## Problem 3

Find the maximum value of

$$
\int_{0}^{1}\left|f^{\prime}(x)\right|^{2}|f(x)| \frac{1}{\sqrt{x}} d x
$$

over all continuously differentiable functions $f:[0,1] \rightarrow \mathbb{R}$ with $f(0)=0$ and

$$
\int_{0}^{1}\left|f^{\prime}(x)\right|^{2} d x \leq 1
$$
Solution. For $x \in[0,1]$ let

$$
g(x)=\int_{0}^{x}\left|f^{\prime}(t)\right|^{2} d t
$$

Then for $x \in[0,1]$ the Cauchy-Schwarz inequality gives

Thus

$$
|f(x)| \leq \int_{0}^{x}\left|f^{\prime}(t)\right| d t \leq\left(\int_{0}^{x}\left|f^{\prime}(t)\right|^{2} d t\right)^{1 / 2} \sqrt{x}=\sqrt{x} g(x)^{1 / 2}
$$

$$
\begin{aligned}
\int_{0}^{1}\left|f^{\prime}(x)\right|^{2}|f(x)| \frac{1}{\sqrt{x}} d x & \leq \int_{0}^{1} g(x)^{1 / 2} g^{\prime}(x) d x=\frac{2}{3}\left[g(1)^{3 / 2}-g(0)^{3 / 2}\right] \\
& =\frac{2}{3}\left(\int_{0}^{1}\left|f^{\prime}(t)\right|^{2} d t\right)^{3 / 2} \leq \frac{2}{3}
\end{aligned}
$$

by $(*)$. The maximum is achieved by the function $f(x)=x$.

Remark. If the condition (*) is replaced by $\int_{0}^{1}\left|f^{\prime}(x)\right|^{p} d x \leq 1$ with $0<p<2$ fixed, then the given expression would have supremum equal to $+\infty$, as it can be seen by considering continuously differentiable functions that approximate the functions $f_{M}(x)=M x$ for $0 \leq x \leq$ $\frac{1}{M^{p}}$ and $\frac{1}{M^{p-1}}$ for $\frac{1}{M^{p}}<x \leq 1$, where $M$ can be an arbitrary large positive real number.