$l^\infty$ spectrum of Toeplitz operator We have the Toeplitz operator $T:l^{\infty}(Z, R^2) \to l^{\infty}(Z, R^2)$. We computed spectrum of $T$ on $l^2$ using its symbol (symbol is continuous function $\varphi(z)$ and eigenvalues of $\varphi(z)$ are real for every $z$, $|z|=1$). What can be said about spectrum of $T$ on $l^{\infty}$? We suspect that it coincides with $l^2$ spectrum.
In our case elements of matrix $T$ are $2 \times 2$ real matrices and $\varphi(z)$ is complex matrix $2 \times 2$.  
 A: That is, indeed, true, though not totally obvious. The point is that for $T$ to act in $\ell^\infty$, the symbol $\varphi$ must be not merely continuous, but in the Wiener space (i.e., to have an absolutely convergent Fourier series) and the matrix-valued function $\psi=\lambda I-\varphi$ is invertible in the Wiener space if and only if it is invertible pointwise on the circle (of course, the inverse is just $\psi^{-1}$; the non-trivial part of the story is that it is in the Wiener space).
A: I am afraid that the proof given in Davies does not work in not commutative case. The series given for $f^{-1}$: $$\sum_{n=0}^\infty (g-f)^{n}g^{-n-1}$$
indeed converges due to estimate on $||g^{-n}||$. But it has nothing to do with $f^{-1}$. It is apparently inspired by a geometric progression with $q=(g-f)g^{-1}$ (then $(1-q)^{-1}=((g-g+f)g^{-1})^{-1}=gf^{-1}$). But in not-commutable case $q^{n}\neq (g-f)^{n}g^{-n}$ (there is no reason for $g-f$ or $f$ to commute with $g^{-1}$).
It seems that if we assume $f\in C^1$ then we may require that both $||g-f||_{\infty}$ and $||g'-f'||_{\infty}$ are small. Hence $||g^{-1}||_{W}\le c(f)$. Consequently $||q||_{W} < 1$ for sufficiently small $||g-f||_{C^{1}}$.
If $a_n$ are Fourier coefficients of $f$ it is enough to ask $\sum ||na_n||<\infty$ for $f \in C^1$. It would be great to avoid this extra condition. 
Is there a counterexample for $f\in W \setminus C^1$?         
