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> 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,
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> I have a long and complicated approach and ask to know a lot of results. Can we make it simple?
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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 $B(H)$.](https://math.stackexchange.com/questions/3825916/spectrum-of-a-positive-operator-in-bh)

**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.

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> 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
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> 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) := $\psi(f)$.
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