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How can we bound this integral:

$${\displaystyle \int_{-1}^{1}2\left[\dfrac{1}{4}-\dfrac{1}{4\left(1-\xi^{2}\right)}\left(1-\dfrac{\xi^{2}}{2}\right)^{2}\right]\left(\hat{f}\left(\xi\right)\right)^{2}e^{-2t}d\xi} $$

I've tried to use Holder inequality but when I compute $\left\Vert 2\left[\dfrac{1}{4}-\dfrac{1}{4\left(1-\xi^{2}\right)}\left(1-\dfrac{\xi^{2}}{2}\right)^{2}\right]\right\Vert _{L_{\xi}^{2}\left(\left[-1,1\right]\right)} $, it is not defined since we have some kinds of $\log\left(\xi\pm1\right) $ in the anti-derivative.

Thank you very much.

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  • $\begingroup$ What is $f$? Or, in other words, in whiuch terms do you want to estimate it? What is the sense of multiple $e^{-2t}$, which does not depend on $\xi$? $\endgroup$
    – Fedor Petrov
    Commented Apr 15, 2016 at 11:01
  • $\begingroup$ You can ignore $e^{-2t}$, just a constant $\endgroup$
    – Truong
    Commented Apr 15, 2016 at 11:09
  • $\begingroup$ Actually I want to find an upper bound of the entire integral. It seems like Holder inequality does not work. $\endgroup$
    – Truong
    Commented Apr 15, 2016 at 11:10
  • $\begingroup$ Or we can bound the function inside the integral before integrating it. $\endgroup$
    – Truong
    Commented Apr 15, 2016 at 11:14
  • $\begingroup$ in general integral diverges at endpoints, you should assume something like ${\hat f}(\pm 1)=0$ $\endgroup$
    – Fedor Petrov
    Commented Apr 15, 2016 at 11:17

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