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 $e^{-\lambda}$.


2 Answers 2


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 $0$.


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"?

  • $\begingroup$ Sorry, I did not notice that Fedja posted his answer while I was writing my own:-) $\endgroup$ Oct 20, 2012 at 19:04
  • $\begingroup$ Thank you for your answer, which I will have to think about more! The Weierstrass approximation theorem I meant (perhaps Stone-Weierstrass) was that a closed subalgebra of the continuous functions on a compact Hausdorff space, which contains a nonzero constant function, and which separates points is the entire algebra of continuous functions. This can be used in my first comment to show that if all the integrals are actually 0, then a must be 0. That's why I was calling the special case in the comment a quantitative Weierstrass approximation theorem; here take a(x)dx to be a measure,instead $\endgroup$ Oct 20, 2012 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.