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I am studying for an exam and am stuck on this practice problem.

Suppose $f:[0,1] \rightarrow \mathbb{R}$ is absolutely continuous and $f' \in \mathcal{L}_{2}$. If $f(0)=0$ does it follow that $\lim_{x\rightarrow 0} f(x)x^{-1/2}=0$?

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up vote 8 down vote accepted

Yes. The Cauchy-Schwarz inequality gives $$f(x)^2=\left(\int^x_0 f^\prime(y)\ dy\right)^2\leq \left(\int^x_0 1\ dy\right) \left(\int^x_0 f^\prime(y)^2\ dy\right)=x\left(\int^x_0 f^\prime(y)^2\ dy\right).$$ Dividing by $x$ and taking square roots we get $$|f(x)/\sqrt{x}|\leq \left(\int^x_0 f^\prime(y)^2\ dy\right)^{1/2}$$ and since $f^\prime$ is square integrable, the right hand side goes to zero as $x\downarrow 0$. This gives the result.

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