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Theorem (Wiener-Levy). Let $A(\mathbb{T})$ be the Fourier-algebra on the unit circle $\mathbb{T}$. Let $f$ be in $A(\mathbb{T})$ and suppose that $F$ is an analytic function on the range of $f$. Then $F\circ f$ is in $A(\mathbb{T})$.

Let $B(\mathbb{R})$ be the Fourier-Stieltjes algebra on the real line (which is the Fourier transform of $M(\mathbb{R})$, the space of all complex Radon Measures on the real line).

Q. Let $f$ be a function in $B(\mathbb{R})$. Let $F$ be an analytic function on the range of $f$. Is $F\circ f$ in $B(\mathbb{R})$?

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    $\begingroup$ The answer is yes even in a more general context. Please look for functional calculus in books about Banach algebras. $\endgroup$
    – Onur Oktay
    Commented Feb 15, 2022 at 21:45

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