TITLE: Showing this limit as $x\to 0$
QUESTION [5 upvotes]: I'm very close to the result and must be missing something basic: $f$ is absolutely continuous on $[\epsilon,1]$ for each $\epsilon \in (0,1)$, $1<p<2$, and, for $0\leq x\leq y\leq 1$, we have (by Hölder's inequality)
$$
\int_x^y |f'(t)|\,dt 
\le \left(\int_x^y t|f'(t)|^p\,dt\right)^{1/p} \left(\int_x^y t^{1/(1-p ) }\,dt\right)^{1-1/p},
$$
where
$\left(\int_0^1 t|f'(t)|^p\,dt\right)^{1/p}=C<\infty$ (by assumption).
By letting $x,y$ approach $0$, I'm trying to show that 
$$
\frac{f(x)}{x^{1-2/p}}\to 0\qquad \text{as}\ x\to 0.
$$
Could someone please explain how this can be obtained rigorously? (I know that by the mean value theorem,  $\int_x^y t^{1/(1-p ) }\,dt =(y-x)s^{1/(1-p )}$
for some $s\in (x,y).$ Of course, this integral behaves like $x^{1+1/(1-p)} = x^{(2-p)/(1-p)}$ which is then raised to the power of $1-1/p$, producing $x^{(p-2)/p}$. But still I haven't proved the result rigorously.)
(Perhaps Hölder's inequality does not actually suffice to exploit the assumption that $\left(\int_0^1 t|f'(t)|^p\,dt\right)^{1/p}=C<\infty$, and I need to use some other inequality?)

REPLY [4 votes]: This, of course, is related to your question Behavior at $0$ of a function that is absolutely continuous on $[\epsilon, 1]$
Given $\epsilon>0$ we must show $\limsup_{t\to0} t^{2/p-1}|f(t)|\le \epsilon$. For $\delta\in (0,1)$, the estimate 
$$\int_t^\delta |f'(x)|\,dx = \int_t^\delta  \left(x^{1/p} |f'(x)|\right) x^{-1/p} \,dx
\le \left(\int_t^\delta x|f'(x)|^p\,dx\right)^{1/p} \left(\int_t^\delta x^{1/(1-p ) }\,dx\right)^{1-1/p}$$
yields
$$|f(t)|\le |f(\delta)|+ C_p t^{1-2/p} \left(\int_t^\delta x|f'(x)|^p\,dx\right)^{1/p} $$
where $C_p$ depends only on $p$. By choosing $\delta$ small enough we can make sure that 
$$
 C_p  \left(\int_0^\delta x|f'(x)|^p\,dx\right)^{1/p} <\epsilon
$$
and therefore 
$\limsup_{t\to0} t^{2/p-1}|f(t)|\le \epsilon$ as desired.