My question is if it is possible to find a compactly supported smooth function $\varphi:\mathbf{R}\to \mathbf{R}$ s.t. the following integration $\int_{\mathbf{R}}\varphi(t)e^{itx}e^{tx}dt$ stays bounded for $x\in \mathbf{R}$.

Some more thoughts: for example, if we take $\varphi$ to be supported on the interval $[1,2]$, and take $x$ to be some large number, then the term $e^{tx}$ hehaves like $e^x$, which might(?) be able pulled out from the integration to obtain $e^{x}\int_{\mathbf{R}}\varphi(t)e^{itx}dt$, thus in order for the last expression to be bounded, we need $\int_{\mathbf{R}}\varphi(t)e^{itx}dt$ to have exponential decay! This is not possible as if it is true, we know $\varphi$ will be real analytic from the Fourier inverse formula, which will then be identically zero from the fact that it has compact support.

I'm now really hoping that before pulling out the term $e^{tx}$, the term $e^{itx}$ will generate enough cancellation!

  • $\begingroup$ If you describe what you have tried others may be able to use that information to provide a hint or answer that is based upon what you know. $\endgroup$
    – Jay
    Dec 18, 2013 at 14:22
  • $\begingroup$ @Jay Thanks, I didn't add this at the very beginning as I thought it might be misleeding. $\endgroup$
    – Shaoming
    Dec 18, 2013 at 15:10

1 Answer 1


I'll add a few details to the answer by Michael Renardy on MathOverflow. The function $$F(z)=\int_{\mathbb R}\varphi(t)e^{itz}\,dz$$ is

  1. Entire (holomorphic in $\mathbb C$), with growth bound $|F(z)|\le Ce^{A|z|}$ coming from $|e^{itz}|\le e^{A|z|}$ where $A=\sup\{|t|:\varphi(t)\ne 0\}$.
  2. Bounded on the real line, because for real $z$ we have $ |e^{itz}|=1 $.
  3. Bounded on the line $\operatorname{Re}z+\operatorname{Im} z=0$, by assumption.

The Phragmén–Lindelöf principle applies to $F$ in the sectors formed by the lines in 2 and 3. Indeed, the exponential growth bound 1 suffices in any sector of opening angle $<\pi$, and we have angles $\pi/4$ and $3\pi/4$. The principle says that $F$ is bounded in each sector; hence, bounded on $\mathbb C$; hence constant by Liouville's theorem. But then $F\equiv 0$, because the Fourier transform tends to $0$ as $z\to\infty$ along the real axis.


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.