Ytukyg

I am learning the Toeplitz operator theory and I encounter a problem.

First, I need to introduce some definitions.

Definition: $T$ is the circle in $mathbb C$, we have $C(T)$, and $mathcal H=L^2(T)$ with its special orthonormal basis ${e_n}_{n=-infty}^infty$, whose elements are also viewed as elements of $C(T)$. For each $fin C(T)$ $(text{or even $fin L^infty(T)$},)$ we let $M_f$ denote the corresponding point wise multiplication operator on $mathcal H$. Let $P$ denote the projection onto the subspace of $mathcal H$ spanned by ${e_n}_{n=0}^infty$. For each $fin C(T)$ $(text{or even $fin L^infty(T)$},)$ the corresponding Toeplitz operator, $T_f$, on $Pmathcal H$ is defined by $T_f=PM_FP$.

Questions:

$1.$ Show that for $fin C(T)$, if $T_f$ is invertible, then $M_f$ on $mathcal H$ is invertible.

I have a hint saying that, first let $c=vertvert T_f^{-1}vertvert^{-1}$, and then show that for each $nin mathbb Z$ and each $zetain Pmathcal H$ we have $vertvert M_fM_{e_n}zetavertvertge cvertvert zetavertvert$, and finally let $mathcal H_n=M_{e_n}Pmathcal H$, and show that $cup_{n=-infty}^0mathcal H_n$ is dense in $mathcal H$.

I am not quite familiar with the theory and can not recover the whole proof. Can someone help me recover the proof?

$2.$ Use the result of $1$ to prove that for each $fin C(T)$ we have $vertvert T_fvertvert=vertvert M_fvertvert$.

As for $2$, I also have no any clue. Any hint and answer to problem $2$ will be helpful and I will be so grateful.

Thanks!

I can try to provide some hints.

I assume you mean that $e_n(z) = z^n$. Then $M_{e_n}$ is an isometry on $L^2(T)$, and it of course commutes with $M_f$, so that for any $zeta in mathcal{H}$ you have $$|M_f z^n zeta | = |M_f M_{e_n} zeta| = |M_{e_n}M_f zeta | = |M_f zeta | geq |PM_f zeta | = |T_f zeta | geq c | zeta |.$$ The point now is that the elements of the form $z^nzeta, zeta in mathcal{H}$ are clearly dense in $L^2(T)$. This is easy to see from Fourier expansions and Parseval's formula. Thus the inequality $|M_f zeta| geq c |zeta|$ actually holds for all $zeta in L^2(T)$. This then easily implies that $f$ is bounded from below, i.e, $|f(z)| > c$ for all $z in T$. Then $1/f in C(T)$, and the inverse of $M_f$ is $M_{1/f}$.

As for the second part, I would use a contradiction argument. It is clear that $|T_f| leq |M_f|,$ because $|T_f zeta| = |PM_f zeta| leq |M_f zeta|$. On the other hand, if it would be so that $|T_f| < |M_f|$, then consider $lambda in mathbb{C}$ such that $|T_f| < |lambda| < |M_f|$. It is easy to show that the spectrum of $M_f$ on $L^2(T)$ is just the range of the function $f$, and that the norm of $|M_f|$ equals the supremum norm of $f$, so we can arrange $lambda$ to be in the range of $f$. Now, since $|lambda| > |T_f|$, we know that the operator $T_f - lambda = T_{f - lambda}$ is invertible. By the first part, $M_{f - lambda}$ is invertible. But this is a contradiction, since $0$ is in the range of $f - lambda$, so $M_{f-lambda}$ should not be invertible.