Let $f:[a,b]$→$\mathbb{R}$ be a continuously differential fuction satisfying f(a)=0. My goal is to show that
$$\int_{a}^{b} |f(x)|^2 dx \le \frac{(b-a)^2}{2} \int_{a}^{b} |f'(x)|^2 dx $$
My attempt is following:
By FTC, $f(x)=f(a)+\int_{a}^{x} f'(t) dt$ = $\int_{a}^{x} f'(t) dt$
However, by M.V.T for integral $\int_{a}^{x} f'(t) dt=(x-a)f'(\zeta_x)$ for some $\zeta_x \in [a, x]$
So, $\int_{a}^{b} |f(x)|^2 dx$=$\int_{a}^{b} |\int_{a}^{x} f'(t) dt|^2 dx$ = $\int_{a}^{b} |(x-a)f'(\zeta_x) dt|^2 dx$ $\le$ $(b-a)^2\int_{a}^{b} |f'(\zeta_x) dt|^2 dx$ $…(*)$
For a convenience, we define a map $g:[a,b]→[a,b]$ by $g(x)=\zeta_x$.
Then the right integral in the inequality $(*)$ becomes $(b-a)^2\int_{a}^{b} |f'(g(x)) dt|^2 dx$.
Consequently, $\int_{a}^{b} |f(x)|^2 dx \le (b-a)^2\int_{a}^{b} |f'(g(x)) dt|^2 dx$.
Hereby, I tried. But I don't know next step.. please inform me how to proceed the problem.
