A problem related to Toeplitz operator 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 $f\in C(T)$ $(\text{or even $f\in L^\infty(T)$}\,)$ 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 $f\in C(T)$ $(\text{or even $f\in L^\infty(T)$}\,)$ the corresponding Toeplitz operator, $T_f$, on $P\mathcal H$ is defined by $T_f=PM_FP$.
Questions:
$1.$ Show that for $f\in C(T)$, if $T_f$ is invertible, then $M_f$ on $\mathcal H$ is invertible.
I have a hint saying that, first let $c=\vert\vert T_f^{-1}\vert\vert^{-1}$, and then show that for each $n\in \mathbb Z$ and each $\zeta\in P\mathcal H$ we have $\vert\vert M_fM_{e_n}\zeta\vert\vert\ge c\vert\vert \zeta\vert\vert$, and finally let $\mathcal H_n=M_{e_n}P\mathcal H$, and show that $\cup_{n=-\infty}^0\mathcal H_n$ 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 $f\in C(T)$ we have $\vert\vert T_f\vert\vert=\vert\vert M_f\vert\vert$.
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!
 A: 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 $z^n\zeta, \zeta \in \mathcal{H}$ 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.
