Timeline for A property of "Schwartz" quadratic forms
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2012 at 14:07 | vote | accept | Vanessa | ||
| Dec 2, 2010 at 21:01 | comment | added | Vanessa | In fact, we can set $u(x)=sup_y{|K(x,y)|}^{1/2}$ $v(y)=sup_x{|K(x,y)|}^{1/2}$ This is a bit cheating since v(y) is not smooth and thus cannot be integrated against a tempered distribution. However, it appears quite certain that one can always choose a Schwartz function larger than any given continuous rapidly decreasing function. | |
| Dec 2, 2010 at 17:42 | comment | added | Vanessa | OK, I have another guess. $$|K(x,y)|<=u(x)v(y)$$ for some rapidly decreasing u and v (I'm not sure how to show this yet but it sounds plausible). Now we can run the argument again with inequalities: $$|h(t)|=|∫f(x−t)K(x,y)g(y−t)dxdy|<=∫|f(x−t)K(x,y)g(y−t)|dxdy<=∫|f(x−t)|u(x)v(y)|g(y−t)|dxdy=[∫|f(x−t)|u(x)dx][∫|g(y−t)|v(y)dy]$$ The 1st factor is rapidly decreasing whereas the 2nd factor of slow growth hence the product is rapidly decreasing | |
| Dec 2, 2010 at 17:39 | history | edited | Vanessa | CC BY-SA 2.5 |
fixed small error
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| Dec 2, 2010 at 15:44 | comment | added | Willie Wong | You are mostly there. Hint: first show that $\tilde{K}_t(y) = \int f(x-t)K(x,y) dx$ is (a) a Schwartz function in $y$, and (b) has its Schwartz seminorms a rapidly decreasing function of $t$. This should follow once you take an upper envelope for $K$. | |
| Dec 2, 2010 at 15:31 | history | edited | Vanessa | CC BY-SA 2.5 |
Some thoughts so far
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| Dec 2, 2010 at 15:23 | answer | added | Willie Wong | timeline score: 2 | |
| Dec 2, 2010 at 11:49 | history | edited | Willie Wong | CC BY-SA 2.5 |
fixed typo in title
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| Dec 2, 2010 at 11:43 | history | edited | Vanessa | CC BY-SA 2.5 |
added a missing condition
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| Dec 2, 2010 at 11:27 | history | edited | Andrey Rekalo | CC BY-SA 2.5 |
TeX
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| Dec 2, 2010 at 11:24 | history | asked | Vanessa | CC BY-SA 2.5 |