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

• 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 $f(A) \ge 0$. Nov 11, 2022 at 13:38
• 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 Nov 11, 2022 at 13:52
• The spectral mapping theorem holds for each function that is continuous in an open neighbourdhood of the spectrum. Nov 11, 2022 at 14:12
• @ Adriano 8 I did not know this result, see theorem 13.9 tqft.net/web/teaching/current/Analysis3/LectureNotes/… Nov 11, 2022 at 14:22
• 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. Nov 11, 2022 at 15:01

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 $$f(A)=\psi(f)\geq0$$.
• What's your "simple" definition of $f(A)$? Nov 11, 2022 at 20:06
• Your $\psi$ is precisely the Gelfand-Naimark representation. Nov 11, 2022 at 22:54