7
$\begingroup$

Please help me to prove $$ \sum\limits_{j=2}^n \frac{1}{j^\alpha (j-1)^\alpha} \int\limits_{j-1}^j \frac{dx}{x^{1-\alpha}(n-x)^\alpha} \leq \int\limits_0^1 \frac{dx}{x^{1-\alpha}(n-x)^\alpha},\quad \alpha\in(0,1/2],\quad n\geq 2. $$ This inequality appears in the study of numerical methods for problems with fractional derivatives.

I tried to let $x=j(j-1)t$ in the integrals on the left in order to remove the coefficients $\frac{1}{j^\alpha (j-1)^\alpha}$. Then we get the equivalent inequality $$ \sum\limits_{j=2}^{n} \int\limits_{1/j}^{1/(j-1)} \frac{dt}{t^{1-\alpha}\bigl(n-j(j-1)t\bigr)^\alpha} \leq \int\limits_0^1 \frac{dx}{x^{1-\alpha}(n-x)^\alpha},\quad \alpha\in(0,1/2],\quad n\geq 2. $$ But I don't see how to prove this inequality, too.

$\endgroup$
  • $\begingroup$ It looks unlikely, if we compare values at $t$ on the left and at $x=t$ on the right: the former is greater. Of course, there is small segment $[0,1/n]$, not covered by left integrals, but is it really so crucial? Have you checked this numerically? $\endgroup$ – Fedor Petrov Feb 4 '16 at 18:45
  • $\begingroup$ Yes, I have checked the inequality numerically. The segment $[0,1/n]$ is small, but the function on the right goes to infinity exactly at this segment (when $x=0$). $\endgroup$ – Mikhail_K Feb 4 '16 at 19:19
  • $\begingroup$ Ah, now I see. Asymptotically both parts are $n^{-\alpha}/\alpha$, so it is quite delicate. $\endgroup$ – Fedor Petrov Feb 4 '16 at 22:51
  • $\begingroup$ You can increase the left-hand side of your inequality by bounding $1/x^{1-\alpha}$ from above on each interval $[j-1,j]$, by convexity, by the corresponding affine function and then computing the resulting integrals. Also, the right-hand side is bounded from below by $\int_0^1 x^{\alpha-1}n^{-\alpha}\,dx=n^{-\alpha}/\alpha$. Thus, you get rid of all integrals, and the resulting (stronger and perhaps simpler) inequality seems still to hold, as numerical experiments suggest. $\endgroup$ – Iosif Pinelis Feb 7 '16 at 3:28

Your Answer

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

Browse other questions tagged or ask your own question.