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

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 $$\phi(g) = (\phi(f)-\lambda)^{-1}$$.
1. 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 $$\lambda \in \sigma(\phi(f))$$.
1. 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.

• Hint: can you find a function $g$ with norm 1 such that $\|(f-\lambda)g\|$ is small?
– Ruy
Commented Oct 7, 2020 at 3:31
• No, I really mean a function. The idea is to work first inside $C(\sigma(A))$.
– Ruy
Commented Oct 7, 2020 at 3:36

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?
• Since $\|B\|=1$, you can choose $\xi$ unital such that $\psi$ is almost unital. Next you scale $\psi$ by a factor near 1 to get a unital vector nearby with all of the required properties.
• If $\|\psi\|$ is arbitrarily close to $1$, and $\| (\phi(f)-\lambda)\psi \|$ is arbitrarily small, then $\| (\phi(f)-\lambda)\psi_0 \|$ is also arbitrarily small for $\psi_0=\psi/\|\psi\|$.