**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 $e^{-\lambda}$.