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The passage below is from a fixed point argument in Wang's paper enter link description here pg 566. At the end of the $I_1$ calculation, it somehow makes the following estimate $$C\|\phi\|_X (e^{-|x|^2/32t} +w^{-1}(x)) \leq C(p,n,l)\|\phi\|_X w^{-1}(x)$$ assuming according to the author that $T>1$. Here $w^{-1}(x) = (1+|x|)^{-l/(p-1)}$. My question is whether this passage is really correct? If $T<1$ I think it works, because $e^{-|x|^2/32t}\leq e^{-|x|^2/32}$ so using that $w(x)e^{-|x|^2/32}<C_1$, $C_1$ constant, $$ (e^{-|x|^2/32t} +w^{-1}(x)) \leq (w^{-1}(x)w(x)e^{-|x|^2/32} +w^{-1}(x)) $$

But the same passage for $T>1$ I can't see how it can be done. Is it something obvious? I appreciate any clarifications.

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2 Answers 2

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This is probably a typo: probably the author meant $T<1$ here, instead of $T>1$. A bit later, in lines 3- and 2- at the bottom of the same page, it is explicitly assumed that $T$ is $<1$ and furthermore small enough to make the standard contraction argument work.

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It is a typo.

Look at the bottom of the same page: where the author says to choose "$T = T(p,n,l,\|\phi\|_X) < 1$ such that ..."

The author is proving local existence here, and it makes no sense to assume a minimum time of existence.

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  • $\begingroup$ Two almost identical answers, but obviously independent of each other, as one was posted seconds after the other! :-) $\endgroup$
    – Iosif Pinelis
    Commented Dec 7, 2023 at 6:14
  • $\begingroup$ Looks like you beat me by a tiny bit according to the timestamps. :-) $\endgroup$
    – Willie Wong
    Commented Dec 7, 2023 at 6:45
  • $\begingroup$ I'm grateful for both comments. In fact, I was confused, because when we read a work we tend to trust what the author writes, so I thought it was something I wasn't seeing. $\endgroup$
    – Ilovemath
    Commented Dec 7, 2023 at 9:56
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    $\begingroup$ @Ilovemath I assume you know this already, but always be ready for typos, mistakes and wrong arguments, whether or not the work is published or ''trustworthy''. $\endgroup$
    – Hollis Williams
    Commented Dec 7, 2023 at 14:09
  • $\begingroup$ @HollisWilliams In general, I'm very careful about trying to understand all the details when reading a paper, but as this work is quite famous, I believed it could really be something that I wasn't calculating correctly. But the lesson remains! $\endgroup$
    – Ilovemath
    Commented Dec 8, 2023 at 2:20

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