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.

  • $\begingroup$ 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

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.

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.