Let $F:\mathbb C^n\longrightarrow \mathbb C$, be an entire function of exponential type, that is $F$ is holomorphic on $\mathbb C^n$ and there exist constants $C, A$ such that $$ \vert F(z)\vert\le C e^{A\vert z\vert}. \tag{1}$$ With $f$ standing for the restriction of $F$ to $\mathbb R^n$, and assuming that $f$ is a bounded function, the Paley-Wiener Theorem implies that the Fourier transform $\hat f$ is compactly supported. Does that imply that $f$ is a constant ?
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No. If $$F(z)=\frac{\sin z}{z}$$ then it is of exponential type $1$, and bounded on the real line. The Fourier transform $\hat{f}$ is the characteristic function of the interval $[-1,1]$.
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$\begingroup$ Many thanks, you are perfectly right. Taking tensor products provides counterexamples in higher dimension. $\endgroup$Bazin– Bazin2025-12-18 15:22:17 +00:00Commented yesterday
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$\begingroup$ In fact for any nonzero bounded, compactly supported function, its Fourier transform is bounded on the real line and of exponential type $1$ but nonconstant (since the delta function is not a function or at least not a bounded function) and thus gives a counterexample. $\endgroup$Will Sawin– Will Sawin2025-12-18 18:41:28 +00:00Commented yesterday