Show that $\lim_{x→0+ } x^{−1/2}f(x)$ exists and determine the value of this limit.

Question

Let $f : [0,1] → \mathbb{R}$ be absolutely continuous, satisfy $f(0) = 0$ and $f′ ∈ L_2([0,1]).$ Show that $\lim_{x→0+ } x^{−1/2}f(x)$ exists and determine the value of this limit.

From absolute continuity we have that $f'$ exists almost everywhere. I wanted to say that $f(1) - \lim_{x \rightarrow 0^+}x^{-1/2}f(x)$ = $\int$$_{0}^1$$\frac{d}{dx}(x^{-1/2}$$f(x))$$dx$ and the show this has $L_1$ norm finite showing the limit exists. But this hasn't worked yet.

Thanks for any help.

Answer 1

Since $f$ is absolutely continuous and $f(0) = 0$, we have

$$f(x) = \int_0^x f'(t)\,dt.$$

Apply the Cauchy-Schwarz inequality, and then use the dominated convergence theorem to show that the limit is $0$.

The Cauchy-Schwarz inequality yields

$$\lvert f(x)\rvert \leqslant \int_0^x \lvert f'(t)\rvert\,dt \leqslant \left(\int_0^x 1^2\,dt\right)^{1/2}\left(\int_0^x \lvert f'(t)\rvert^2\,dt\right)^{1/2} = \sqrt{x}\left(\int_0^x \lvert f'(t)\rvert^2\,dt\right)^{1/2},$$

so we have

$$x^{-1/2}\lvert f(x)\rvert \leqslant \lVert f'\cdot \chi_{[0,x]}\rVert_{L^2},$$

and by the dominated convergence theorem, $f'\cdot \chi_{[0,x]} \xrightarrow{L^2} 0$ as $x\to 0$.