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?)


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.

  • $\begingroup$ Thank you! I need to understand why $\left(\int_t^\delta x^{1/(1-p ) }\,dx\right)^{1-1/p}\leq C_pt^{1-2/p}$? $\endgroup$ – Aubrey Mar 9 '15 at 3:22
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    $\begingroup$ If raise both sides to $p/(p-1)$, evaluate the integral, and move the power of $t$ to the left, the result should be something that is evidently bounded for $0\le t\le \delta$. $\endgroup$ – user147263 Mar 9 '15 at 3:23
  • $\begingroup$ OK, at first I thought that could be an immediate consequence of the mean value theorem. $\endgroup$ – Aubrey Mar 9 '15 at 3:24

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