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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?

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    $\begingroup$ 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? $\endgroup$
    – N. S.
    Commented Aug 22, 2019 at 22:10
  • $\begingroup$ @N.S. I think your suggestion is true, but I'm struggling to see it. I edited my post, thank you. $\endgroup$ Commented Aug 23, 2019 at 2:24
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    $\begingroup$ $$\int_\mathbb{R} \lambda^2 d \nu_u= \int_I \lambda^2 d \nu_\mu+\int_{\mathbb{R}\backslash I} \lambda^2 d \nu_u$$ 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 $$\int_\mathbb{R} \lambda^2 d \nu_u= \int_I \lambda^2 d \nu_u$$ $\endgroup$
    – N. S.
    Commented Aug 23, 2019 at 3:04
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    $\begingroup$ 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? $\endgroup$
    – N. S.
    Commented Aug 23, 2019 at 3:06
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    $\begingroup$ 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 $$C:=sup_{\lambda \in I} \{\lambda ^2\}?$$ $\endgroup$
    – N. S.
    Commented Aug 23, 2019 at 3:17

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