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It depends on what you mean by "comes from the Fourier transform of a continuous function." 1) If you mean "Any $L^2$ function with Fourier coefficients support on $E$ must be continuous," then a set $E$ will have this property if and only if it is finite. The fact that finite sets have this property is obvious. Conversely, assume that $E$ is infinite and ...

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The Fourier transform of a function with compact support is automatically holomorphic. This support can be deduced from growth conditions on the former, at least in the case when it is a euclidean ball and the function is square integrable. This is the celebrated Payley-Wiener theorem, upon which there is a comprehensive Wikipedia article for starters.

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Since $M^{1,1}(\mathbb R)$ is a Banach algebra, $P(f(t))$ is in $M^{1,1}(\mathbb R)$ whenever $\|f\|_{M^{1,1}}$ is smaller than the radius of convergence of the series $P$. Here you have to use a sub-multiplicative norm $\|\;\|_{M^{1,1}}$. On a Banach algebra you can always change to an equivalent sub-multiplicative norm.

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The Schwartz space itself, $\mathscr S(\mathbb R^n)$, can be viewed as a subspace of smooth functions defined on the sphere $\mathbb S^n$, the unit sphere of $\mathbb R^{n+1}.$ We recall that $$\mathscr S(\mathbb R^n)=\{\phi\in C^\infty(\mathbb R^n),\forall \alpha, \beta\in \mathbb N^n, x^\alpha\partial_x^\beta \phi\in L^\infty(\mathbb R^n)\}.$$ It is easy ...

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