Timeline for Condition for boundedness in stationary phase theorem
Current License: CC BY-SA 3.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Feb 17, 2015 at 15:04 | comment | added | teagut | @WillieWong What I'm interested in is if $f(x,y)$ depends on parameter $y$ and is integrated over $x$. Then what I think the theorem says is if $f(x,y)$ (and its derivatives) is uniformly bounded, then $C$ is bounded and is not a function of $y$ (i.e. not $C(y)$). Then the question is what is the necessary condition for $C$ to be bounded.? | |
| Feb 17, 2015 at 9:07 | comment | added | Willie Wong | The statement can be formalized as "$\forall B\subset C^{k+1}(X)$ a bounded subset, there exists $C$ such that for every $f \in B$ with $\mathrm{Im} f \geq 0$..." In particular the constant $C$ can depend on the choice of $B$. // When you have many quantifiers in your statement, and you ask about "necessary" conditions, you should specify which of the statements you want to remove from the hypotheses. This is especially the case since for every fixed $f$ the estimate is true. // Are you in particular looking for a sequence of $f_k$ such that the best corresponding $C_k$ diverges? | |
| Feb 16, 2015 at 22:55 | history | edited | teagut | CC BY-SA 3.0 |
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| Feb 16, 2015 at 22:50 | history | asked | teagut | CC BY-SA 3.0 |