# Question on proof of the spectral mapping theorem for self-adjoint operators

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.