How to show the following :
Given $f:[a,b] \rightarrow \mathbb R$, $f(a) = f(b) = 0$, f and f' continuous
in the space $\ell_2$ with norm $\|f\| = \sqrt{(\int_a^b f^2)dx}$.
Show $\|f\| \leq \|f'\|\frac{b-a}{\sqrt{2}}$.
Cauchy - Schwarz (which was suggested as a method) gives $|\int_a^b f'| \leq \sqrt{\int_a^b f'^2} * \sqrt{b-a}$
But this result seems useless since $|\int_a^b f'|$ is 0
Somehow my solution does not require both zeros of $f$. I hope it's still correct and I think that one doesn't need bot zeros of $f$.
Let $x\in[a,b]$ be given. Since $$|f(x)|^2=(|f(a)+\int_a^x f'|)^2 \leq (\int _a^x |f'(s)| ds)^2\leq (\int_a^x |f'(s)|^2 ds) (x-a)$$ we can integrate this term and receive $$\int_a^b |f(x)|^2 dx \leq \int_a^b (\int_a^x |f'(s)|^2 ds) (x-a) dx \leq ||f'||^2 (b-a)^2/2$$ Taking the root gives the result. Note that we estimated the integral containing $f'$ with choosing $x=b$. This is allowed since the integrand is non-negative. Note that $$\int_a^b (x-a) dx= \frac{1}{2} (x^2-ax)|_a^b=\frac{1}{2}(b-a)^2$$
If there is an error please tell me