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.



On the dimension of the range of the resolution of the identity
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 $PAP < \lambda P \leq PBP$.)


