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 in the usual way)
• 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.

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.