Lef $f'$ be integrable and $f(0)=0$. Show that $|f(x)|\leq\sqrt{\int\limits_0^1|f'(t)|^2dt}$

Question

Lef $f$ be a fuction such that $f'$ is integrable in $[0,1]$, $f(0)=0$. Show that $$|f(x)|\leq\sqrt{\int\limits_0^1|f'(t)|^2dt}$$ forall $x\in[0,1]$

I did $f(x)^2\leq\int\limits_0^1(f'(t))^2dt$ but then I got no idea.

Answer 1

Using Cauchy-Schwarz for the first inequality,

$$ \left|\,f(x)\,\right| = \left|\,f(x) - f(0)\,\right|= \left|\int_0^x 1 \cdot f'(t) \ dt \right| \leq \sqrt{ \int_0^x 1^2 \ dt } \cdot \sqrt{ \int_0^x |f'(t)|^2 \ dt } \leq \sqrt{ \int_0^1 |f'(t)|^2 \ dt }$$

The second inequality follows as $0 \leq x \leq 1$ and $|f'(x)|^2 \geq 0$.