In our functional analysis lecture we have defined the continuous functional calculus via this theorem

Theorem (Continuous Functional Calculus) Let $\DeclareMathOperator{\H}{\mathscr{H}}\DeclareMathOperator{\lsk}{\langle}\DeclareMathOperator{\rsk}{\rangle} (\H, \lsk \cdot, \cdot \rsk)$ be a complex Hilbert space and $T \in L(\H)$ a bounded linear operator. If $T$ is self-adjoint, there exists a unique mapping $\Phi: \mathcal{C}(\sigma(T)) \to L(\H)$ with $\Phi(\mathbb{1}) = I$ and $\Phi(\text{id}) = T$, which is a continuous involutive homomorphism between algebras, which we call continuous function calculus of $T$ and write $f(T) = \Phi(f)$.

Now we want to prove the following

Theorem Let $T \in L(\H)$ be self-adjoint and $f \in \mathcal{C}(\sigma(T))$. Then the spectral mapping theorem holds for all $f \in \mathcal{C}(\sigma(T))$, i.e. $$\sigma(f(T)) = f(\sigma(T)).$$

Proof. The spectral mapping theorem holds for polynomials. Let $\mu \not\in f(\sigma(T))$. Then $g: (f - \mu)^{-1} \in \mathcal{C}(\sigma(T))$ and $g(f - \mu) = (f - \mu)g = 1$. Hence we get $$ g(T) ( f(T) - \mu \text{id}) = (f(T) - \mu \text{id}) g(T) = \text{id},$$ implying $\mu \in \rho(f(T))$ showing "$\subset$".

My Question Why is $f - \mu$ invertible? I think we have \begin{align} & \mu \not\in \sigma(f(T)) = \{ f(\lambda): \lambda \in \sigma(T) \} \\ \iff & \mu \in \{ f(\lambda): \lambda \in \sigma(T) \} ^{\complement} = \{ f(\lambda): T - \lambda \text{id} \text{ not invertible} \} ^{\complement} \\ \iff & \mu \in \{ f(\lambda): T - \lambda \text{id} \text{ invertible} \} \end{align} But I don't know how we can conclude anything about $f(x) - \mu x$.


We have that $g \in C(\sigma(T))$ is invertible iff $0 \notin g(\sigma(T))$. Hence, if $\mu \notin f(\sigma(T))$, then $0 \notin (f - \mu)(\sigma(T))$ and $(f-\mu)$ is invertible.


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