TITLE: Example of a singular element which is not a topological divisor of zero
QUESTION [6 upvotes]: We know that every topological divisor of zero in a commutative Banach algebra is singular. I need an example of a singular element which is not a topological divisor of zero.

REPLY [7 votes]: Let $\mathbb{D}$ be the open complex unit disk, $\overline{\mathbb{D}}$ the closed unit disk, and $\partial\mathbb{D}$ the unit circle.
Recall that the disk algebra $A(\mathbb{D})$ is the algebra of continuous functions on $\overline{\mathbb{D}}$ which are holomorphic on $\mathbb{D}$. By continuity+compactness, and the maximum modulus principle, we have
$$
\|f\|_\infty =\sup_{z\in\mathbb{D}}|f(z)|=\max_{z\in\overline{\mathbb{D}}}|f(z)|=\max_{z\in \partial\mathbb{D}}|f(z)|
$$
for every $f\in A(\mathbb{D})$. This norm turns $A(\mathbb{D})$ into a Banach algebra, with unit the constant function equal to $1$.
Now consider $f_0(z)=z$, which is clearly a singular element of $ A(\mathbb{D})$ since invertible elements can't vanish. Moreover, for every $f\in  A(\mathbb{D})$, we have
$$
\|f_0f\|_\infty=\max_{z\in \partial\mathbb{D}}|zf(z)|=\max_{z\in \partial\mathbb{D}}|f(z)|
=\|f\|_\infty.$$
It follows that $f_0$ is not a topological divisor of zero. Indeed, if a sequence $f_n$ is such that $f_0f_n$ converges to $0$ (i.e. $\|f_0f_n\|_\infty\longrightarrow 0 $), then $f_n$ must converge to $0$.
Remark: if your embed isometrically $A(\mathbb{D})$ into $B(A(\mathbb{D}))$ by mutiplication, this example is actually a non surjective (whence singular) isometric (whence not a a topological divisor of zero) shift. You could do the same thing with the larger Hardy space $H^\infty(\mathbb{D})$.
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