Behavior at $0$ of a function that is absolutely continuous on $[\epsilon, 1]$ The function $f$ on $[0,1]$ is absolutely continuous on $[\epsilon,1]$ for $0<\epsilon<1.$ I further have that $$\int_0^1x|f'(x)|^pdx<\infty.$$
I'm trying to show that
$$
\lim_{x\to 0}f(x)\ \text{exists and is finite}\qquad \text{if}\ p>2,
$$
$$
\frac{f(x)}{|\log x|^{1/2}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p=2,\ \text{and}
$$
$$
\frac{f(x)}{x^{1-\frac{2}{p}}}\to 0\ \text{as}\ x\to 0\qquad \text{if}\ p<2.
$$
(Of course, for $0<\epsilon\leq x\leq 1$ we have that
$$
f(\epsilon)=f(x)-\int_{\epsilon}^x f'(x)dx,
$$
and, for $s>-1$, $r\geq1$, Hölder's inequality gives
$$
\int_0^1x^{s-(s-1)/r}|f'(x)|^{p/r} dx<\infty;
$$
in particular,
$$
\int_0^1x^a|f'(x)|^{p/2} dx<\infty\qquad \text{for}\ a>0.)
$$
I'd greatly appreciate any suggestions!
 A: Hölder's inequality should be used to estimate $\int_\epsilon^1 |f'(x)|\,dx$.
$$
\int_\epsilon^1 |f'(x)|\,dx  = \int_\epsilon^1  \left(x^{1/p} |f'(x)|\right) x^{-1/p} \,dx
\le \left(\int_\epsilon^1 x|f'(x)|^p\,dx\right)^{1/p} \left(\int_\epsilon^1 x^{1/(1-p ) }\,dx\right)^{1-1/p} \tag{1}
$$
Since the first factor in the product on the right of (1) is bounded by $C = \left(\int_0^1 x|f'(x)|^p\,dx\right)^{1/p} $, we have
$$
\int_\epsilon^1 |f'(x)|\,dx \le C \left(\int_\epsilon^1 x^{1/(1-p)  }\,dx\right)^{1-1/p} \tag{2}
$$


*

*When $p>2$, the   integral on the right of (2) converges.

*When $p=2$ it behaves like $|\log \epsilon|$, which is then raised to the power of $1-1/p = 1/2$. 

*When $1< p<2$ it behaves like $x^{1+1/(1-p)} = x^{(2-p)/(1-p)}$ which is then raised to the power of $1-1/p$, producing $x^{(p-2)/p}$.

*When $p=1$, the above calculations need an adjustment: 
$$
\int_\epsilon^1 |f'(x)|\,dx  = \int_\epsilon^1  \left(x  |f'(x)|\right) x^{-1} \,dx
\le \epsilon^{-1} \int_\epsilon^1 x|f'(x)|   \,dx 
$$
hence $f(\epsilon)\le C/\epsilon$.


This yields the desired conclusions about the behaviour of $f(\epsilon)$ as $\epsilon\to 0$.
