TITLE: $\int_0^1 f(x)^2\le 1$ and $\int_0^1 f'(x)^2\le 1$ $\Rightarrow$ $\left|f(x)\right|\le \sqrt3$
QUESTION [6 upvotes]: Let $f:[0, 1]\rightarrow \mathbb{R}$ be a function that is continous on $[0,1]$ and derivable on $(0, 1)$. If $\int_0^1 f(x)^2\le 1$ and $\int_0^1 f'(x)^2\le 1$, show that $\left|f(x)\right|\le \sqrt3$ for all $x\in [0, 1]$.
Using Cauchy-Schwarz inequality I've shown that $\left|f(x)-f(0)\right|\le \sqrt{x}$ and $\left|f(x)^2-f(0)^2\right|\le 2$. Any other ideas?

REPLY [6 votes]: Finally I think I have a proof.
By mean value theorem for integral, there exists $\xi\in [0, 1]$ such that $f(\xi)^2=\int_0^1 f(x)^2 dx\le 1$. Suppose $x> \xi$ (if $x\le \xi$ we can give a similar proof), by Cauchy-Schwarz inequality,
$$\left| \frac{f(x)^2-f(\xi)^2}{2} \right |=\left| \int_{\xi}^x f(t)f'(t)dt \right|\le \left( \int_{\xi}^x f(t)^2 dt \right)^{1/2}\left( \int_{\xi}^x f'(t)^2 dt \right)^{1/2}\le 1.$$
We conclude that $f(x)^2\le 3$ for all $x\in [0, 1]$.