# Under which assumptions is $e^{ig(t)}f(t)$ of exponential type if $f$ is of exponential type?

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!

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 $$g=\log F$$.
• 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 $g$) Dec 11, 2021 at 18:22
• @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. Dec 11, 2021 at 18:44