# Proof of continuous functional calculus via BLT

Background: I'm working on a proof of the spectral theorem as given by Halmos. As I've figured out, the first step, which is omitted in the proof, is to define a continuous functional calculus, i.e. define how the expression $$f(A)$$ in case of a continuous function $$f$$ and a self-adjoint operator $$A$$ is to be interpreted.

Notation: Let $$A$$ be a bounded linear operator on a Hilbert space $$H$$ with spectrum $$\sigma(A)$$. For self-adjoint operators $$A$$ we have $$\sigma(A) \subset \mathbb{R}$$. $$\mathcal{L}(H)$$ is the (Banach) space of bounded linear operators on $$H$$ and $$\mathcal{C}(X,Y)$$ the space of continuous functions from $$X$$ to $$Y$$.

Definition: Continuous Functional Calculus
Let $$A \in \mathcal{L}(H)$$ be a self-adjoint operator. Then there exists a unique bounded linear map from $$\mathcal{C}(\sigma(A); \mathbb{R})$$ to $$\mathcal{L}(H)$$, denoted as $$f \mapsto f(A)$$, such that when $$f(\lambda)=\lambda^{m}$$, we have $$f(A)=A^{m}$$. This mapping, denoted as $$f \mapsto f(A)$$, where $$f\in \mathcal{C}(\sigma(A);\mathbb{R})$$, is commonly referred to as the (real-valued) continuous functional calculus for the operator $$A$$. The map furthermore has the following properties:

1. Multiplicativity: For all $$f,g$$, we have $$(fg)(A)=f(A)g(A)$$, where $$fg$$ denotes the pointwise product of $$f$$ and $$g$$.
2. Self-adjointness: For all $$f$$, the operator $$f(A)$$ is self-adjoint.
3. Non-negativity For all $$f$$, if $$f$$ is non-negative, then $$f(A)$$ is a non-negative operator.
4. Norm and spectrum properties: For all $$f$$, we habe $$\|f(A)\|=\|f\|_{\infty}$$ and $$\sigma(f(A))=\{f(\lambda) :\, \lambda \in \sigma(A)\}$$

Proof:
Step 1: Constructing a polynomial functional calculus.
For each real-valued polynomial $$p$$ with $$p(\lambda)=\sum\limits_{i=0}^{n}\alpha_{i}\lambda^{i}$$ it's straight forward to define the operator $$p(A)$$ to be given by $$p(A)=\sum\limits_{i=0}^{n}\alpha_{i}A^{i}$$. We then have $$\|p(A)\|\leq \sum\limits_{i=0}^{n}|\alpha_{i}|\|A\|^{i} < \infty$$ and hence $$p(A)$$ is a bounded linear operator. Properties 1.-3. follow by straight forward calculation and property 4. is given by the spectral mapping theorem (for polynomials). Furthermore we have $$||p(A)\|=\|p\|_{\infty}$$. We thus define the map $$\phi$$ by: $$\phi : \mathcal{P}(\sigma(A),\mathbb{R}) \rightarrow \mathcal{L}(H), p \mapsto \phi(p):=p(A).$$

It's easily verified that $$\phi$$ is linear and by $$||p(A)\|=\|p\|_{\infty}$$ we have that's it*s not only bounded but isometric.

Step 2: Extending to continuous functional calculus.
By the theorem of Stone-Weierstrass, $$\mathcal{P}(\sigma(A),\mathbb{R})$$ is dense in $$\mathcal{C}_{c}(\sigma(A),\mathbb{R})$$ and since $$\sigma(A)$$ is compact we have $$\mathcal{C}_{c}(\sigma(A),\mathbb{R})=\mathcal{C}(\sigma(A),\mathbb{R})$$. Furthermore $$\mathcal{L}(H)$$ is a Banach space i.e. complete. Hence, by the BLT-theorem, there exists a unique, norm preserving extension of $$\phi$$ to $$\mathcal{C}$$. We denote this extension by $$\phi^{\ast}$$. Properties 1.-3. hold for polynomials and by taking limits they also hold for $$f \in \mathcal{C}(\sigma(A),\mathbb{R})$$. The first part of 4. follows by the norm preserving property of the extension. We thus get out continuous functional calculus by: $$\phi^{\ast} : \mathcal{C}(\sigma(A),\mathbb{R}) \rightarrow \mathcal{L}(H), f \mapsto \phi^{\ast}(f):=f(A).$$ If $$f$$ is not a polynomial, $$f(A)$$ is the limit of the sequence $$\{p_{n}(A)\}$$ with $$\{p_{n}\}$$ being a sequence of polynomials converging to $$f$$. Otherwise $$f(A)$$ is given by the *polynomial functional calculus$. Question: 1. Is my proof correct? 2. Am I right that $$f(A)$$, in case if $$f$$ begin not a polynomial, is defined as the limit in the uniform operator topology (convergence in norm)? • Your functional calculus might have well definedness issues if i understand it correctly. For example if$\sigma (A) = \{1\}$, then the polynomials$P_1 := z \mapsto z^2 + 3 z^3$and$P_2: z \mapsto 4 z$are the same as functions in$C(\sigma ( A))$. But you would assign$P_1(A) = A^2 + 3A^3 $and$P_2(A)= 4A$. So you would need to show that$P_2(A) = P_1(A)$in cases like these. – jd27 Commented Dec 10, 2023 at 20:29 • One way to circumvent this issue would be to define an equivalence relation on the polynomials, where two polynomials are equivalent if their restrictions to$\sigma (A)$agree with each other. – jd27 Commented Dec 10, 2023 at 20:31 • What im trying to say is: A real valued polynomial does not define a unique elment in$C(\sigma(A))$. So you are not defining a map from a subset of$C(\sigma(A)) \to \mathcal{L}(H)$with your polynomial functional calculus. – jd27 Commented Dec 10, 2023 at 20:35 • I understand your point. However I was following this proof (top of page 157) which seems to do exactly the same. What am I missing? Commented Dec 10, 2023 at 20:50 • I think the author is being sloppy with his definition of "polynomials on$\sigma(A)$", i think what he means here is the equivalence class construction i have described above (where two polynomials "are the same" if they agree on$\sigma (A). – jd27 Commented Dec 10, 2023 at 20:56 ## 1 Answer Let me provide the quotient construction that was discussed in the comments: Let $$p : x \mapsto \sum_{i=1}^n \alpha_i x^i$$ be a real polynomial of one variable. Define $$p(A):= \sum_{i=1}^n \alpha_i A^i$$. Then the map $$p \mapsto p(A)$$ is linear (in $$p$$) and it can be shown that $$\|p(A) \| = \sup_{ \lambda \in \sigma (A) } |p(\lambda)|$$. Now define an equivalence relation on the set of all real polynomials of one variable as follows: $$p \sim q$$ if $$q|_{\sigma (A)} = p|_{\sigma(A)}$$. Denote the set of all equivalence classes by $$\mathcal{P}(\sigma(A))$$. For a real polynomial $$p$$ denote by $$[p]$$ the equivalence class of $$p$$ in $$\mathcal{P}(\sigma(A))$$. Every $$[p] \in \mathcal{P}(\sigma(A))$$ defines a unique element $$\hat p \in C(\sigma (A))$$, by the formula $$\hat p (x) = p(x)$$. Under this identification $$\mathcal{P}(\sigma(A))$$ is a subspace of $$C(\sigma (A))$$. Now define the polynomial functional calculus $$\Phi$$ by: \begin{align*} \Phi : \mathcal{P}(\sigma(A)) &\longrightarrow L (\mathcal{H}) \\ [g] &\longmapsto q(A), \ q \in [g] \end{align*} To show that $$\Phi$$ is well defined (does not depend on the chosen representant of $$[g]$$): Let $$p,q \in [g]$$. Then $$p$$ and $$q$$ agree on $$\sigma (A)$$. Therefore $$\| p(A)- q(A) \|= \| (p-q) (A) \|= \sup_{\lambda \in \sigma(A)}| p(\lambda) - q(\lambda)| =0.$$ And so $$p(A) = q(A)$$, which implies that $$\Phi([g])$$ does not depend on the chosen representant $$q \in [g]$$ in the definition. The added complexity of the above quotient construction is why some authors prefer to define a functional calculus on $$C(\mathbb{R})$$ instead of $$C(\sigma(A))$$. • I'm still puzzled on howp(A)$would look like in your example from above. Or is it not possible to explicitly write down$p(A)$? Commented Dec 11, 2023 at 13:21 • @bayes2021 i dont really understand the question.$p(A)$is defined in the second paragraph of my answer for any polynomial, so the definition is exactly the same as in my example above. Its just that the map$p \mapsto p(A)$is not the polynomial functional calculus. – jd27 Commented Dec 11, 2023 at 17:23 • What I mean is the following: Take the polynomials from you example, i.e.$P_{1}: z \mapsto z^{2}+3z^{3}$and$P_{2}: z \mapsto 4z$. With$\sigma(A)=\{1\}$, they are both in the same equivalence class. But$P_{1}(A)=A^{2}+3A^{3}$and$P_{2}(A)=4A$. Since this contradicts$\| P_{1}(A)-P_{2}(A)\|=0$it seems that I've still not fully understood the details. Commented Dec 11, 2023 at 20:02 • @bayes2021 No it does not contradict$\|P_1 (A) - P_2 (A) \| =0$, see the last big equation in my answer, in fact we have shown that$A^2 +3 A^3 = 4A$in this case. – jd27 Commented Dec 11, 2023 at 20:08 • @bayes2021 to understand this, think about matrices. A self adjoint matrix$A$can be written as$ A= S^{-1} D S$, where$D$is the identity matrix in the example (since$1$is the only spectral value = eigenvalue). Now$A^2 + 3 A^3 = S^{-1} D^2S +3 S^{-1} D^3 S = 4 S^{-1} D S = 4A$, since$D^3 = D^2 = D\$.
– jd27
Commented Dec 11, 2023 at 20:12