I want to show that the maximal ideal space of the Wiener algebra $W$ is $ \{ M_z : z \in \mathbb{T} \}$ where $M_z = \{ g \in W : g(z)=0 \}$
Could you please help me?
2026-03-31 08:12:38.1774944758
Show that the maximal ideal space of the Wiener algebra $W$ is $ \{ M_z : z \in \mathbb{T} \}$
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Hints:
The Wiener algebra is a commutative Banach algebra.
To see that the $M_z$ is a maximal ideal, write it as the kernel of a character.
To see that every maximal ideal $M$ is of the form $M_z$ for some $z$, consider the image of the identity function under the quotient map $\phi\colon W\to W/M\cong\mathbb C$.