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2 votes
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Inequality with decreasing rearrangement and non-decreasing function

$\newcommand\la\lambda\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer to this question is yes. Indeed, let $h:=f^*/g$. Then $h$ is a nonincreasing function and the inequality in question ...
6 votes
Accepted

Inequality with decreasing rearrangement function

No (if $c$ cannot depend on $f^*$ or $g$). Indeed, let $h:=f^*/g$. Then $h$ can be any positive function and the inequality in question can be rewritten as $$lhs:=\Big(\int_0^\infty h(s)^{p'}ds\Big)^{...
6 votes

How was Claim 5 in "A non-linear generalisation of the Loomis–Whitney inequality and applications" thought up?

As Dan Romik points out, this technique is relatively old folklore by now. Terry Tao calls this the "tensor power trick" in a blog post dedicated to the subject; the two elementary ...
6 votes

How was Claim 5 in "A non-linear generalisation of the Loomis–Whitney inequality and applications" thought up?

An instance of this idea of killing an unwanted factor in an inequality by considering an inequality for $k$-th powers and then taking the limit as $k\to\infty$ appears in the proof of the Kraft-...
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6 votes
Accepted

Is the extension (dual restriction) operator on any smooth hypersurface a solution to some PDE?

Your question was basically answered by David Roberts in the comments, but let me write a few more words. Given a constant coefficient linear differential operator of degree $N$ $$ L = \sum_{|\alpha| \...
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0 votes

How to reduce the compact support to the case of small diameters in Tao's "A sharp bilinear restriction estimate for paraboloids"

Stupid me! I posted my own answer anyway for other fledglings like me or for anyone to double check my proof! ($f_{1,i}$ is $f_1$ restricted to a small compact subset indexed as $i$. Similarly to $f_{...
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