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Consider the Banach $^* $-algebra $\ell^1(\mathbb Z)$ with multiplication given by convolution and involution given by $a^*(n)=\overline{a(-n)}$.

I would like to find nice necessary and sufficient conditions for an element $b\in\ell^1(\mathbb Z)$ to be positive, that is, to be of the form $a^* * a$ for some $a\in\ell^1(\mathbb Z)$.

By now, I have found two necessary conditions. Namely, if $b\in\ell^1(\mathbb Z)$ is positive, then $$b(-n)=\overline{b(n)}$$ and $$\lvert b(n)\rvert\leq b(0)$$ for every $n\in\mathbb Z$.

Edit: As t3suji states in his comment below both conditions follow from the more general fact that $a$ is a positive-definite function.

Question: Is this condition also sufficient for positivity? If not, what to I have to add?

Good references would also be great.

Motivation: In the end I want to investigate the (failure of) the Gelfand–Naimark theorem for the above non-C*-algebra.

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Another (almost obvious) relation: The $n\times n$ matrix $A$ with $a_{ij}=b(i-j)$ has to be Hermitian positive definite (or rather non-negative) for any $n$. This includes both of your conditions. –  t3suji Sep 1 '10 at 20:43
    
In what sense do you hope/expect/know that Gelfand-Naimark failes for $\ell^1(\mathbb Z)$? Certainly there is an injective, contractive, involution-respecting homomorphism from $\ell^1(\mathbb Z)$ to $B(H)$ for some Hilbert space $H$ (indeed, $H=\ell^2(\mathbb Z)$ works). This isn't an isometry, of course, but in the C-star world, that's proved by spectral methods, isn't it? –  Matthew Daws Sep 1 '10 at 21:00
    
@Matthew Daws: I want to apply the GNS-construction to all pure states on$\ell^1(\mathbb Z)$ , and then take the direct sum of all the irreps I ontain this way. I will find that this rep is not isometric. To do this I have to understand what positive functionals on $\ell^1(\mathbb Z)$ are. And therefore I have to understand the postive elements in $\ell^1(\mathbb Z)$. –  Rasmus Bentmann Sep 1 '10 at 21:11
    
I haven't thought about this too hard, but it seems to me that the failure of GNS to be isometric for $\ell^1(\mathbb Z)$ is less to do with it having too few positive functionals, but the fact that (as Matthew mentions) the GNS construction involves taking completions. The point is that an injective homomorphism between C*-algebras is an isometry, but if you replace C* by Banach there is no reason to expect this will work. –  Yemon Choi Sep 1 '10 at 21:14
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An injective star-homomorphism between star-Banach algebras still doesn't need to be an isometry. If I recall correctly, $\ell^1(\mathbb Z)$ and $C^∗({\mathbb Z})=C({\mathbb T})$ have *exactly the same positive linear functionals. It seems to me that all your approach will do, is provide a complicated way of checking that the norm on $\ell^1(\mathbb Z)$ does not satisfy the $C^∗$-condition... –  Yemon Choi Sep 1 '10 at 21:37

2 Answers 2

up vote 2 down vote accepted

Although I am not sure that answering this question will help all that much with your original motivating question/problem, I may as well post a link to Bochner's theorem (see these remarks on Wikipedia).

The passage I have in minds says:

Positive-definiteness arises naturally in the theory of the Fourier transform; it is easy to see directly that to be positive-definite is a necessary condition ... to be the Fourier transform of a function $g$ on the real line with $g(y) \geq 0$.

The converse result is Bochner's theorem, stating that a continuous positive-definite function on the real line is the Fourier transform of a (positive) measure.

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There are four facts which clarify things a little bit:

1) The inclusion $\ell^1 {\mathbb Z} \subset C(S^1)$ preserves the spectrum. (That is Wiener's Theorem)

2) If $f = \sum_i a_i z^{i}\in \ell^1 {\mathbb Z}$ has non-negative coefficients, then $\|f\| = \|f\|_{C(S^1)}$.

3) If $f(z)>0$ for all $z \in S^1$, then $f$ has a square-root in $\ell^1 {\mathbb Z}$ by holomorphic functional calculus.

4) $2 - z-z^{-1}$ is non-negative, but has no square-root in $\ell^1{\mathbb Z}$.

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@Andreas: Regarding 4): Isn't $2-z-z^{-1}=(1-z^{-1})(1-z)=(1-z)^*(1-z)$? –  Rasmus Bentmann Sep 4 '10 at 13:58
    
That is true. What is meant is that the positive square-root (which exists in $C(S^1)$) does not lie in $\ell^1 {\mathbb Z}$. Otherwise, I think that Riesz proved that every non-negative Laurent polynomial is a hermitian square (or maybe sum of such squares). –  Andreas Thom Sep 4 '10 at 21:25

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