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Consider $K(x, y)$, $f(x)$ Schwartz functions and $g(y)$ a tempered distribution. Suppose $$K(x, y) = K(y, x)$$ Define

$$h(t) = \int f(x - t) K(x, y) g(y - t) dx dy$$

It appears to me $h(t)$ is a Schwartz function. I'd be glad if anyone can prove / disprove it.

Thx!

Some thoughts so far:

The assumption of symmetry on K is probably redundant. If we take K to be $$K(x, y) = u(x)v(y)$$ for u, v Schwartz, we get $$h(t) = [\int f(x - t)u(x)dx][\int g(y - t)v(y)dy]$$ Now, the 2nd factor is probably smooth of slow (at most polynomial) growth whereas the 1st factor is Schwartz. $h(t)$ is thus Schwartz. Linear combinations of such "factorized" kernels are dense in the entire Schwartz space, but I still don't see how to deduce the claim from this.

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  • $\begingroup$ 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 $K$. $\endgroup$
    – Willie Wong
    Dec 2, 2010 at 15:44
  • $\begingroup$ 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 $\endgroup$
    – Vanessa
    Dec 2, 2010 at 17:42
  • $\begingroup$ 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. $\endgroup$
    – Vanessa
    Dec 2, 2010 at 21:01

1 Answer 1

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Let $H(x,y) = f(-x)\otimes g(-y)$ be a tempered distribution on the product space. Then it is well-known that the function $\tilde{h}(t,s) = H*K(t,s) = \int H(t-x,s-y)K(x,y)$ is in $C^\infty \cap \mathcal{S}'$. (See, e.g. pg 25 of Stein-Weiss, Introduction to Fourier Analysis on Euclidean Spaces).

Noting that $h(t) = \tilde{h}(t,t)$, it suffices to show that $h$ has rapid decay. This follows from the fact that the convolution of a tempered distribution with a Schwartz function has at most polynomial growth (ibid) so that $\sup_t |\tilde{h}(t,s)| \leq A (1+ |s|)^M$ for some $A$ and $M$. So a small bit of computation tells you that $\tilde{h}(t,s)$ is bounded by the product of a rapidly decreasing envelope in $t$ and a polynomial in $s$. Therefore on its diagonal it is still rapidly decreasing.

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  • $\begingroup$ 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? $\endgroup$
    – Vanessa
    Dec 2, 2010 at 15:53
  • $\begingroup$ 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 $Kg$... $\endgroup$
    – Willie Wong
    Dec 2, 2010 at 16:13
  • $\begingroup$ 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. $\endgroup$
    – Vanessa
    Dec 2, 2010 at 17:06
  • $\begingroup$ 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 $\int (1 + |x|^2)^{-a} (1 + |x-t|^2)^{-a} dx \lesssim (2 + t^2)^{-a}$. $\endgroup$
    – Willie Wong
    Dec 2, 2010 at 18:19
  • $\begingroup$ Ack, the right hand side should be $-a + 1$ in the exponent, not $-a$. $\endgroup$
    – Willie Wong
    Dec 2, 2010 at 18:21

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