Limit

Beta · **Joined:** Thu, 30 Oct 2008 10:47:09 UTC **Posts:** 182

Suppose f is absolutely continuous on $[\epsilon,1], 0<\epsilon<1$ , and $\int_0^1x|f'(x)|^p dx<\infty$ . Prove that
a) $\lim_{x\to 0}f(x)$ exists and finite for p>2

,
b) $\lim_{x\to 0} \frac{|f(x)|}{|\log(x)|^{1/2}}=0$ , for p=2

.

How to relate the limit to given integral?

Shadow · **Posted:** Mon, 1 Aug 2011 19:51:31 UTC

What is p in this case? Is it for all p, or just some p?

outermeasure · **Posted:** Mon, 1 Aug 2011 23:03:05 UTC

Beta wrote:

Suppose f is absolutely continuous on $[\epsilon,1], 0<\epsilon<1$ , and $\int_0^1x|f'(x)|^p dx<\infty$ . Prove that
a) $\lim_{x\to 0}f(x)$ exists and finite for p>2

,
b) $\lim_{x\to 0} \frac{|f(x)|}{|\log(x)|^{1/2}}=0$ , for p=2

.

How to relate the limit to given integral?

I think you want $\forall \epsilon \in (0,1)$ , rather than only some $\epsilon\in (0,1)$ , that $f\vert[\epsilon,1]$ is absolutely continuous.

If p>2, then from measurability of f' and integrability of $x\lvert f'\rvert^p$ , you know that $x^{2+\delta}\lvert f'(x)\rvert^p$ is (uniformly) bounded for all $\delta>0$ (OK, technically speaking you want a representative of f' so that blahblahblah). Thus by choosing small enough $\delta$ you can make sure $2+\delta<p$ and hence $\lvert f'(x)\rvert\leq\dfrac{C}{x^q}$ with q<1, so dominated convergence gives...

For p=2, use a similar idea but dominate with $\dfrac{1}{x\lvert\log x\rvert^q}$ instead of $\dfrac{1}{x^q}$ .

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Limit

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