Q:

how to prove $\frac{f(x)}{\sqrt{|\log x|}}\rightarrow 0$ as $x\rightarrow 0$

If $f$ is absolutely continuous on $[\epsilon,1]$ for $0<\epsilon <1$ and $\int_0^1 x|f'(x)|^p dx <\infty.$
How to prove $\frac{f(x)}{\sqrt{|\log x|}}\rightarrow 0$ as $x\rightarrow 0$ if $p=2$?
How to prove $\frac{|f(x)|}{x^{1-2/p}}\rightarrow 0$ as $x\rightarrow 0$ if $p<2$?
I don't know how to deal with the condition. I only got that 
$f(x)-f(\epsilon)=\int_\epsilon^x f'(t)dt$,  $|f(x)-f(\epsilon)|\leq\int_\epsilon^x |t^{1/p}f'(t)\cdot t^{-1/p}|dt\leq (\int_\epsilon^x t|f'|^pdt)^{1/p}\cdot (\int_\epsilon^x t^{-q/p})^{1/q} $

A:

You are very close th the actual solution. The trick is, fix small $0<\epsilon<1$ and notice that for small x, we can write
$$ \int_{x}^{\epsilon} |f'(t)| \, dt = \int_{0}^{1} t^{1/p}|f'(t)| 1_{(0,\epsilon)}(t) \cdot t^{-1/p} 1_{(x,1)}(t) \, dt. $$
Then you can apply the Holder inequality as you tried, which gives you a bound
$$ \limsup_{x\to0+} \frac{|f(x)|}{g(x)} \leq C_{p} \left( \int_{0}^{\epsilon} t|f'(t)|^{p} \right)^{1/p}, $$
where $g(x) = |\log x|^{1/2}$ if $p=2$ and $g(x) = x^{1-2/p}$ if $1<p<2$ and $C_{p} > 0$ is a generic constant depending only on $p$. Therefore taking $\epsilon \downarrow 0$, the claim follows.

