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I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(\lambda)$, $E_B(\lambda)$ its corresponding spectral resolutions, i.e., $$A=\int_{[m_A,M_A)}t\;dE_A(t)\qquad\text{and}\qquad B=\int_{[m_B,M_B)}t\;dE_B(t).$$ If $A\geq B$ (in the sense of positive operators) then $\mathrm{dim}(\mathrm{rg}E_A(\lambda))\leq \mathrm{dim}(\mathrm{rg}E_B(\lambda))$ for all $\lambda\in\mathbb{R}$.

I think for each $\lambda$ we can define the linear operator $T_\lambda:\mathrm{rg}(E_B(\lambda))\to\mathrm{rg}(E_A(\lambda)),\;x\mapsto E_A(\lambda)x$ and prove that this opeator is surjective but I do not know how to do it.

Can someone give me an idea? I will be grateful.

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Suppose the range of $E_A(\lambda)$ has strictly larger dimension than the range of $E_B(\lambda)$, for some $\lambda$. Then we can find a vector $v$ in the first range which is orthogonal to the second range, i.e., is in the range of $I-E_B(\lambda)$. Let $P$ be the orthogonal projection onto the (one dimensional) span of $v$. Since $A\geq B$, also $PAP \geq PBP$. But $PAP \leq \lambda P < PBP$, contradiction.

(I am taking $E_B(\lambda)$ to be the spectral projection of $B$ for the interval $(-\infty, \lambda]$; if you want it to be $(-\infty, \lambda)$ then you would have $PAP < \lambda P \leq PBP$.)

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  • $\begingroup$ You mean $v\in\mathrm{rg}(I-E_B(\lambda))$? $\endgroup$
    – Andrés Felipe
    Commented Sep 11, 2020 at 19:23
  • $\begingroup$ There is also a second part in the question which follows similarly. If $X$ is the Hilbert space and $E_AE_B(X) \neq E_A (X)$ (I omit $\lambda$), there is a norm-one $x \in E_A(X)$ orthogonal to $E_AE_B(X)$. Then $0=(x, E_AE_By)=(x,E_By)=(E_Bx,y)$ for all $y$ gives $E_Bx=0$. Then $x \in (I-E_B)(X)$ and then $(Bx,x)>\lambda$ and $(Ax,x) \geq (Bx,x) >\lambda$, in contrast with $x \in E_A(X)$. $\endgroup$
    – Giorgio Metafune
    Commented Sep 11, 2020 at 19:29
  • $\begingroup$ @AndrésFelipe you are right, corrected. $\endgroup$
    – Nik Weaver
    Commented Sep 11, 2020 at 19:32
  • $\begingroup$ It is not clear to me why $PAP\leq \lambda{P}$. Do I have to use the integral representation of A? $\endgroup$
    – Andrés Felipe
    Commented Sep 11, 2020 at 19:38
  • $\begingroup$ @AndrésFelipe the cheap way to do this is to assume, without loss of generality, that $A$ is a multiplication operator, $A = M_f$ on some $L^2(X)$, and the range of $E_A(\lambda)$ is the set of vectors supported on $\{x: f(x) \leq \lambda\}$.Then $\lambda P - PAP = (\lambda I - A)P = M_{\lambda - f)P$ is multiplication by a positive function on that range, so it is positive. $\endgroup$
    – Nik Weaver
    Commented Sep 11, 2020 at 19:57

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