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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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  • $\begingroup$ Since the algebra structure isn't taken into account of the definition, one can simply make any Banach space into a Banach algebra with zero multiplication? $\endgroup$ – Narutaka OZAWA Jan 12 '16 at 5:03
  • $\begingroup$ How do you mean zero multiplication? There would be no question of invertibility in that case, isn't it? $\endgroup$ – Arundhathi Jan 12 '16 at 6:15
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The answer to your immediate question is no: there is no other possible solution for $\phi$. To see this, consider that $$\phi(-1) + i\phi(1) = 2$$ implies $${\rm Re}(\phi(-1)) + {\rm Re}(i\phi(1)) = 2.$$ If $\|\phi\|_\infty = 1$ then both terms on the left are at most 1, and equal 1 if and only if $\phi(-1) = 1$ and $\phi(1) = -i$. So these values are forced. By a similar argument, the only unitaries in $l^1$ are the elements of the form $\alpha e_n$ where $(e_n)$ is the standard basis and $|\alpha| = 1$.

However, there is an easy counterexample. Consider $\mathbb{C}^2$ with the $l^1$ norm, i.e., $\|(a,b)\| = |a| + |b|$. Give it the product $(a,b)\cdot(c,d) = (ac, ad + bc)$. This is a Banach algebra because $$\|(a,b)\cdot(c,d)\| = |ac| + |ad + bc| \leq (|a| + |b|)(|c| + |d|) = \|(a,b)\|\|(c,d)\|.$$ The element $(1,0)$ is the unit, and it is unitary, but so is $(0,1)$, which is not invertible.

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    $\begingroup$ Nice example. I'd just like to add (for the benefit of the OP) that this algebra can be regarded as the set of all $\pmatrix{a & b \\ 0 & a}$ ($a,b \in {\mathbb C}$) or alternatively as a semidirect product / split extension ${\mathbb C} \rtimes {\mathbb C}$. $\endgroup$ – Yemon Choi Jan 11 '16 at 17:15
  • $\begingroup$ Yes, but you just mean as an algebra, right? I don't think the operator norm duplicates the $l^1$ norm I'm using. $\endgroup$ – Nik Weaver Jan 11 '16 at 19:54
  • $\begingroup$ Oh, absolutely: the operator norm is not going to be the same as the $\ell^1$-norm. The one you define is somehow the natural way to do things if one replaces ${\mathbb C}$ by a general Banach algebra $\endgroup$ – Yemon Choi Jan 11 '16 at 20:12

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