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Giving some motivation is hard here, so I'll just ask the question. I want an element $a=(a_n)\in\ell^1(\mathbb Z)$ such that:

  • $\|a\|>1$
  • a is power bounded (turn $\ell^1(\mathbb Z)$ into a Banach algebra for the convolution product)
  • we have also that $\|a^m\|_\infty \rightarrow 0$.

I'm sure a clever use of the Fourier transform would work. For example, the third condition is ensured if, letting $f\in C(\mathbb T)$ be the Fourier transform of $a$, we have that $|f|<1$ almost everywhere. The 2nd condition implies that $|f|\leq1$, but of course this isn't if and only if.

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By "the usual way" you mean "convolution"? – Nate Eldredge Aug 31 '10 at 20:01
Yep! Thanks! I've edited the question... – Matthew Daws Aug 31 '10 at 20:19
In the third condition, does the sup norm mean the sup norm of the Fourier transform as a function on the unit circle (i.e. the spectral radius)? – Yemon Choi Aug 31 '10 at 22:29
To nitpick, why "almost"? – Jonas Meyer Aug 31 '10 at 22:48
@Yemon: Ah, sorry, no, I mean the norm in $\ell^\infty(\mathbb Z)$. – Matthew Daws Sep 1 '10 at 8:16
up vote 5 down vote accepted

I'm a bit uncertain what is meant in the third condition: is this the supremum norm of the Gelfand/Fourier transform of $a^m$, or the norm of $a*a*\dots*a$ in $\ell^\infty(\mathbb Z)$?

In the first interpretation, it would clearly suffice to find an element satisfying the first two conditions, and then multiply it by an appropriate scalar betwen $0$ and $1$. In the second interpretation, as Matthew says, it would suffice to find $a$ satisfying conditions 1) and 2) with the additional property that the Fourier transform of $a$ has modulus $<1$ at all but finitely many points of the unit circle.

Anyway, it turns out that we can find (trigonometric) polynomials satisfying conditions 1) and 2), which should by the previous remarks be enough. I don't know where the first examples are in the literature, but you can find explicit quadratic examples in this paper of D. J. Newman:

MR0241980 (39 #3315) Newman, D. J. Homomorphisms of $l_{+}$. Amer. J. Math. 91 1969 37--46

which I personally think is a gem.

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Yep, that works! Thanks! For the record, Newman shows that the trig polynomial $(1+z-z^2)/\sqrt5$ works. In my language, $a=5^{-1/2}(\cdots,0,1,1,-1,0,\cdots)$ works. Clearly $\|a\|=3/\sqrt5>1$, Newman shows that $a$ is power-bounded, and the Fourier transform is $(1-\frac45\cos^2\theta)^{1/2}$, which is indeed $<1$ almost everywhere. – Matthew Daws Sep 1 '10 at 8:32

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