Let's first recall the definition of Spectral set for a bounded operator $T$ on a hilbert space $H$. Let $X\subseteq \Bbb{C}$ be compact such that $\sigma(T)\subseteq X$. Define $$\mathcal{R}(X):=\left\{\frac{p}{q}:\ p(z),q(z)\in\mathbb{C}[z],\ q(z)\ne0\ \forall z\in X\right\}$$ which is a subalgebra of $C(X)$ endowed with sup norm.
Then for any $p/q\in\mathcal{R}(X)$, $\sigma(q(T))=q(\sigma(T))\subseteq\Bbb{C}^*$, so $q(T)$ is invertible. We define $\rho:\mathcal{R}(X)\to B(H)$ by $\rho(p/q)=p(T)q(T)^{-1}$. We say $X$ is spectral set for $T$ if this $\rho$ is well defined and $\lVert\rho\rVert\le 1$.
We have to use Von Neumann's inequality to prove the following-
$T\in B(H)$ is contraction if and only if $X=\overline{\Bbb{D}}$ is spectral set for $T$.
If part is obvious, since $z/1\in\mathcal{R}(X)$ with sup norm $1$, hence $\lVert T\rVert=\lVert\rho(z/1)\rVert\le 1$. I am stuck with the only if part. I have proved that $\rho$ is well defined.
Let $p/q=p_1/q_1$ in $X$. Then by Von Neumann's inequality $$\lVert p(T)q_1(T)-p_1(T)q(T)\rVert\le \text{sup}\{|p(z)q_1(z)-p_1(z)q(z)|:\ z\in\overline{\Bbb{D}}\}=0$$ This shows that $$ p(T)q_1(T)-p_1(T)q(T)=0$$ hence $$ p(T)q(T)^{-1}=p_1(T)q_1(T)^{-1}\implies \rho(p/q)=\rho(p_1/q_1)$$ This proves $\rho$ is well defined.
But I am not able to prove that $\lVert\rho\rVert\le1$. Can anyone provide a hint or way-out for this part? Thanks for your help in advance.
