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Suppose $f \in C^{\infty}_c((-1,1))$ and assume that there exists constants $a,b>0$ such that $$ \bigg|\int_{\mathbb R} f(t) \,e^{\tau t^2+i\tau t}\,dt\bigg| \leq a\,e^{-b|\tau|},$$ for all $\tau \in \mathbb R$.

Does it follow that $f=0$?

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2 Answers 2

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The expression inside absolute value sign in the LHS is an entire function of $\tau$ (since it can be differentiated under the integral sign and derivative exists for all complex $\tau$). Your inequality shows that it tends to zero as $\tau\to\infty$ therefore it is zero by Liouville's theorem. Then Fourier inversion says that $f=0$.

To address some comments. The function is entire of exponential type. Since the indicator is negative for $\arg\tau\in(0,\pi)$ is must be $0$ by Phragmen-Lindelof (for all complex $\tau$).

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    $\begingroup$ But why does it tend to zero for non-real $\tau$? $\endgroup$ Jun 30, 2022 at 7:27
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    $\begingroup$ @FedorPetrov Fragmen-Lindelof. The exponential decay on the line plus the exponential growth outside immediately give an exponential decay in a small angle around the real line, after which everything becomes clear... $\endgroup$
    – fedja
    Jun 30, 2022 at 10:33
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    $\begingroup$ @FedorPetrov I'm more concerned about the "Fourier inversion" part: what Alexandre meant here still escapes me :-( $\endgroup$
    – fedja
    Jun 30, 2022 at 11:59
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    $\begingroup$ @fedja Denote $t^2+it$ by a new variable? $\endgroup$ Jun 30, 2022 at 13:48
  • $\begingroup$ @FedorPetrov How? it is $\tau$ that runs over the complex plane, not $t$. $\endgroup$
    – fedja
    Jul 2, 2022 at 12:25
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Edit: I'm undeleting this now, though it may not be of much interest any more. I thought the argument was incomplete, but actually the missing step is quite easy, given a fact about density of polynomials on compact subsets of $\mathbb C$ (Mergelyan's theorem).

Thank you to fedja for pointing this out; see his comments to the linked question.


I think I now know how to provide all the details in Alexandre's strategy.

Consider $F(z)=\int f(t)e^{zt^2+izt}\, dt$. This is an entire function of exponential type. Its indicator function $h(\alpha)=\limsup_{r\to\infty} (1/r)\log |F(re^{i\alpha})|$ satisfies $h(0), h(\pi)\le -b$, by assumption. This is impossible for an $F\not\equiv 0$ since always $h(\alpha)+h(\alpha+\pi)\ge 0$ (see, for example, Levin's book for this).

So $F\equiv 0$ and also $$ F^{(n)}(0) = \int_{-1}^1 f(t) t^n(t+i)^n\, dt = 0 $$ for all $n\ge 0$. Since $\gamma(t)=t(t+i)$ is injective on $[-1,1]$, this implies that $f=0$.

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    $\begingroup$ Shouldn’t you use the complex version of the Stone-Weierstrass theorem? In that case I am not sure we have a unit algebra here since the conjugation of t(t+i) polynomials are not there $\endgroup$
    – Ali
    Jun 30, 2022 at 16:54

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