This question stems out of: which sequences $(a_n)_{n\in\mathbb{Z}}$ of complex numbers have the property that if there exists a continuous function $f$ on the circle with Fourier coefficients $b_n$, then there also exists a continuous function with Fourier coefficients $a_nb_n$? First we note that $(a_n)$ must be bounded, or else we could take a subsequence $a_{m_n}$ with $|a_{m_n}|\geq 2^n$, and define $f(x)=\sum_{n\geq 0}\frac{1}{a_{m_n}} e^{im_n x}$ and then the hypothesis would give us a continuous function whose Fourier coefficients do not tend to zero, which cannot happen.

The natural question arises: then does $(a_n)$ being bounded suffice? What can be said about the $(a_n)$ that do work? One thing we can note is that any such sequence $a=(a_n)$ determines a continuous linear operator $T_a:C(S^1)\to C(S^1)$ taking $f$ to the function $T_a(f)$ with $n$th Fourier coefficient $a_n\hat{f}(n)$. Thus, we get a commutative Banach subalgebra $A\subseteq \mathcal{L}(C(S^1),C(S^1))$ whose elements are the $T_a$ for sequences $a$ such that the hypothesis holds. There is a Banach algebra homomorphism $F:A\to \ell^{\infty}(\mathbb{Z})$ taking $T_a$ to $a$, which is contractive.

In many of the basic cases, we even have $\|T_a\|=\|a\|$, so this prompts the questions: is $F$ an isomorphism? If $F$ isometric? Is $A$ even a $C^*$-algebra?

There are (non-continuous) $S^1$-actions by isometries on $A$ and $\ell^{\infty}(\mathbb(Z))$ underlying everything here, where $e^{iz}$ acts by taking a sequence $(x_n)$ to $(e^{inz}x_n)$, so perhaps there’s some nice way to use some $S^1$-equivariant generators of $\ell^{\infty}(\mathbb{Z})$ to see that $F$ is surjective, if it is at least. Since $\ell^{\infty}(\mathbb{Z})\simeq C(\beta \mathbb{Z})$, perhaps there’s some way to utilize Stone-Weierstrass here. This would let us show the image of $A$ is dense so long as for any infinite set $S$, the sequence $(\delta_{n\in S})$ is in the image of $A$.