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May 29, 2023 at 16:42 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 16:40 vote accept Denis Serre
May 29, 2023 at 16:30 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 16:24 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 16:06 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 15:36 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 15:31 comment added Terry Tao I simplified the argument to avoid this degradation of exponents by taking more advantage of the triangle inequality in the far field case.
May 29, 2023 at 15:30 history edited Terry Tao CC BY-SA 4.0
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May 29, 2023 at 10:32 comment added Fedor Petrov @DenisSerre Is it so much obvious that this constant is not $\infty$?
May 29, 2023 at 8:28 comment added Denis Serre @FedorPetrov As I mentioned in the question, the case $n=2$ is obvious, the integral being a constant.
May 29, 2023 at 8:10 comment added Fedor Petrov (for $n=2$ we may bound $R_1^{-1}R_2^4$ from below by $R_2^3$)
May 29, 2023 at 7:49 history edited Denis Serre CC BY-SA 4.0
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May 29, 2023 at 5:15 comment added Fedor Petrov The exponent of $R_1^{(n^2-3)/(n-1)}$ seems to be actually $(n-3)/(n-1)$, but still non-negative for $n>2$.
May 29, 2023 at 0:31 history answered Terry Tao CC BY-SA 4.0