Question Let H a Hilbert space and A ∈ B(H) be self-adjoint and $f$ is a positive continuous function on $\sigma(A)$, Show $f(A)$ is a positive operator,

I have a long and complicated approach and ask to know a lot of results. Can we make it simple?

1- A being self-adjoint, its spectrum $\sigma(A)$ is compact. Since f is continuous on $\sigma(A)$, we can approximate it by a sequence of polynomial functions on $\sigma(A)$. We can assume that f is a polynomial

2- Since A is self-adjoint and f polynomial, a spectral analysis theoreme shows that $\sigma(f(A))=f(\sigma(A))$ and since f is positive on $\sigma(A)$, then $\sigma(f(A))$ is positive

3- I remind this spectral analysis theorem Let A ∈ B(H) be self-adjoint and let $f:\sigma (A)\to \mathbb {C}$ continuous . Then f(A) is self-adjoint iff f is real-valued.

We deduce that f(A) is self-adjoint and since $\sigma(f(A))$ is positive then , f(A) is a positive operator see for example Spectrum of a positive operator in $B(H)$.

Addition Martin Argerami asked You don't say how you define the functional calculus.

Let K be a compact of $\mathbb R$.

  1. We denote by C(K; $\mathbb K$) the $\mathbb K$-vector space of continuous functions from K to $\mathbb K$. We equip this space with the uniform convergence

  2. $\mathcal { P}(K)$ denotes the vector subspace of C(K; $\mathbb K$) consisting of the polynomials of K[X] restricted to K.

Let T ∈ L(H) be a self-adjoint operator and P ∈ $\mathcal { P}$(σ(T)). be Q, R ∈ K[X] two extensions from P to K. Then, we have Q(T) = R(T). We can therefore define the operator P(T) ∈ L(H) by setting P(T) := Q(T), where Q ∈ K[X] is any extension from P to K

Let T ∈ L(H) be a self-adjoint operator. Then, the map Φ defined from $\mathcal { P}$(σ(T)) to L(H) by Φ(P) := P(T), extends uniquely to C(σ(T); $\mathbb K$) in an isometric linear map $\psi$ . Thus, for any f ∈ C(σ(T); $\mathbb K$), we can define f(T) ∈ L(H) by setting f(T) := $\psi(f)$.

  • 1
    $\begingroup$ If $A$ is normal, then $A \ge 0$ if and only if $\sigma(A) \subset [0,\infty)$. This is a classic result. Now use the Spectral Mapping thm: $\sigma(f(A)) = f(\sigma(A)) \subset [0,\infty)$ to deduce that $f(A) \ge 0$. $\endgroup$ Nov 11, 2022 at 13:38
  • $\begingroup$ This is exactly my proof except that the Spectral Mapping thm is valid for a polynomial function and I explained why f(A) is also self-adjoint $\endgroup$ Nov 11, 2022 at 13:52
  • $\begingroup$ The spectral mapping theorem holds for each function that is continuous in an open neighbourdhood of the spectrum. $\endgroup$
    – Adriano
    Nov 11, 2022 at 14:12
  • $\begingroup$ @ Adriano 8 I did not know this result, see theorem 13.9 tqft.net/web/teaching/current/Analysis3/LectureNotes/… $\endgroup$ Nov 11, 2022 at 14:22
  • $\begingroup$ Your approach requires that if $f\ge 0$ on $\sigma(A)$ there is a sequence of polynomials $p_n$ nonnegative on $\sigma(A)$ and convergent uniformly to $f$ on $\sigma(A).$ This can be achieved by the Weierstrass theorem. $\endgroup$ Nov 11, 2022 at 15:01

2 Answers 2


Let $g:=\sqrt f$ and $B:=g(A).$ Then, $B^*=B$ and $A=B^2.$


Your map $\psi:C(\sigma(A))\to B(H)$ is not only linear and isometric, but multiplicative and $*$-preserving. In particular, it preserves positivity. So if $f\geq0$, then $f(A)=\psi(f)\geq0$.

  • $\begingroup$ Martin Argerami H is a Hilbert space. we are therefore looking for a simple proof, which does not require very fine arguments $\endgroup$ Nov 11, 2022 at 17:42
  • $\begingroup$ What's your "simple" definition of $f(A)$? $\endgroup$ Nov 11, 2022 at 20:06
  • $\begingroup$ I edited to explain $\endgroup$ Nov 11, 2022 at 22:23
  • $\begingroup$ Your $\psi$ is precisely the Gelfand-Naimark representation. $\endgroup$ Nov 11, 2022 at 22:54
  • $\begingroup$ thanks i didn't know that $\endgroup$ Nov 11, 2022 at 23:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .