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

Question

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) := $\psi(f)$.

Answer 1

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

Answer 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 $f(A)=\psi(f)\geq0$.