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I really need this as tool for other threads!

Given a Hilbert space $\mathcal{H}$.
Also a Borel space $\Omega$.

Consider a spectral measure: $$E:\mathcal{B}(\Omega)\to\mathcal{P}(\mathcal{H}):\quad E(\sum_kA_k)\varphi=\sum_kE(A_k)\varphi$$

Regard the total measure: $$|\langle E\varphi,\chi\rangle|(A):=\sup_\mathcal{A}\sum_{A\in\mathcal{A}}|\langle E(A)\varphi,\chi\rangle|$$

Then for every Borel function: $$\omega\in\mathcal{B}(\Omega):\quad\left(\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|}\right)^2\leq\int|\omega|^2\operatorname{d\|E\varphi\|}\cdot\|\chi\|^2$$ How can I proof this estimate?

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If $P$ is a positive operator, then $[x,y]_P = (Px,y)$ is a pseudo inner product, meaning that $(Px,x)=0$ might be true even if $x\ne 0$. For any pseudo inner product, Cauchy-Schwarz still holds: $$ |[x,y]_P| \le [x,x]_P^{1/2}[y,y]_P^{1/2}. $$ Therefore, if $P$ is a positive Borel operator measure on a space $\Omega$, \begin{align} \left|\sum_{n}a_n(P(E_n)x,y)\right| & \le \sum_{n}|a_n|(P(E_n)x,y)| \\ & \le \sum_{n}|a_n|(P(E_n)x,x)^{1/2}(P(E_n)y,y)^{1/2} \\ & \le \left(\sum_{n}|a_n|^2(P(E_n)x,x)\right)^{1/2}\left(\sum_{n}(P(E_n)y,y)\right)^{1/2} \\ & \le \left(\sum_{n}|a_n|^2(P(E_n)x,x)\right)^{1/2}(P(\Omega)y,y)^{1/2} \end{align} In other words, if $f$ is a simple function on $\Omega$, $$ \left|\int f(\omega)d(P(\omega)x,y)\right| \le \left(\int |f(\omega)|^2 d(P(\omega)x,x)\right)^{1/2}\|P(\Omega)\|^{1/2}\|y\|. \tag{$\dagger$} $$ Because $P(\Omega)=I$ for a spectral measure, the above gives you what you want. But you don't need to assume a spectral measure; a positive operator measure is good enough. Therefore, if $f \in L^2(\Omega;dP((\cdot)x,x))$, there exists a unique vector--call it $\int f(\omega) dP(\omega)x$--such that $$ \int f(\omega)d(P(\omega)x,y)=\left(\int f(\omega)dP(\omega)x\;,\;y\right), \;\;\; y \in \mathcal{H}. $$ And, if one assumes the normalization $\|P(\Omega)\|=1$, $$ \left\|\int f(\omega)dP(\omega)x\right\|^2 \le \int |f(\omega)|^2 d(P(\omega)x,x) $$ Variation Integral: Because you don't want to use Radon-Nikodym, then start with a Borel set $E$ and any finite or countable Borel partition $\{E_n\}$ of $E$, and use the same arguments leading to $(\dagger)$. Start by choosing unimodular constants $\alpha_n$ such that $\alpha_n\mu_{x,y}(E_n)=|\mu_{x,y}(E_n)|$. Then \begin{align} \sum_{n}|\mu_{x,y}(E_n)| & = \sum_{n}\alpha_n\mu_{x,y}(E_n) \\ & = \sum_{n}\mu_{\alpha_n x,y}(E_n) \\ & \le \sum_{n}\mu_{x,x}(E_n)^{1/2}\mu_{y,y}(E_n)^{1/2} \\ & \le \left(\sum_{n}\mu_{x,x}(E_n)\right)^{1/2}\left(\sum_{n}\mu_{y,y}(E_n)\right)^{1/2} \\ & = (\mu_{x,x}(E))^{1/2}(\mu_{y,y}(E))^{1/2} \end{align} Therefore the variation of the complex measure $\mu_{x,y}$ is bounded by $$ |\mu_{x,y}|(E) \le \mu_{x,x}(E)^{1/2}\mu_{y,y}(E)^{1/2} $$ For any simple function $\sum_{n}\alpha_n E_n$ with disjoint Borel sets $E_n$, \begin{align} \int |f|d|\mu_{x,y}| & = \sum_{n}|\alpha_n||\mu_{x,y}|(E_n) \\ & \le \sum_{n}|\alpha_n|\mu_{x,y}(E_n)^{1/2}\mu_{y,y}(E_n)^{1/2} \\ & \le \left(\sum_{n}|\alpha_n|^2\mu_{x,x}(E_n)\right)^{1/2}\left(\sum_{n}\mu_{y,y}(E_n)\right)^{1/2} \\ & \le \left(\int |f|^2d\mu_{x,x}\right)^{1/2}\|y\| \end{align}

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  • $\begingroup$ @Freeze_S : Where do you see $(\dagger)$ failing to extend to all bounded Borel functions from simple functions? $\endgroup$ – DisintegratingByParts Mar 20 '16 at 15:05
  • $\begingroup$ @Freeze_S : I started deleting comments to appease the comment police. I've replaced the Radon-Nikodym argument that you didn't like with the same technique I used before with the generalized Cauchy-Schwarz. It gives you want you want for simple functions, which can be easily extended using only positive measures to give you the desired inequality. $\endgroup$ – DisintegratingByParts Mar 20 '16 at 19:19
  • $\begingroup$ Oh that's a good one thanks!! I really like your new argument!!! Even better than my answer.. :) Besides don't get me wrong I love Radon-Nikodym! :D $\endgroup$ – C-Star-W-Star Mar 20 '16 at 22:52
  • $\begingroup$ @Freeze_S : By the way, have you ever seen the Riesz Representation proof of Radon-Nikodym? That's another gem of von Neumann. $\endgroup$ – DisintegratingByParts Mar 21 '16 at 1:38
  • $\begingroup$ Yep extremely clever proof of him - like it a lot. :) $\endgroup$ – C-Star-W-Star Mar 21 '16 at 1:39
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Precalculus

For simple functions: $$\sigma\in\mathcal{B}(\Omega):\quad\int\sigma\operatorname{dE}:=\sum_k\sigma_kE(A_k)\quad(\sigma=\sum_k\sigma_k1_k)$$

It holds the relation: $$\|\{\int\sigma\operatorname{dE}\}\varphi\|^2=\sum_{kl}\sigma_k\overline{\sigma_l}\langle E(A_k)\varphi,E(A_l)\varphi\rangle\\ =\sum_k|\sigma_k|^2\|E(A_k)\varphi\|^2=\int|\sigma|^2\operatorname{d\|E\varphi\|^2}$$

Especially one obtains: $$\|\{\int\sigma\operatorname{dE}\}\varphi\|^2\leq\|\sigma\|_\infty\|\varphi\|\implies\|\{\int\sigma\operatorname{dE}\}\|\leq\|\sigma\|_\infty$$

For bounded functions:* $$\omega\in\mathcal{B}(\Omega):\quad\int\omega\operatorname{dE}:=\lim_n\int\sigma_n\operatorname{dE}\quad(\|\omega\|_\infty<\infty)$$

So the relation remains: $$\|\{\int\omega\operatorname{dE}\}\varphi\|^2=\lim_n\|\{\int\sigma_n\operatorname{dE}\}\varphi\|^2\\=\lim_n\int|\sigma_n|^2\operatorname{d\|E\varphi\|^2}=\int|\omega|^2\operatorname{d\|E\varphi\|^2}$$

*Convergence: Operatornorm!

Integration

By Radon-Nikodym: $$\langle E(A)\varphi,\chi\rangle=\int_A\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}}\operatorname{d|\langle E\varphi,\chi\rangle|}=:\int_A(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$

For simple functions: $$\langle\{\int\sigma\operatorname{dE}\}\varphi,\chi\rangle=\langle\{\sum_k\sigma_kE(A_k)\}\varphi,\chi\rangle\\ =\sum_k\sigma_k\langle E(A_k)\varphi,\chi\rangle=\int\sigma(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$

For bounded functions:* $$\langle\{\int\omega\operatorname{dE}\}\varphi,\chi\rangle=\lim_n\langle\{\int\sigma_n\operatorname{dE}\}\varphi,\chi\rangle\\ =\lim_n\int\sigma_n(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}=\int\omega(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}$$

*Convergence: Uniform!

Estimate

For the derivative:* $$|\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}}|\equiv1\implies|(\dfrac{\operatorname{d\langle E\varphi,\chi\rangle}}{\operatorname{d|\langle E\varphi,\chi\rangle|}})^{-1}|\equiv1$$

For bounded functions: $$0\leq\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|}=\int|\omega|(\ldots)^{-1}(\ldots)\operatorname{d|\langle E\varphi,\chi\rangle|}=\langle\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi,\chi\rangle$$

By Cauchy-Schwarz: $$|\langle\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi,\chi\rangle|\leq\|\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi\|\cdot\|\chi\|$$

And one arrives at: $$\|\{\int|\omega|(\ldots)^{-1}\operatorname{dE}\}\varphi\|^2=\int|\omega|^2\cdot1\operatorname{d\|E\varphi\|^2}=\int|\omega|^2\operatorname{d\|E\varphi\|^2}$$

For unbounded functions: $$(\int|\omega|\operatorname{d|\langle E\varphi,\chi\rangle|})^2=\lim_n(\int|\omega_n|\operatorname{d|\langle E\varphi,\chi\rangle|})^2\\ \leq\lim_n\int|\omega_n|^2\operatorname{d\|E\varphi\|^2}\cdot\|\chi\|^2=\int|\omega|^2\operatorname{d\|E\varphi\|^2}\cdot\|\chi\|^2$$

*Choice: Everywhere!

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