Assume that \begin{align*} \pi_p &= \frac{2}{p} \int_0^1 \left[ u^{1-p} + (1-u)^{1-p} \right]^{1/p} \mathrm{d}u \end{align*}
How to prove that if $\frac{1}{p} +\frac{1}{q}=1$ then $\pi_p=\pi_q$?
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- 2$\begingroup$ What do you mean by $\pi q=\pi q$? The obvious version of that forces $p=q=2$. And what do you mean by $\pi_p$? $\endgroup$JoshuaZ– JoshuaZ2025-05-16 12:54:52 +00:00Commented May 16 at 12:54
- $\begingroup$ you can view π as a function. the problem is to prove that π(p)=π(q) is always right as long as 1/p+1/q =1 $\endgroup$chensw– chensw2025-05-16 13:17:32 +00:00Commented May 16 at 13:17
- 2$\begingroup$ I am surprised by such a strong negative reaction. Surely, there were problems with TeX, language, and a lack of explanation of where the problem comes from ... and yet. Also, now most of these problems are resolved. $\endgroup$Iosif Pinelis– Iosif Pinelis2025-05-16 13:55:26 +00:00Commented May 16 at 13:55
- 2$\begingroup$ Apparently, this problem was of significant interest to Joseph Keller and Ravi Vakil. $\endgroup$Iosif Pinelis– Iosif Pinelis2025-05-16 14:37:29 +00:00Commented May 16 at 14:37
- 3$\begingroup$ @AlexM. : I think the paper by Keller and Vakil is a research one, but the techniques there are elementary, which may be why they chose The Monthly. On the other hand, I think it is nowhere stated that the MO research must be done by non-elementary means. In fact, a great many MO answers use only elementary methods -- which I think is actually good. So, I don't think that I can agree with you. $\endgroup$Iosif Pinelis– Iosif Pinelis2025-05-16 15:43:19 +00:00Commented May 16 at 15:43
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1 Answer
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Using the substitution $u=x^p$ and the symmetry of your integrand wrt the reflection $u\leftrightarrow1-u$, one sees that your $\pi_p$ is the "value of $\pi$ in $\ell_p$" as expressed in formula (3) of this paper, where it is shown that indeed $\pi_p=\pi_q$ if $p>1$ and $\frac1p+\frac1q=1$.
answered May 16 at 14:12