5
$\begingroup$

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?)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.