Let $\mathcal{A}(\mathbf{T})$ be the Banach algebra of continuous complex-valued functions on the unit circle with absolutely convergent Fourier series. Then Wiener's lemma states that if $f \in \mathcal{A}(\mathbf{T})$ and $f(z) \neq 0$ everywhere on $\mathbf{T}$, then $1/f \in \mathcal{A}(\mathbf{T})$. Said differently, the lemma states that the algebra $\mathcal{A}(\mathbf{T})$ is inverse-closed in $\mathcal{C}(\mathbf{T})$, the algebra of continuous functions on the unit circle.
Hulanicki's lemma is a general result about inverse-closedness. It states that if $A \subseteq B$ are Banach $*$-algebras with common involution and multiplicative identity and $B$ is symmetric, then the following are equivalent:
$A$ is inverse-closed in $B$.
$r_A(a) = r_B(a)$ for all self-adjoint $a \in A$, where $r$ denotes the spectral radius with respect to the algebra in the subscript.
I have been told that Hulanicki's lemma is a generalization of Wiener's lemma, but I can't see how Wiener's lemma follows from Hulanicki's lemma. If I could show that $r_{\mathcal{A}(\mathbf{T})}(f) = r_{\mathcal{C}(\mathbf{T})}(f)$ for all $f=\overline{f}$ in $\mathcal{A}(\mathbf{T})$, then that would solve things, but this doesn't seem easy. Actually, establishing this formula is part of one proof I have seen of Wiener's lemma.
Does anyone know in what sense Hulanicki's lemma can be seen as a generalization of Wiener's lemma, if this is possible at all?