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Suppose that $f(t)$ is a square-integrable, band-limited function, i.e. the Fourier transform $\hat f$ has compact support.

Problem: Under which assumptions on a function $g(t)$ is the map $h(t) := e^{ig(t)}f(t)$ band-limited, i.e. $\hat h$ has compact support.

My idea would be to use the Paley-Wiener theorem which says $f$ extends to an entire function of exponential type (in particular, it has order less or equal than 1). Now one has to find assumptions on $g$ so that $h$ is again of exponential type (maybe $g$ must be a linear function then?).

Thanks in advance for any help!

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A function is entire and of exponential type if and only if its Fourier transform has bounded support (This is Paley-Wiener theorem). Since your $f$ belongs to this class, then, assuming that $g$ is also entire, $h=e^gf$ will belong to this class if and only if $g$ is linear.

If you do not want to assume that $g$ is entire, since $h$ and $f$ are entire, $e^g$ is meromorphic, with poles contained in the zero set of $f$, and of exponential type in the sense that $T(r,e^g)=O(r),$ where $T$ is the Nevanlinna characteristic. This is a complete characterization: take any meromorphic function $F$ of exponential type whose poles belong to the zero set of $f$, and take $g=\log F$.

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  • $\begingroup$ Great, thanks a lot for the answer! I have a short follow-up question and was wondering if you have a quick answer to it: Suppose we further want that the modulus of $f$ and $h$ agree on the real line, i.e. $|f(x)| = |h(x)|$ for all real $x$. Does this assumption restrict $g$ even more (without making a prior holomorphy assumption on $g$) $\endgroup$
    – J. Swail
    Commented Dec 11, 2021 at 18:22
  • $\begingroup$ @J. Swail: not much more: it only means that $g$ is pure imaginary on the real line. It restricts $g$ but not too much. The most important restriction on $g$ comes from the condition that poles of $e^g$ are among zeros of $f$. For example, if you know that all zeros of $f$ are real, and $g$ is pure imaginary then $g$ must be linear. $\endgroup$
    – Alexandre Eremenko
    Commented Dec 11, 2021 at 18:44

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