$p^2=p\in \bar{I}$, I ideal of Banach algebra $\Rightarrow p\in I$

Question

Let $I\subset A$ be a ideal of a Banach algebra $A$. Assume $p\in \overline I$ and $p^2=p$.

Show: $p\in I$.

Can someone give me a little hint how to solve this, please?

Answer 1

This is recorded as Lemma 2.7 in

Niels Jakob Laustsen, Thomas Schlumprecht, and András Zsák, The lattice of closed ideals in the Banach algebra of operators on a certain dual Banach space, J. Operator Theory 56 (2006), no. 2, 391–402.

I assume that $A$ is unital, if not, we can pass to the unitisation.

Suppose that $(t_n)_{n=1}^\infty$ is a sequence in $I$ that converges to an idempotent $p$. Since $pAp$ is a Banach algebra with identity $p$ and $pt_np \to p^3=p$, there exists $n$ so that $pt_np$ is invertible in $pAp$. Consequently, $p = (psp)(pt_np)$ for some $s\in A$. This proves that $p\in I$.