I'm reading in Rynne and Youngson's Linear Functional Analysis book, and I have a question about Lemma 6.57. It says:
Let $\mathcal H$ be a complex Hilbert space, and let $\mathcal S$ be the real Banach space of all self-adjoint operators in $B(\mathcal H$) (the space of bounded linear operators). If $S\in\mathcal S$ then there exists $\Phi\in B(C_{\mathbb R}(\sigma(S)),\mathcal S)$ such that $\Phi(p)=p(S)$ whenever $p$ is a polynomial in $C_{\mathbb R(\sigma(S))}$.
Notation: $\sigma(S)$ denotes the spectrum of $S$, $C_{\mathbb R}(\sigma(S),\mathcal S)$ denotes the space of continuous functions from $\sigma(S)$ to $\mathcal S$ (endowed with the uniform metric), $B(X,Y)$ is the space of bounded linears operators for a normed space, and $B(X)=B(X,X)$.
The proof proof starts by considering the linear subspace $\mathcal P\subset C_{\mathbb R}(\sigma(S))$ of polynomials, and defining $\phi(p)=p(S)$ (which will later be extended to $\Phi$).
My problem is that I don't see why this is well-defined. What I basically want to show is that if we have a polynomial $p$ that vanishes on $\sigma(S)$, then $p(S)$ must also vanish.
The claim is easily verified if we have a basis of eigenvectors for $S$. Also it's clear that as soon as $\sigma(S)$ has infinitely many values, there cannot be two polynomials that restrict to the same function on $\sigma(S)$. However I don't see why it's true when $\sigma(S)$ is finite.
We have the equality $\sigma(p(S))=p(\sigma(S)).$ This is true for any bounded operator in a Hilbert space. Hence $\|p(S)\|=\displaystyle\sup_{t\in\sigma(S)}|p(t)|$ as $p(S)$ is normal. For normal operators the operator norm and the spectral radius coincide.