Assume that $f\colon[0,1]\to\mathbb{R}$ is a differentiable function with $f'$ square integrable (if that's too weak or unpractical, assume that $f$ is continuously differentiable) such that $f(0)=f(1)=0$. Is it the case that
$$
\left(\int_0^1 |f|\right)^2 \leq \frac{1}{12}\int_0^1 {f'}^2
$$
? It holds for the functions $f$ I have tried (e.g., the obvious $f(x)=x(1-x)$, and things like $f(x)=\sin(\pi x)$). Moreover, it is not difficult to see that $1/12$ is the best one can hope for, since it's achieved for $f(x)=x(1-x)$.
It seems to reek of Cauchy-Schwarz and/or integration by parts, but I can't see how to prove it. It feels I am one simple trick short.
Best Answer
Fix a $c \in (0,1)$. For $0 \leq t\leq c$, since $f(0) =0$ we have $$|f(t)| = |\int_0^t f'(s) ds|\leq \int_0^t |f'(s)| ds.$$ Therefore, by Fubini theorem, we have $$ \int_0^c |f(t)| dt \leq \int_0^c \int_0^t |f'(s)| ds dt = \int_0^c |f'(s)| (c-s) ds.$$ Similarly, we get $$ \int_c^1 |f(t)| dt \leq \int_c^1 |f'(s)| (s-c) ds.$$ Hence, by summing two inequalities, we obtain $$\int_0^1 |f(t)| dt \leq \int_0^1 |f'(s)| |c -s| ds.$$ Applying Cauchy-Schwartz, we have $$(\int_0^1 |f(t)| dt)^2 \leq (\int_0^1 f'(s)^2 ds )(\int_0^1(c-s)^2 ds) =(c^2 -c + \frac13) \int_0^1 f'(s)^2 ds,$$ for any $c\in (0,1)$. Since $c^2-c + \frac13 \geq \frac1{12}$ for any $c\in (0,1)$ and attains at $c = \frac12$. By taking $c =\frac12$, we get the desired inequality.