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Consider the operator

$$(Tx)_n= \frac{x_{n+1}}{3} - \frac{2x_{n-1}}{3} $$ for $n \ge 0$ and

$$(Tx)_n = \frac{2x_{n+1}}{3} - \frac{x_{n-1}}{3} $$ for $n <0$.

I am wondering whether this operator is compact on $\ell^{\infty}(\mathbb Z)$?

My conejecture is no, but I find it difficult to show.

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Let $e_n$ be, for $n \geq 2$, the sequence which is zero everywhere, except at $n$ where it is one. Then $(e_n)$ is bounded but one easily checks that $\|Te_n-Te_m\| \geq 1/3$ for $n \neq m$, so that $(Te_n)_n$ has no convergent subsequence. Hence $T$ is not compact.

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