Proving the spectral mapping theorem via the continuous functional calculus - significant progress

Question

In my functional analysis class, I was asked to prove the spectral mapping theorem in a specific way. Before the problem, I will give the necessary background on the continuous functional calculus

Let $A$ be a self-adjoint operator on a Hilbert space $H$. Then there is a unique map $\phi : C(\sigma(A)) \rightarrow L(H)$ - - - ($C(\sigma(A))$ denotes the set of continuous complex functions on the spectrum of $A$ which is real since $A$ is self-adjoint and $L(H)$ denotes the bounded linear operators on $H$) - - - such that

(a) $\phi$ is an algebra *-homormorphism, meaning \begin{gather*} \phi(f+g) = \phi(f)+\phi(g) \\ \phi(fg) = \phi(f)\phi(g) \\ \phi(\lambda f) = \lambda \phi(f) \\ \phi(1) = I \\ \phi(\bar{f}) = \phi(f)^* \end{gather*} (b) $\phi$ is continuous with $\|\phi(f)\|_{L(H)} \leq C \|f\|_{\infty}$.

(c) For $f(x)=x$ we have $\phi(f)=A$.

(d) If $A\psi = \lambda \psi$, then $\phi(f)\psi = f(\lambda) \psi $.

(e) $\sigma(\phi(f)) = \{ f(\lambda) \mid \lambda \in \sigma(A) \}$ and this the spectral mapping theorem.

(f) If $f \geq 0$ then $\phi(f) \geq 0$.

(g) We can strengthen (b) and get $\|\phi(f)\|_{L(H)} = \|f\|_{\infty}$.

Now, based on the continuous functional analysis, we are asked to solve the problem

Let $A$ be a bounded self-adjoint operator on the Hilbert space $H$ and $f$ a continuous function on $\sigma(A)$.

1. If $\lambda \notin \text{Ran} \; f$, let $g = (f-\lambda)^{-1}$. We are asked to prove $\phi(g) = (\phi(f)-\lambda)^{-1}$.

2. Let $\lambda \in \text{Ran} \; f $. We are asked to prove there are $ \psi \in H $ with $\|\psi\|=1$ and $\| (\phi(f)-\lambda)\psi \|$ arbitrarily small so that $\lambda \in \sigma(\phi(f))$.

3. We are asked to conclude (e) above, the spectral mapping theorem, which says $\sigma(\phi(f)) = \{ f(\lambda) \mid \lambda \in \sigma(A) \}$.

I think I understand the continuous functional calculus, but I honestly have no idea about proving the spectral theorem using the given steps. All three of 1,2,3 elude me and I have no idea where to start. I understand the properties of the functional calculus, but cannot seem to apply them to get the desired conclusions of 1,2,3. I would appreciate all help on this and I thank all helpers.

************* Progress: managed to do step 1 quite directly which shows one direction of inclusion. For step 2, I thought I could prove it first for polynomial functions $P$ and use Weierstrass's approximation theorem but I get something weird. I tried to look at $$A\psi = \lambda \psi + \epsilon $$ and I want something like $$P(A)\psi = P(\lambda) \psi + \tilde{\epsilon}$$ but I cannot seem to work it out. I start with $P(A)=A^n$ and get a weird function of $\epsilon$, so basically what I have is $$ A^n\psi = \lambda^n \psi + +G(A)\epsilon $$ and I do not know how this $G$ behaves for the uniformly converging $P_n$ in Weierstrass' approximation theorem. Can someone just show me part 2? I have 1 and can prove 3 based on 1 and 2. I do believe $$ G(A)\epsilon = \lambda^{n-1}\epsilon+\lambda^{n-2}A\epsilon + \dots + A^{n-1}\epsilon $$ if it helps. I thank all helpers.

Answer 1

Since $\lambda$ is in the range of $f$, there is some $x$ such that $f(x)=\lambda$. Therefore $f-\lambda$ vanishes on $x$, so given any positive $\epsilon$, there is a neighborhood $V$ of $x$ where $|f-\lambda|<\epsilon$.

Take any function $g:\sigma(A)\to [0,1]$ vanishing off $V$, such that $g(x)=1$ and observe that $\|(f-\lambda)g\|<\epsilon$.

If $B=\phi(g)$ then $\|(\phi(f)-\lambda)B\|<\epsilon$, so for any unit vector $\xi$ in $H$, the vector $\psi=B(\xi)$ satisfies the last condition in (2) so it remains to choose $\xi$ so that $\psi$ is unital. Can you pick up from here?