Let $H$ be a $\mathbb R$-Hilbert space and $(H_\lambda)_{\lambda\ge0}$ be a spectral decomposition of $H$ (see below). Now, let $$\mathcal D\left(A_\varphi\right):=\left\{x\in H:\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,x\rangle_H<\infty\right\}$$ and $$\langle A_\varphi x,y\rangle_H:=\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,y\rangle_H\;\;\;\text{for all }x\in\mathcal D\left(A_\varphi\right)\text{ and }y\in H\tag1$$ (the integral has to be understood as a Lebesgue-Stieltjes integral) for Borel measurable $\varphi:[0,\infty)\to\mathbb R$.

If I got it right, the spectral theorem states that if $(\mathcal D(A),A)$ is a nonnegative self-adjoint operator on $H$, $(H_\lambda)_{t\ge0}$ can be chosen such that $A_\varphi=A$, where $\varphi(\lambda):=\lambda$ for $\lambda\in[0,\infty)$. However, I've always seen the identity $$A_1=\operatorname{id}_H,\tag2$$ but this seems to be wrong. By $(1)$ and the definition of the Lebesgue-Stieltjes measure, $$\langle A_1x,y\rangle_H=\lim_{\lambda\to\infty}\langle\pi_\lambda x,y\rangle_H-\langle\pi_0x,y\rangle_H\tag3.$$ Clearly $\lim_{\lambda\to\infty}\langle\pi_\lambda x,y\rangle_H=\langle x,y\rangle_H$, so it seems like $(1)$ should be replaced by $$\langle A_\varphi x,y\rangle_H:=\varphi(0)\langle\pi_0 x,y\rangle_H+\int_0^\infty\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,y\rangle_H\;\;\;\text{for all }x\in\mathcal D\left(A_\varphi\right)\text{ and }y\in H\tag4.$$ On the other hand, we could also extend $H_\lambda$ for $\lambda<0$ by setting $H_\lambda:=\left\{0\right\}$ (and hence $\pi_\lambda=0$) for all $\lambda<0$. With this definition we could define $$\langle A_\varphi x,y\rangle_H:=\int_{\mathbb R}\varphi(\lambda)\:{\rm d}\langle\pi_\lambda x,y\rangle_H\;\;\;\text{for all }x\in\mathcal D\left(A_\varphi\right)\text{ and }y\in H\tag5$$ by setting $\varphi(\lambda)=0$ for all $\lambda<0$. With $(4)$ we would again obtain $(2)$.

So, I'm highly confused. Which is the usual definition when one is interested in the spectral measure for a nonnegative self-adjoint operator? (Please note that I'm only interested in Lebesgue-Stieltjes integration; not in Riemann-Stieltjes integration.)


$(H_\lambda)_{\lambda\ge0}$ is called spectral decomposition of $H$ if

  1. $H_\lambda$ is a closed subspace of $H$ for all $\lambda\ge0$;
  2. $(H_\lambda)_{\lambda\ge0}$ is nondecreasing and right-continuous, i.e. $$\bigcap_{\mu>\lambda}H_\mu=H_\lambda\;\;\;\text{for all }\lambda\ge0;$$ and
  3. $\bigcup_{\lambda\ge0}H_\lambda$ is dense.

Let $\pi_\lambda$ denote the orthogonal projection of $H$ onto $H_\lambda$ for $\lambda\ge0$. It can be shown that

  1. $[0,\infty)\ni\lambda\mapsto\pi_\lambda$ is nondecreasing, i.e. $$\langle\pi_\lambda x,x\rangle_H\le\langle\pi_\mu x,x\rangle_H\;\;\;\text{for all }x\in H,$$ and right-continuous (with respect to the strong operator topology)


  1. $[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,x\rangle_H=\left\|\pi_\lambda x\right\|_H^2$ is bounded (by $\left\|x\right\|_H^2)$, nondecreasing and right-continuous for all $x\in H$
  2. $[0,\infty)\ni\lambda\mapsto\langle\pi_\lambda x,y\rangle_H=2^{-1}\left(\langle\pi_\lambda(x+y),x+y\rangle_H-\langle\pi_\lambda x,x\rangle_H-\langle\pi_\lambda y,y\rangle_H\right)$ is right-continuous and of bounded variation for all $x,y\in H$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.