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Unitary elements of a Banach space have been defined in this paper as follows:

Let $A$ be a Banach space and $a\in A, \|a\|=1$. Let $S_{a}=\{f\in A':\|f\|=1=f(a)\}$. Then $a$ is said to be (geometrically) unitary if $A'=\text{ span }S_{a}$.

Here, $A'$ is the dual space of $A$.

We note that this property is true for unitary operators on a Hilbert space. Unitary operators on Hilbert spaces are invertible, and their spectra lie on the unit circle.

Now, suppose $A$ is a Banach algebra. Can we say that if $a\in A$ is geometrically unitary, then it is invertible?

I have been trying to find a counter-example, but have been unsuccessful so far. I considered the Banach algebra $l^{1}(\mathbb{Z})$, with convolution as the multiplication. The dual of $l^{1}(\mathbb{Z})$ is $l^{\infty}(\mathbb{Z})$. An element $f$ of $l^{1}(\mathbb{Z})$ is invertible iff $f(z)\neq 0 \, \forall z\in \mathbb{T}$, where $\mathbb{T}$ is the unit circle in $\mathbb{C}$.

Let us take, for example, $f=(\cdots,0,\frac{1}{2},0,\frac{i}{2},0,\cdots)$, where the central $0$ is in the $0^{th}$ position. Then $f$ is of norm $1$ and can be shown to be not invertible.

We now consider elements $\phi$ of $S_{f}\subseteq l^{\infty}(\mathbb{Z})$. $\phi$ must satisfy the following:

$ \|\phi\|_{\infty}=1\\$ and $\phi(-1)\frac{1}{2}+\phi(1)\frac{i}{2}=1$.

One possible solution is $\phi=(\cdots,1,x,-i,\cdots)$, where $x$ and the dots can be any scalar less than or equal to $1$.

Is there another possible solution? In that case, we would have a unitary element that is not invertible.

Alternately, is it indeed true that unitary elements described in this geometric fashion are invertible?

I'd be grateful for help with this.

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