## Using a quadratic kernel instead of a linear kernel in the Laplace transform

Suppose $f$ is a bounded continuous function on $[0,\infty)$ such that $\int_0^\infty f(t) \exp(-xt) \: dt \rightarrow 0$ as $x \rightarrow 0^+$. Does it follow that $\int_0^\infty f(t) \exp(-xt^2) \: dt \rightarrow 0$ as $x \rightarrow 0^+$? Is the reverse implication true?

I suspect that the answer is "no" in both cases, so here's my real (although vague) question: is there a notion of regularity for $f$ (along the lines of the notion of almost-periodicity) such that the two limit-assertions imply each other when $f$ is regular?

-

There exists a Tauberian theorem of the following form. Suppose that $a\in L^1_{loc}(\mathbb{R}_{>0})$ and for any $x>0$ the integral

$$G(x):=\int_0^\infty e^{-x^2t^2} a(t) dt$$

exists and satisfies

$$\sup_{x>0}|G(x)|<\infty,$$

and

$$\lim_{x\searrow 0} G(x) = A\in\mathbb{R}.$$

The function $G(x)$ is called the Gauss-Weierstrass transform of $a$. Suppose also that the integrals

$$L(x)=\int_0^\infty e^{-xt} a(t) dt$$

exist for any $x>0$. Then

$$\lim_{x\searrow 0} L(x)= A= \lim_{x\searrow 0} G(x).$$

Let us point out that if $a\in L^1(\mathbb{R}_{>0})$, then the integrals $G(x)$ and $L(x)$ exist for any $x$ and

$$A=\int_0^\infty a(t) dt.$$

For a proof see Chap.I, Sec 14, Thm. 21 of the book

S. Bochner, K. Chandrasekharan: Fourier Transforms, Ann. Math. Studies, vol. 19, Princeton University Press, 1949.

Addendum It seems that one of Wiener's Tauberian theorem. (see J. Korevaar: Tauberian Theory. A Century of developments, Theorem 5.1, Chap II).

-
 @Liviu: Thanks, but I am confused. What is $s$? Also, does $L_{loc}^1$ mean locally integrable in the sense of being integrable over every finite interval, or is some sort of uniformity required? – James Propp Mar 3 at 4:34 That $s$ was mant to be an $x$ I fixed the typo. $L^1_{loc}$ means integrability on compacts, no uniformity assumed. – Liviu Nicolaescu Mar 3 at 10:24