Functionals in Banach-*-algebras

Question

I would like to know if every bounded linear functional $\varphi :A \to \mathbb{C}$ of a Banach-*-algebra $A$ is a linear combination of positive ones, i.e. of functionals $\varphi:A \to \mathbb{C}$ satisfying $\varphi(a^*a)\ge 0$. For $C^*$-algebras this comes from a theorem of Gelfand-Naimark, the Riesz-Markow representation theorem which identifies these functionals with complex Borel measure, as well as the Hahn-Jordan-decomposition of measures.

Otherwise, is there maybe such a result for functionals of the type $\varphi(a^*)=\overline{\varphi(a)}$?

Best regards, Dominik

Answer 1

The answer is no. If a Banach $^*$-algebra has two element $a$ and $b$ such that $a^*a=-b^*b\neq 0$, then every positive linear functional $\varphi $ on $A$ ncessarily vanishes on $a^*a$, and hence so will be the case for all linear combinations of positive linear functionals.

However, by Hahn-Banach there exists a continuous linear functional that does not vanish on $a^*a$.

I'll let you try to cook up a concrete such algebra but I'll be glad to help if necessary.

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EDIT: Beware of the spoiler!!

Here is a nice example of a Banach algebra with the above properties I just thought of. Consider the Banach algebra $C([-1, 1])$ formed by all continuous, complex valued functions on $[-1,1]$, made into a $^*$-algebra via the involution given by $$f^\dagger (x) = \overline {f(-x)}.$$ If $a$ and $b$ are the elements of this algebra given by $a(x) = x$, and $b(x)=|x|$, then $$(a^\dagger a)(x) = \overline{-x}x=-x^2,$$ while $$(b^\dagger b)(x) = \overline{|-x|}|x|=|x|^2=x^2.$$ Therefore $a^\dagger a=-b^\dagger b$.

Answer 2

Consider the algebra $\ell^1(\mathbb{Z})$ with convolution and standard conjugation $\{a_n\}^*=\{\bar{a}_{-n}\}.$

A sequence $\{a_n\}$ is called positive definite, if it satisfies $$\sum_{n,m}a_{n-m}z_n\bar{z}_m\ge 0$$ for any sequence $\{z_n\}$ with finitely many nonzero terms.

In particular positive functionals defined in the OP question correspond to positive definite sequences.

By Herglotz's theorem a sequence $\{a_n\}$ is positive definite if and only if it is of the form $$a_n=\int_{-\pi}^\pi e^{inx}\,d\mu(x) \qquad (*)$$ for a positive measure on $(-\pi,\pi].$

Linear combinations of positive definite sequences are thus of the form $(*),$ where $\mu$ is a linear combination (with complex coefficients) of four positive measures on $(-\pi,\pi].$

There exist sequences $\{a_n\}\in \ell^\infty(\mathbb{Z}),$ which are not of the form $(*).$ For example let $$a_n=\begin{cases} 1 & n\ge 0\\ 0 & n<0 \end{cases} $$ If $\{a_n\}$ is of the form $(*),$ by the F. and M. Riesz theorem, the measure $\mu$ is absolutely continuous. Thus $d\mu(x)=h(x)\,dx$ for $h\in L^1(-\pi,\pi).$ By $(*)$ we have $\hat{h}(n)=a_n.$ Now the Riemann-Lebesgue lemma implies $a_n\to 0,$ when $|n|\to \infty,$ which gives a contradiction.

Answer 3

General Banach-$^*$ algebras allow self-adjoint nonzero nilpotent elements, unlike $C^*$-algebras, which suggests the simplest example: Let $$A=\mathbb C[\epsilon]/(\epsilon^2)=\{a+b\epsilon|a, b\in\mathbb C\}$$ be the ring of dural numbers over $\mathbb C$.

Equip it with $\mathcal l^1$-norm $\|a+b\epsilon\|=|a|+|b|$ and involution $(a+b\epsilon)^{*}=\bar a + \bar b\epsilon$.

We claim $x=a+b\epsilon=yy^*$ for some $y$ iff $x=0$ or ($a>0$ and $b\in\mathbb R$). Indeed $$(c+d\epsilon)(c+d\epsilon)^*=c\bar c + (c\bar d + d \bar c)\epsilon$$ If $a=c\bar c =0$, then $c=0$ and $x=0$. Otherwise $a=c\bar c>0$, and $b=c\bar d + d\bar c\in\mathbb R$ is clear. On the other hand, if $x=a+b\epsilon$ with $a>0, b\in\mathbb R$, then we can easily find the decomposition $x=(\sqrt a + \frac{b}{2\sqrt a}\epsilon)^2$.

Now a positive functional must have $f(1+r\epsilon) = f(1) + rf(\epsilon)\ge 0$ for any $r\in\mathbb R$. From here, we can easily deduce $f(\epsilon)=0$. Hence the positive functionals cannot span the space of all functionals.

The same argument can be applied to any Banach-$^*$ algebra with self-adjoint nontrivial nilpotent element.