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 $\sigma(f(A))=\{f(\lambda) :\, \lambda \in \sigma(A)\}$

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$.


  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)?
  • $\begingroup$ 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. $\endgroup$
    – jd27
    Commented Dec 10, 2023 at 20:29
  • $\begingroup$ 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. $\endgroup$
    – jd27
    Commented Dec 10, 2023 at 20:31
  • $\begingroup$ 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. $\endgroup$
    – jd27
    Commented Dec 10, 2023 at 20:35
  • $\begingroup$ 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? $\endgroup$
    – bayes2021
    Commented Dec 10, 2023 at 20:50
  • $\begingroup$ 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 $\sigma (A)$. $\endgroup$
    – jd27
    Commented Dec 10, 2023 at 20:56

1 Answer 1


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 $C(\sigma(A))$.

  • $\begingroup$ I'm still puzzled on how $p(A)$ would look like in your example from above. Or is it not possible to explicitly write down $p(A)$? $\endgroup$
    – bayes2021
    Commented Dec 11, 2023 at 13:21
  • $\begingroup$ @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. $\endgroup$
    – jd27
    Commented Dec 11, 2023 at 17:23
  • $\begingroup$ 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. $\endgroup$
    – bayes2021
    Commented Dec 11, 2023 at 20:02
  • $\begingroup$ @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. $\endgroup$
    – jd27
    Commented Dec 11, 2023 at 20:08
  • $\begingroup$ @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 $D^3 = D^2 = D$. $\endgroup$
    – jd27
    Commented Dec 11, 2023 at 20:12

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