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

This is a modification of a previous question.

Question: Suppose $a(s)\in C^\infty([0,1])$ and $H(s,x)\in C^\infty([0,1]\times [0,1])$ with $H(s,x)>0$, $\forall s,x\in [0,1]$. Suppose, $$\sup_{\lambda>1} \bigg|\lambda \int_0^1 \exp\big( \lambda\int_s^1 H(r,1/\lambda) \: dr \big) a(s) ds \bigg|<\infty.$$ Must $a(1)=0$? Must $a\equiv 0$?

Comments: If $H(s,x)$ is independent of $x$, then $a$ must be zero. Indeed, define the entire function $$F(\lambda)=\int_0^1 \exp\big( \lambda\int_s^1 H(r) \: dr \big) a(s) ds.$$ By assumption $F$ is bounded for $\lambda>0$. $F$ is clearly bounded on the imaginary axes, and clearly tends to zero as $\lambda\rightarrow -\infty$. A Phragmen-Lindelof theorem (applied to each quadrant) shows that $F$ is bounded. As a bounded, entire function it must be constant. Since $F$ tends to zero as $\lambda\rightarrow -\infty$, we have $F\equiv 0$. From here, it is easy to conclude that $a\equiv 0$. A similar proof, using Laurent series, applies to the case when $H$ is real analytic in $x$. So the main purpose is to understand the case when $H$ is not real analytic.

For comments on the choice of title for this post, and for an example showing that a particular stronger result is not true, see the previous question.

-