# Integral with respect to spectral measure

Let $$A:D(A)\subset H \to H$$ be a self-adjoint unbounded operator on complex Hilbertspace $$H$$ with corresponding spectral measure $$E:\mathcal{B}(\mathbb{R})\to\mathcal{L}(H)$$.
I want to show that an element $$u\in H$$ with $$\Vert u \Vert =1$$ and $$E(I)u=u$$ for some open interval $$I$$ is $$u \in D(A)$$, that is $$\int_\mathbb{R} \lambda^2 d\Vert E(\lambda)u \Vert^2 < \infty.\quad (*)$$

I can think of two approaches: measure theory or generalisation of Riemann–Stieltjes integral.

From m.t. point of view the integral in (*) means $$\int_\mathbb{R} \lambda^2 d \mu_u$$ where $$\mu_u$$ is a measure on Borel sets defined by $$\mu_u(S)=\langle E(S)u,u\rangle.$$

In order to calculate this integral I can use monotone convergence theorem. Since it's only matter of construction, let's assume we found a sequence $$f_k:\mathbb{R}\to \mathbb R$$ with $$f_k \uparrow \lambda^2$$, so we may write $$\int_\mathbb{R} \lambda^2 d \mu_u=\lim \int_\mathbb{R} f_k d \mu_u .$$ But that isn't really helpfull...

If we try second approach, then (*) is defined as $$\int_\mathbb{R} \lambda^2 d\Vert E(\lambda)u \Vert^2=\lim_{\text{ partition of } \mathbb{R} \to 0} \sum t_i^2( \Vert E(\lambda_i)u \Vert^2-\Vert E(\lambda_{i-1})u \Vert^2 ),$$ where $$t_i\in (\lambda_{i-1}, \lambda_i)$$ and $$E(\lambda):=E((-\infty,\lambda])$$. Again it's not obvious for me that this leads anywhere.

Any help, hints or suggestions are greatly appreciated!

Edit 1: Let's try again. @N.S. comment suggests, it suffices to show $$E(I)u \in D(A)$$, that is $$\int_\mathbb{R} \lambda^2 d \mu_{E(I)u}< \infty.$$ We note $$\mu_{E(I)u}(S)=\langle E(I\cap S)u,u \rangle=\mu_u(I\cap S)=:\nu_u(S)$$ for any measurable set $$S$$. Coincidentally $$\nu_u$$ is also a measure!

Question 1: Is $$\int_\mathbb{R} \lambda^2 d \mu_{E(I)u}= \int_\mathbb{R} \lambda^2 d \nu_u=\int_{I} \lambda^2 d \mu_u \leq sup_{\lambda \in I} \{\lambda ^2\} \mu_u(I) <\infty$$ true? If so, how is second "=" justifiable?

• I'm probably making a mistake here, bur doesn't the condition $E(I)u=u$ imply that $$\int_\mathbb{R} \lambda^2 d\Vert E(\lambda)u \Vert^2 =\int_\mathbb{I} \lambda^2 d\Vert E(\lambda)u \Vert^2 ?$$ Is $I$ finite interval? Aug 22, 2019 at 22:10
• @N.S. I think your suggestion is true, but I'm struggling to see it. I edited my post, thank you. Aug 23, 2019 at 2:24
• $$\int_\mathbb{R} \lambda^2 d \nu_u= \int_I \lambda^2 d \nu_\mu+\int_{\mathbb{R}\backslash I} \lambda^2 d \nu_u$$ Now, for each Borel set $A \subseteq \mathbb{R}\backslash I$ you have by definition $\nu_{\mu}(A)=0$. This shows that $$\int_{\mathbb{R}\backslash I} \lambda^2 d \nu_u=0$$ and hence $$\int_\mathbb{R} \lambda^2 d \nu_u= \int_I \lambda^2 d \nu_u$$ Aug 23, 2019 at 3:04
• Nest, for each Borel set $S \subseteq I$ you have $\mu_u(S)=:\nu_u(S)$ and hence $$\int_I \lambda^2 d \nu_u=\int_I \lambda^2 d \mu_u$$ BUT, is $I$ bounded? Aug 23, 2019 at 3:06
• I think that the above completes the proof. But doesn't the claim follow then trivially from from $$\int_\mathbb{R} \lambda^2 d\Vert E(\lambda)u \Vert^2 \leq \int_\mathbb{R}C^2 d\Vert E(\lambda)u \Vert^2=C^2 \int_\mathbb{R} d\Vert E(\lambda)u \Vert^2$$ where $$C:=sup_{\lambda \in I} \{\lambda ^2\}?$$ Aug 23, 2019 at 3:17