Zero Sets of Gelfand transform in Banach Algebras

Question

I am reading the book Fourier and Fourier-Stieltjes Algebras on Locally Compact Groups (2018) and I am stuck on the proof that the sets: $$\left[ I(E) = \{a \in A \mid \hat{a}(\phi) = 0, \, \forall \phi \in E\} \right]$$ and $$\left[ j(E) = \{a \in A \mid \hat{a} \, \text{has compact support disjoint from} \, E\} \right]$$ are closed ideals of $A$. Here, $E$ is a closed subset of $\sigma(A)$, and $A$ is a semisimple commutative Banach algebra. The function $\hat{a}$ is the Gelfand transform.As far as the proof of the ideal property goes, I am fine, but my problem is that I don't understand in which topology and how to prove that the sets are closed. I thought that infinity norm but i stucked.

Answer 1

The closedness of $I(E)$ (in norm topology) can be easily verified by observing that $$I(E) = \bigcap_{\phi \in E} \ker \phi,$$ and $\ker \phi$ is closed for all $\phi \in \mathrm{Spec} A$ (the Gelfand spectrum). The other ideal $j(E)$ usually is not closed and I was not able to find the opposite statement in the book you are reading. Moreover, the authors of the book are interested in conditions which imply $I(E) = \overline{j(E)}$ (where bar means closure). Obviously, if $j(E)$ was closed, there would be no need in taking the closure.

Let me give an example where $j(E)$ is not closed. Let $A = C(X)$ - the algebra of continuous functions on $X$, where $X$ is a compact Hausdorff space. We identify $\mathrm{Spec} A$ with $X$ in the usual way. Let $x \in X$ be a point and let $E = \{x\}$. I claim that if $x$ is not isolated and $E$ is a $G_\delta$-set then $j(E)$ is not closed. If $X$ is metrizable, any non-isolated point $x \in X$ shall suffice (since all closed subsets in $X$ are $G_\delta$ provided $X$ is metrizable).

At first, I claim that $\overline{j(E)} = I(E)$ (here we don't need additional conditions on $x$ and $E$, only the fact that $E = \{x\}$). Since $I(E)$ is closed and, obviously, it contains $j(E)$, we conclude $\overline{j(E)} \subset I(E)$. To prove the opposite inclusion consider $f \in I(E)$ (i.e. $f \in A$ such that $f(x) = 0$). Fix arbitrary $\varepsilon > 0$ and find $U$, a neighborhood of $x$, such that $|f(y)| < \varepsilon$ for $y \in U$. Then find a function $\phi:X \rightarrow [0,1]$ such that $\phi$ is identically zero in some neighborhood of $x$ and $\phi(y) = 1$ for all $y \in X \setminus U$ (use Urysohn's lemma). Clearly, $\|\phi f - f\| < \varepsilon$ and $\phi f \in j(E)$. The $\varepsilon > 0$ was arbitrary, so $f \in \overline{j(E)}$.

From the previous paragraph we only need to prove that $j(E) \ne I(E)$ to show that $j(E)$ is not closed. As $E$ is a $G_\delta$-set, there is $f \in A$ such that $f(y) = 0$ if and only if $y = x$. Clearly, $\mathrm{supp}f = X$ as $x$ is not isolated, therefore, $f \in I(E)$ and $f \notin j(E)$.