# Quantitative Weierstrass Approximation and Paley-Wiener for the Laplace Transform

Question: Suppose $a(x,y)\in C^\infty([0,1]\times [0,1])$ and suppose $$\sup_{\lambda>1} \bigg|\lambda\int_0^1 e^{\lambda x} a(x,1/\lambda)dx\bigg|<\infty.$$ Is $a(x,0)=0$, $\forall x\in[0,1]$?

Comments: Here are some side comments about a special case. Suppose $a$ is independent of $y$, then the result is true. Indeed, something slightly stronger is true. If $$\sup_{\lambda>1} \bigg|\int_0^1 e^{\lambda x}a(x)dx\bigg|<\infty,$$ then $a\equiv 0$. The above can be seen as a quantitative version of the Weierstrass approximation theorem. Indeed, if $\int_0^1 e^{\lambda x} a(x) dx=0$, $\forall \lambda\geq 1$, then the Weierstrass approximation theorem shows $a(x)=0$.

This quantitative Weierstrass approximation can be proved in several ways. None of the ways I know of seem to generalize to the more general question. Here is a sketch of one such way. Let $F(\lambda)=\int_0^1 e^{\lambda x} a(x) dx$. $F$ extends to an entire function. For $\lambda>0$, $F$ is bounded, by our assumption. For $\lambda$ purely imaginary, $F$ is bounded. As $\lambda$ tends to $-\infty$ along the negative reals, $F(\lambda)$ tends to $0$. Also $|F(\lambda)|\leq C exp(|\lambda|)$, $\forall \lambda$. Since $F$ is bounded on the coordinate axes, and because of the above bound, the a Phgragmen-Lindelof theorem (applied to each quadrant) shows $F$ is a bounded entire function, and therefore constant. Since $F(\lambda)$ tends to $0$ as $\lambda$ tends to $-\infty$, $F=0$, and it follows that $a\equiv 0$. Perhaps an argument like this might work for the more general question if $a(x,y)$ were real analytic in $y$, but I really want to know the answer for $a(x,y)\in C^\infty$.

In another direction, if we let $b(x)=a(1-x)$ then the above special case can be re-written as $$\bigg|\int_0^1 e^{-\lambda x} b(x) dx\bigg| \leq C e^{-\lambda}, \quad \lambda>1.$$ If this holds, then $b\equiv 0$. In this case, we can see this as a Paley-Weiner theorem for the Laplace transform: for a (nice) function on $[0,\infty)$ to be supported outside $[0,1]$, it is necessary and sufficient that its Laplace transform fall off like $e^{-\lambda}$.

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Unless I misunderstand the question, the answer is that $a(x,0)$ can be pretty much anything it wants.

Take any smooth $f(x)$ supported on $[0,1-\delta]$. Put $$a(x,y)=f(x)-[y(e^{1/y}-1)]^{-1}\int_{0}^{1}f(t)e^{t/y}\\,dt.$$ It looks like a $C^\infty$ function to me because the exponent $e^{\delta/y}$ in the denominator coming from the extra length in the support of the constant function is stronger than any inverse power of $y$ coming from differentiations for small $y$. However, now the integral in question is identically $0$.

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The answer to your question is "no" in a very strong sense: I will construct $a$ such that $a(x,0)\neq 0$ and $$\int_0^1e^{\lambda x}a(x,1/\lambda)dx\equiv 0.$$ Begin with $a(x)$ infinitely smooth, not identically $0$, supported by $[0,1]$. Set $$F(\lambda,t)=(1-\lambda t)\int_0^1e^{\lambda x}a(x)dx,$$ where $t\geq 1$ is a parameter.
This is an entire function of exponential type with indicator diagram $[0,1]$, it decreases in both imaginary directions faster than any negative power of $|\lambda|$, and has a zero at the point $1/t$. Now consider its Laplace transform: $$f(z,t)=\int_0^\infty e^{-\lambda z}F(\lambda,t)d\lambda,$$ where the integral is over a ray. This function $f$ is analytic in $C\backslash[0,1]$, zero at infinity. One can easily see that the boundary values from above and from below on $[0,1]$ make two smooth functions. Then let $a(x,t)$ be the difference of these two boundary functions, divided by $2\pi i$. Then we have $$F(\lambda,t)=\int e^{x\lambda}a(x,t)dt,$$ by the Laplace (Borel) inversion formula, and $F(\lambda,1/\lambda)\equiv 0$. The reference for Laplace and Borel transforms is the book of Levin, Distribution of roots of entire function.