# Is this operator compact

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.

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.