Let $(e_n)_{n \in \mathbb{Z}}$, where $e_n := (2\pi)^{-1/2}z^n$, be an orthonormal basis for $L^2(\mathbb{T})$ ($\mathbb{T}$ is the unit circle in $\mathbb{C}$). Let $H^2$ denote the closed subspace of $L^2(\mathbb{T})$ spanned by $(e_n)_{n \geq 0}$. Let P denote the orthogonal projection of $L^2(\mathbb{T})$ on $H^2$. For each $f \in L^{\infty}(\mathbb{T})$ define the Toeplitz operator $T_f$ in ${\bf B}(H^2)$ by \begin{align*} T_fx = P(fx), \ x \in H^2. \end{align*}
I am trying to show two things:
The map $f \to T_f$ is norm decreasing. I have shown that the map is linear. However, I am not sure how to show that the map is norm decreasing. My main issue is the definition of norm decreasing. So, for this map to be norm decreasing, am I trying to show that if given $f,g \in L^{\infty}(\mathbb{T})$ such that $\left\vert\left\vert{f}\right\vert\right\vert \leq \left\vert\left\vert{g}\right\vert\right\vert$, then $\left\vert\left\vert{T_f}\right\vert\right\vert \geq \left\vert\left\vert{T_g}\right\vert\right\vert$? If this is the case, I am not seeing how the conclusion would follow.
I am also trying to show that $T_f^* = T_{\overline{f}}$. I started by considering for every $x \in H^2$ \begin{align*} (T_fx|x) = (x|T^*_fx). \end{align*} However, I am not sure how to proceed. More so, would $T^*_fx = P^*(fx)$, and since P is an orthogonal projection, P is self adjoint. Hence, $T^*_fx = P^*(fx) = P(fx)$?
Thank you in advance.
To show that $f\mapsto T_f$ is norm-decreasing, what you want to show is that $\|T_f\|\leq \|f\|$. (What you were trying to show is not true, for example if $f=0$ and $g=1$.)
To show that $T_f^*=T_{\bar f}$, it suffices to show that for every $x,y\in H^2$, we have $$\langle T_fx,y\rangle=\langle x,T_{\bar f}y\rangle.$$ To show this, it helps to know that $T_f=PM_f$, where $M_f:L^2(\mathbb T)\to L^2(\mathbb T)$ is multiplication by $f$, that $P$ is self-adjoint, that $Px=x$ for all $x\in H^2$, and that $M_f^*=M_{\bar f}$.