Let $\mathbb{T}$ be the unit circle of $\mathbb{C}$, Let $H^2 \subset L^2(\mathbb{T})$ be the Hardy space (i.e. $H^2 = \text{span}\{(2π)^{−1/2}z^n \in L^2 | n \in \mathbb{Z}\}$). Let $f \in L^\infty$ and let $T_{f}(x)$ be a $H^2 \rightarrow H^2$ operator defined as
$$T_f(x)=P(fx), x∈H^2$$
I need to show that the operator $f \mapsto T_f$ is norm decreasing, i.e. $\|T_f\| \le \|f\|$.
I have tried to decompose $f$ into it's orthonormal components, i.e. let $f = g + g^\perp$ with $g \in H^2$. But then I got $P(fx) = P(gx + g^\perp x) = gx$ and so $LHS = \underset{\|x\| = 1}\max \|gx\|$ and $RHS = \|g + g^\perp\| \le \|g\| + \|g^\perp\| \le \|g\| = \underset{z \in \mathbb{T}}\max\|g(z)\|$. Unfortunately it seems that I getting close to the opposite of what I am trying to show. Is there a way to go about solving this problem?
\begin{align*} \|T_{f}(x)\|_{L^{2}}=\|P(fx)\|_{L^{2}}\leq\|P\|\|fx\|_{L^{2}}\leq\|fx\|_{L^{2}}\leq\|f\|_{L^{\infty}}\|x\|_{L^{2}}, \end{align*} so $\|T_{f}\|\leq\|f\|_{L^{\infty}}$. Note that the projection has norm less than $1$.