# Odd element of L^1 group algebra of the integers

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

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