5
$\begingroup$

It is quite easy to show that if $\mu$ is positive finite Borel measure on, say $[0,1]$, and for all $n \in \mathbb{N}$

$$\int_{[0,1]} e^{-nx}\mu(dx)=0$$

holds true, then $\mu=0$. Does this still hold if $\mu$ is signed finite Borel measure?

For positive measure, by applying Holder's inequality it can be shown that $\int_{[0,1]} e^{-rx} \mu(dx)=0$ holds for all positive rational $r$, hence by DCT holds for all $r\geq 0$. By taking differentiation under the integral sign and set $t=0$ we find $$\int_{[0,1]}f(x)\mu(dx)=0$$ holds for all polynomial $f(\cdot)$ and hence for continuous $f(\cdot)$ by Weistrass's approximation theorem. Now Lusin's theorem for Radon measure implies the above equality also holds for measurable $f(\cdot)$, especially for indicating functions so we reach the conclusion.

But for signed measure, it seems Holder's inequality does not work out so I tried to show $\int_{[0,1]}e^{-nx}\mu_+(dx)=0$ and $\int_{[0,1]}e^{-nx}\mu_{-}(dx)=0$ where $\mu=\mu_+-\mu_-$ is the Hahn decomposition. But it did not work out so well.

[Possible solution]Below is one possible solution I gave.

Let $T:\mathbb{R}\supset A \to B\subset \mathbb{R}$ be diffeomorphism,($\mathcal{C}^1$-diffeomorphism seems to be sufficient but smoothness isn't really what matters here), then $$\int_A f(x) \mu(dx)=\int_B f\circ T^{-1} (y) (\mu\circ T^{-1})(dy)$$ holds for all $f$ that is $\mu$-measurable and $\mu\circ T^{-1}$ is the image measure induced naturally by $T$. Now let $T={x \mapsto e^{-nx}}$, and $A=[0,1]$, then $B=[e^{-1},1]$, and $$\int_{[e^{-1},1]}x^n \tilde{\mu}(dx)=0$$ where $\tilde{\mu}=\mu\circ T^{-1}$. Now the conclusion follows immediately by similar argument as stated above.

Improve this question
$\endgroup$
8
  • $\begingroup$ We can use Stone-Weierstass theorem in order to prove that $\int_{[0,1]}f(x)\mu(dx)=0$ holds for all $f$ in the closure of the vector space generated by $x\mapsto e^{-nx}$, $n\in\mathbb N$. We thus have for all $f$ continuous that $\int_{[0,1]}f(x)d\mu^+(x)=\int_{[0,1]}f(x)d\mu^-(x)$ and we can conclude. $\endgroup$
    – Davide Giraudo
    Aug 7, 2013 at 11:52
  • $\begingroup$ @DavideGiraudo Do I understand any continuous function can be 'approximate' in some sense by the linear combination of the family $\{x \mapsto e^{-nx}\}_{n \in \mathbb{N}}$? $\endgroup$
    – zzr
    Aug 7, 2013 at 12:12
  • $\begingroup$ Yes (actually uniformly), that what I mean. $\endgroup$
    – Davide Giraudo
    Aug 7, 2013 at 12:17
  • $\begingroup$ exponentials can be uniformly approximated by polynomials but I am not very sure if the converse holds true? Say,how can $x\mapsto x^2$ be approximated by exponential family $\{x \mapsto e^{-nx}\}$? If the converse holds then it is clear that $\{x \mapsto e^{-nx}\}$ can be applied G-S process to form an CONS in $L^2$ but I am not sure yet. $\endgroup$
    – zzr
    Aug 7, 2013 at 12:31
  • $\begingroup$ @RoyHan: it is a consequence of the quoted theorem. But you can also use the classic Weierstrass approximation theorem (uniform approximation by polynomials) applied to the continuous function $f(-\log t )$ on $[1/e, 1]$. If $P$ is uniformly close to $f\circ(-\log)$ on $[1/e, 1]$, then $P(e^{-x})$ is uniformly close to $f(x)$ on $[0,1]$. $\endgroup$
    – Pietro Majer
    Aug 7, 2013 at 12:37

1 Answer 1

Reset to default
5
$\begingroup$

This is an old Theorem of M. Lerch, 1903. For more info see Theorem 6.2, Chap.II, $\S 6$ of

D.V. Widder: The Laplace Transform, Princeton University Press 1941 (or Dover 2010).

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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