Absolutely continuous function whose derivative is in $L^2([0,1])$ etc., evaluate $\lim_{x\to0^+}\frac{f(x)}{\sqrt{x}}$

Question

Suppose $f$ is absolutely continuous on $[0,1]$, $f(0)=0$, and $f'(x)\in L^2([0,1])$. Show that $$\lim_{x\to 0^+}\frac{f(x)}{\sqrt{x}}=0$$

So far I've got the following: Since $f$ is AC by the FTC we get $f(x)-f(0)=\int_0^x f'(t)\,dt=f(x)$ since $f(0)=0$. Then $$\left|\frac{f(x)}{\sqrt x}\right|\leq \frac{\int_0^x |f'(t)|\,dt}{\sqrt x}\overset{\text{Holders}}{\leq}\frac{||f'||_2||\chi_{(0,x)}||_2}{\sqrt x}=\frac{||f'||_2\sqrt x}{\sqrt x}=||f'||_2$$ so the limit is bounded above by $||f'||_2$. We need it to be zero. Since we are on a finite measure space, $f'$ is in $L^p$ for all $p<2$. Trying these gives a bound that goes to infinity in $x$. (e.g. trying $||f'||_{3/2}$ instead of $||f'||_2$ in the above argument)

I also tried to write it in way that could take advantage of the Lebesgue Differentiation Theorem.

Lastly, applying L'Hospitals to the limit tells you its the same as $\lim_{x\to0^+} 2\sqrt xf'(x)$ so if $f'$ were bounded (or say, even continuous since we are on compact set) we would be done.

What am I missing?

Answer 1

Do what you did. Except don't plug in the $L^2$ norm:

$$\int_0^x|f'(t)|\,dt\le\sqrt x\left(\int_0^x|f'(t)|^2\,dt\right)^{1/2}.$$

Now (prove and) use the fact that $\int_0^x|f'|^2\to0$ as $x\to0$.

Answer 2

After you use Holder, it is $\|f'\|_{L_2([0,x])}$ and it goes to $0$ when $x$ goes to $0$, because $f'$ is in $L_2([0,1])$.