I'm working through the proof of the following theorem from C* algebra by Murphy. For context, $T_{\varphi}: H^2(T) \longrightarrow H^2(T)$ is given by $T_{\varphi}(f) = p(\varphi f)$ for $p: L^2(T) \longrightarrow H^2(T)$ the orthogonal projection of $L^2(T)$ onto $H^2(T)$.
I don't understand how the collection $(\epsilon_n f)_{n \in \mathbb{Z}}$ is $L^2$ norm dense in $L^2(T)$ for seemingly every $f \in H^2(T)$. What if $f=0$?
While I understand that $||M_{\varphi}(g)||^2 = \langle M_{\varphi}^{*}M_{\varphi}(g),g\rangle \geq \langle g,g\rangle/M^2$ which implies $||M_{\varphi}^{*}M_{\varphi}(g)|| \geq ||g||/M$, so $M_{\varphi}^{*}M_{\varphi} \geq I/M^2 > 0$, and thus by applying the Gelfand representation theorem to $C^{*}(M^{*}_{\varphi}M_{\varphi})$ (using normality of $M_{\varphi}$ to conclude this algebra is abelian), we conclude that $M_{\varphi}^{*}M_{\varphi} \in \text{Inv}(B(L^2(T)))$, why does this imply $M_{\varphi}$ is also invertible? I don't see why normality implies this.
I think Murphy means that $\{\varepsilon_nf:n\in\mathbb{Z}\wedge f\in H^2\}$ is dense in $L^2$. This is all you need, because the goal is to show that $\|M_\varphi(g)\|\ge \|g\|/M$ for all $g\in L^2(\mathbb{T})$.
For 2, use the spectral theorem: if $A$ is a unital $C^*$-algebra and $a\in A$ is normal, then $\sigma(f(a))=f(\sigma(a))$ for any $f\in C(\sigma(a))$. So $\sigma(aa^*)=\{|\lambda|^2:\lambda\in\sigma(a)\}$. If $aa^*$ is invertible, then $0\not\in \sigma(aa^*)$, and the previous set equality implies that $0\not\in \sigma(a)$. This means that $a$ is invertible.