Find necessary and sufficient conditions for $T$ to be compact Let $b \in l^{1} (\mathbb{Z})$ and define an operator $T \in L(l^{2} (\mathbb{Z}))$ (space of sequences $\mathbb{Z}\to\mathbb{C}$ with the norm $\sum_{n = -\infty}^{\infty} |a_n|^{2}$) by $(Tx)_n = \sum_{n=-\infty}^{\infty} b_{n-m} x_m$. Find necessary and sufficient conditions on $b$ for $T$ to be compact.
I suspect that $T$ is compact only if $b\equiv 0$, since $Te_{n} = \sum_{m=-\infty}^{\infty} b_{m-n} e_m$, so the bounded sequence $e_{n}$ has an image without a convergent subsequence (I think? since $Te_n$ all have the same norm and $\langle Te_n | Te_k \rangle = \sum_{m=-\infty}^{\infty} b_{n-m}b_{k-m}^{*} $, and it seems to me that for sufficiently far away n and k this should be very small since $b \in l^{1} (\mathbb{Z})$ so the sequence cant converge ?)
Any help or hint would be very appreciated. Thanks!
Edit: Since $ \langle Te_n | Te_m \rangle $ can be rewritten az $\sum_{m=-\infty}^{\infty} b_{n-k+m}b_{m}^{*}$, it seems that it would be sufficient to show that if $S$ is the 'shift right' operator in $l_{2}$ then $S^{n} v$ weakly converges to $0$ for any $v \in l_{2}$, but im unsure if this is true.
 A: Let $\mathbb T$ be the unit circle and let $Z:\mathbb T\to {\mathbb C}$ be the function defined by $Z(z)=z$, for all $z\in \mathbb T$.
Consider the operator (Fourier transform)
$$
  \mathfrak F:\ell ^2({\mathbb Z})\to L^2(\mathbb T)
  $$
given on the canonical orthonormal basis $\{e_n\}_{n\in {\mathbb Z}}$ by
$$
  \mathfrak F(e_n) = Z^n, \quad\forall n\in {\mathbb Z}.
  $$
It is then easy to see that $\mathfrak F$ is a unitary operator and that
$$
  \hat T:=  \mathfrak F T\mathfrak F^*
  $$
is the so called multiplication operator given by
$$
  \hat T(f)|_z = \hat b(z) f(z), \quad\forall f\in L^2(\mathbb T),\quad\forall z\in  \mathbb T,
  $$
where
$\hat b$
is the function on $\mathbb T$ defined by
$$
  \hat b(z) = \sum_{n\in {\mathbb Z}}b_nz^n.
  $$
If $T$ is compact,  then so is $\hat T$,  but this cannot happen for a nonzero $b$,  since the spectrum of $\hat T$ coincides with
the range of $\hat b$, which is a closed interval rather than a  sequence converging to zero as it would be expected for a
compact operator.
