Timeline for Under which assumptions is $e^{ig(t)}f(t)$ of exponential type if $f$ is of exponential type?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 12, 2021 at 11:16 | vote | accept | J. Swail | ||
| Dec 11, 2021 at 18:44 | comment | added | Alexandre Eremenko | @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 18:22 | comment | added | J. Swail | 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 16:34 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
added 118 characters in body
|
| Dec 11, 2021 at 16:26 | history | edited | Alexandre Eremenko | CC BY-SA 4.0 |
added 44 characters in body
|
| Dec 11, 2021 at 16:15 | history | answered | Alexandre Eremenko | CC BY-SA 4.0 |