f(0)=0$.

We have $x+f(x)>x$, therefore $f(x+f(x))>f(x)$, although I really don't know how to handle the integral.

$\mathbf{Edit}$: using the inequality posted by Vladimir below, we can take $g(x)=f(x+f(x))-f(x)$, and we get $g'(x) \ge (f'(x))^2 >0$, therefore $g$ is strictly increasing, with $g(0)=0$.However, by integrating the inequality, we have:

$\int_{0}^{1}g'(x)dx \ge \int_{0}^{1}(f'(x))^2dx \ge g(x)$, in other words, $g(1) \ge g(x)$, for all $x$ in the domain,which contradicts the monotonicity of $g$.

• $f(0)=0$ and $f$ increasing implies $x+f(x)>x!!$ – mfl Jul 2 '14 at 9:09
• Oh, sorry, I typed wrong. – Alexander Smith Jul 2 '14 at 9:13
• Do you really mean $\int_0^1(f'(t))^2dt$ which is constant, or $\int_0^x(f'(t))^2dt$, (on the LHS of your second condition) ? – Omran Kouba Jul 2 '14 at 9:42
• The integral is constant. – Alexander Smith Jul 2 '14 at 9:54

Hint: $$(f(x+f(x)))'-f'(x)=f'(x+f(x))(1+f'(x))-f'(x)\ge (f'(x))^2.$$