Timeline for A property of "Schwartz" quadratic forms
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 8, 2012 at 14:07 | vote | accept | Vanessa | ||
| Dec 2, 2010 at 18:22 | comment | added | Willie Wong | (or if you are working in $d$ dimensions, $-a + d$. ) | |
| Dec 2, 2010 at 18:21 | comment | added | Willie Wong | Ack, the right hand side should be $-a + 1$ in the exponent, not $-a$. | |
| Dec 2, 2010 at 18:19 | comment | added | Willie Wong | No, which is why I said computation is required. You need to use the fact that for a function $K(x,y)$ that is Schwartz, its Schwartz seminorms in $x$ for fixed $y$ is a rapidly decreasing function of $y$. This allows you to say that first $\int K(x,y)f(x-t) dx$ is Schwartz in the variables $t$ and $x$, and then also the claim of the decay of $\tilde{h}$ on the diagonal. Here you need some quantitative estimates on how the Schwartz seminorms behave under convolution and evaluation, which can be derived from the estimate $\int (1 + |x|^2)^{-a} (1 + |x-t|^2)^{-a} dx \lesssim (2 + t^2)^{-a}$. | |
| Dec 2, 2010 at 17:06 | comment | added | Vanessa | SORRY, you are right. OK, so h~(t,s) is smooth of slow growth in both variables and Schwartz in t for any fixed s. So what? $e^{-(t-s)^2}$ also has these properties and yet is has no rapid decay on the diagonal. | |
| Dec 2, 2010 at 16:13 | comment | added | Willie Wong | Did you change notation half way through? Please re-read your question: $f$ is Schwartz, $g$ is a tempered distribution. So $Kf$ is Schwartz, but not $Kg$... | |
| Dec 2, 2010 at 15:53 | comment | added | Vanessa | Sounds good. I apologize for my stupidity but I didn't entirely follow it through. h~ is smooth of slow growth: agreed. h(t)=h~ (t,t) : agreed. It suffices to show h has rapid decay: agreed. OK, now we consider h~ (t,s) for fixed s. Since h~(s,t)=∫f(x−t)K(x,y)g(y−s)dxdy for fixed s we get essentially the convolution of f with a Schwartz function Kg. Thus we get a smooth function of slow growth. Why does it even have a supremum? Maybe you meant to consider supremum w.r.t. s for fixed t? | |
| Dec 2, 2010 at 15:23 | history | answered | Willie Wong | CC BY-SA 2.5 |