Showing that a function cannot be at once Gabor, smooth, and compactly supported.

Question

From a textbook on Harmonic Analysis:

Attempt: Let $g$ be Gabor, smooth, and compactly supported. Then $g$ is Schwartz, and hence $\widehat{g}$ is also Schwartz, and hence Theorem 9.11 is contradicted.

Is this correct?

Answer 1

Yes, your reasoning is correct.

One can also avoid $\hat g$ here: if $g$ is $C^1_c$, then both $\int x^2|g(x)|^2\,dx$ and $\int |g'(x)|^2\,dx$ converge.