# projection-valued measure and bounded self-adjoint operator

Let $E$ be a projection-valued measure on Borel sets of $\mathbb{R}$ and assume $E([-R,R]) = \text{id}$ for some $R$. For each $x\in \mathcal{H}$, define the Boreal measure $\nu_x(B) = \left\lVert E(B)x\right\rVert^2$. How to show that there exists a bounded self-adjoint operator $T$ in $\mathcal{H}$ such that $$\langle Tx,\, x\rangle = \int_\mathbb{R} \lambda \;dv_x(\lambda)$$

• You could just take $T =\int_\mathbb{R} f dE$ where $f(t) = t$ for $t \in [-R,R]$ and zero elsewhere. – Demophilus Nov 17 '17 at 11:34

Define $$b(x,y) = \int_{-R}^{R}\lambda d_{\lambda}\langle E(\lambda)x,y\rangle.$$ By polarization, $\mu_{x,y}(S)=\langle E(S)x,y\rangle$ is a complex measure that can be written in terms of finite positive measures \begin{align} \mu_{x,y}(S)=\langle E(S)x,y\rangle&=\frac{1}{4}\sum_{n=0}^{3}i^n\langle E(S)(x+i^ny),x+i^ny\rangle \\&= \frac{1}{4}\sum_{n=0}^{3}i^n\mu_{x+i^ny,x+i^ny}(S). \end{align} So the variation of $\mu_{x,y}$ is bounded by $$|\mu_{x,y}|(S) \le \frac{1}{4}\sum_{n=0}^{3}\mu_{x+i^ny,x+i^ny}(S) = \mu_{x,x}(S)+\mu_{y,y}(S)$$ Therefore, $$|b(x,y)| \le R\mu_{x,x}[-R,R] + R\mu_{y,y}[-R,R] \le R(\|x\|^2+\|y\|^2)$$ Replacing $x$, $y$ by $\frac{1}{\|x\|}x$, $\frac{1}{\|y\|}y$, respectively, gives the bound $|b(x,y)| \le R\|x\|\|y\|$. So there is a unique bounded linear operator $B$ such that $b(x,y)=\langle Tx,y\rangle$. $T$ is selfadjoint because $b$ is symmetric. And $\|T\| \le R$. Finally, $$\langle Tx,x\rangle = b(x,x) = \int_{-R}^{R}\lambda d_{\lambda}\langle E(\lambda)x,x\rangle,\;\; x\in \mathcal{H}.$$ The selfadjont operator $T$ is uniquely determined by the above.