Bounding $\int_0^1 f(x) dx $ under the condition $\int_0^1 f'(x)^2 dx \le 1$

Solution 1:

A try for a solution:

We have by integration by parts: $$J(f)=\int_0^1 (1-t)f^{\prime}(t)dt$$ and Cauchy-Schwarz gives that $|J(f)|\leq \frac{1}{\sqrt{3}}$. Let $f_0(x)=\sqrt{3}(x-\frac{x^2}{2})$. Then $f_0^{\prime}(x)=\sqrt{3}(1-x)$, and $\int_0^1 f_0^{\prime}(t)^2 dt=1$. We have $J(f_0)=\frac{1}{\sqrt{3}}$, and we are done.

Solution 2:

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