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 $e^{-\lambda}$.
 A: 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 $0$.
A: 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.
And please tell me, what do you call "Weierstrass theorem"? 
