TITLE: Bounding $\int_0^1 f(x) dx $ under the condition $\int_0^1 f'(x)^2 dx \le 1$
QUESTION [13 upvotes]: Any tips on how to solve this?

Problem 1.1.28 (Fa87) Let $S$ be the set of all real $C^1$ functions $f$ on $[0, 1]$ such that $f(0) = 0$ and
$$\int_0^1 f'(x)^2 dx \le 1 \;. $$
Define
$$J(f) = \int_0^1 f(x) dx \; .$$
Show that the function $J$ is bounded on $S$, and compute its supremum. Is there a function $f_0 \in S$ at which $J$ attains its maximum value? If so, what is $f_0$?

I tried using Cauchy-Schwartz and got a bound of $\frac23$ but it doesn't seem strong enough.

REPLY [8 votes]: Another way for proving the upper bound is the following. We have, using Cauchy-Schwarz twice, that $$\begin{align}
\left|\int_{0}^{1}f\left(x\right)dx\right|= & \left|\int_{0}^{1}\int_{0}^{x}f'\left(t\right)dtdx\right| \\
\leq & \left(\int_{0}^{1}\left(\int_{0}^{x}f'\left(t\right)dt\right)^{2}dx\right)^{1/2} \\ \leq & \left(\int_{0}^{1}x^{2}\int_{0}^{x}f'^{2}\left(t\right)dtdx\right)^{1/2} \\
\leq & \left(\int_{0}^{1}x^{2}dx\right)^{1/2} \\ = & \frac{1}{\sqrt{3}}.
\end{align}$$