# Zero Sets of Gelfand transform in Banach Algebras

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.

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 $$f \notin j(E)$$.
• Thanks a lot! I have a question. Why does it require Ε to be a $G_{\delta}$ set? Commented Sep 8 at 22:22
• @ΘάνοςΚ. I am not sure if it is really necessary to require it, maybe the statement holds for an arbitrary non-isolated point $x$. I used the assumption that $E$ is $G_\delta$ to derive that there exists $f$ that vanishes exactly at $x$ and nowhere else. For this function (if $x$ is not isolated) it is easy to see that $f \notin j(E)$. In general such function may not exist. Commented Sep 8 at 22:28