If $f=f(x)$ is absolutely continuous on $[0,1]$ such that $f' \in L^2([0,1])$ and $f(0)=0$. Prove then that
$$\lim_{x \rightarrow 0^+}\frac{f(x)}{x^{\frac{1}{2}}}=0$$
The hint says to use Fundamental theorem of calculus but I don't see where to use that.
I know if $f' \in L^2([0,1])$ then
$$\bigg(\int_{[0,1]} \vert f'(x) \vert^{2} dx\bigg)^{\frac{1}{2}} < + \infty.$$
I don't know what to do from here though. Any hints would be appreciated. Am I merely using L'Hôpital's rule, since I get $0/0$?
For $x > 0$, using the fundamental theorem of calculus: $$ \begin{align*} \left| \frac{f(x)}{\sqrt{x}} \right| &= \left| \frac{\int_0^x f^\prime(t)dt}{\sqrt{x}} \right| \\ &\leq \frac{\int |\mathbb{1}_{[0,x]}(t) f^\prime (t)| dt}{\sqrt{x}} \\ &= \frac{\|1_{[0,x]} f^\prime\|_1}{\sqrt{x}} \\ &\leq \frac{x \|f^\prime\|_2}{\sqrt{x}} \\ &= \sqrt{x} \|f^\prime\|_2\\ &\to 0 \end{align*}$$ as $x \to 0^+$, where the last inequality is Hölder's inequality.