Dgdrxrt

Let $H$ be a $mathbb R$-Hilbert space and $H_lambda$ be a closed subspace of $H$ for $lambdainmathbb R$. Assume $H_lambdasubseteq H_mu$ for all $lambda,muinmathbb R$ with $lambdalemu$ and let $pi_lambda$ denote the orthogonal projection of $H$ onto $H_lambda$ for $lambdainmathbb R$. Moreover, let $$varrho_x(lambda):=left|pi_lambda xright|_H^2;;;textfor lambdainmathbb R$$ and $$operatorname P_x(lambda):=pi_lambda x;;;textfor lambdainmathbb R$$ for $xin H$.

Fix $xin H$. Are we able to show that the variation of $varrho_x$ and $operatorname P_x$ on any interval coincides?

In order to prove the desired claim, let $kinmathbb N$ and $lambda_0,ldots,lambda_kinmathbb R$ with $lambda_0lecdotslelambda_k$. We need to show that $$A:=sum_i=1^kleft|varrho_x(lambda_i)-varrho_x(lambda_i-1)right|=sum_i=1^kleft|operatorname P_x(lambda_i)-operatorname P_x(lambda_i-1)right|_H=:B.$$ Since $varrho_x$ is nondecreasing, $A=varrho_x(lambda_k)-varrho_x(lambda_0)$. Moreover, we easily see that $$left|operatorname P_x(lambda_i)-operatorname P_x(lambda_i-1)right|_H^2=varrho_x(lambda_i)-varrho_x(lambda_i-1)tag1$$ for all $iinleft1,ldots,kright$.

So, it seems like the claim is wrong, if I'm not missing something. I would at least like show that the variation of $operatorname P_x$ is bounded by the variation of $varrho_x$, i.e. $Ble A$, but that seems to be wrong too.

Remark: The question came up to my mind as I was considering the construction of the spectral measure for self-adjoint operators on a Hilbert space. In the construction of that measure, once is basically integrating against a right-continuous function of bounded variation of the form $operatorname P_x$, but one is usually defining the domain of the integration to be $L^2(varrho_x)$. So, it seems like one is thinking that $varrho_x$ and $operatorname P_x$ (which gives rise to a Lebesgue-Stieltjes vector measure) have the same variation.