Is there a real $C^{1}$ function on $[0, 1]$ such that $f(0) = 0$, $\int_{0}^{1}f'(x)^{2}\, dx \leq 1$ and $\int_{0}^{1}f(x)\, dx = 1$?
I initially was thinking of something like $\pi\sin(\pi x)/2$ or $ce^{x}$ but those satisfy 2 of the 3 conditions.
There is no such function, because we have
$$\int_0^1 \lvert f(x)\rvert\, dx \leqslant \int_0^1 \int_0^x \lvert f'(t)\rvert\,dt\,dx \leqslant \int_0^1 \sqrt{x}\left(\int_0^x \lvert f'(t)\rvert^2\,dt\right)^{1/2}\, dx \leqslant \int_0^1\sqrt{x}\,dx = \frac23$$
by the Cauchy-Schwarz inequality.
