# operator theory – On the dimension of the range of the resolution of the identity

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.,
$$A=int_{(m_A,M_A)}t;dE_A(t)qquadtext{and}qquad B=int_{(m_B,M_B)}t;dE_B(t).$$
If $$Ageq B$$ (in the sense of positive operators) then $$mathrm{dim}(mathrm{rg}E_A(lambda))leq mathrm{dim}(mathrm{rg}E_B(lambda))$$ for all $$lambdainmathbb{R}$$.

I think for each $$lambda$$ we can define the linear operator $$T_lambda:mathrm{rg}(E_B(lambda))tomathrm{rg}(E_A(lambda)),;xmapsto E_A(lambda)x$$ 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.