Prove $\int_{0}^{1}|f(x)|^2\,dx\le\int_{0}^{1}|f{'}(x)|^2\,dx$ with $f(0)=0$

Question

Let $f$ has following properties: 1)f has continuous derivative in$[0,1]$; 2)$f(0)=0$; then proof $$\int_{0}^{1}|f(x)|^2\,dx\le\int_{0}^{1}|f{'}(x)|^2\,dx$$

I think the condition "$f(0)=0$"may be used to add $f(0)$ or $f^2(0)$ in any place. And then maybe I can use the Lagrange Mean Value Theorem to substitude $f(x)-f(0) $with the $f{'}(\theta)x$. But I can't proof it. So How can I do with this problem? thank you.

By the way, a question that has been asked is quite similar to it: just this

Answer 1

In fact we have a better inequality: $\int_0^{1}|f(x)|^{2}dx =\int_0^{1}|\int_0^{x} f'(t)dt|^{2}dx\leq \int_0^{1} \int_0^{x} |f'(t)|^{2}dt xdx \leq \frac 1 2\int_0^{1} |f'(t)|^{2}dt $.

I have used Cauchy - Schwarz inequality and the fact that $\int_0^{x}|f'(t)|^{2}dt \leq \int_0^{1}|f'(t)|^{2}dt$.