Let $f \in H^1(a,b)$, $f(c)=0$ ($a<c<b$). Prove that $|f(x)|^2 \le\| f\|_2\|f'\|_2$ almost everywhere.

Question

Let $f \in H^1(a,b)$, $f(c)=0$ ($a<c<b$). Prove that $|f(x)|^2 \le\| f\|_2\|f'\|_2$ almost everywhere.

I have proved that $\left[f(x) \right]^2=2\displaystyle\int_c^x f(t) f'(t) dt$, and I tried to apply it to solve but I have no idea to solve this. Thank you a lot.

Answer 1

I would think that the inequality you are trying to show is missing a factor of two on the right-hand side: Indeed, with $c = 0$ and $x$ large enough, $f(x) = \exp(x)$ should serve as a counterexample.

Answer 2

Counterexample:

Let $f(x) = \begin{cases} 0 & -1\leq x\leq 0 \\ x & 0\leq x\leq 1\end{cases}$, then $f\in H^1(-1,1)$ and $f(0)=0$. By direct computation we have $\|f\|_2 = \sqrt{\frac13}$ and $\|f'\|_2 = 1$. For $x\in \bigl(\sqrt[4]{\frac13},1\bigr)$, $f^2(x) > \|f\|_2\|f'\|_2$.