Let $H$ be a $\mathbb R$-Hilbert space and $\pi_\lambda$ be an orthogonal projection on $H$ for $\lambda\in\mathbb R$ such that $(\pi_\lambda)_{\lambda\in\mathbb R}$ is nondecreasing, i.e. $$\langle\pi_\lambda x,x\rangle_H\le\langle\pi_\mu x,x\rangle_H\;\;\;\text{for all }x\in H\text{ and }\lambda\le\mu\tag1,$$ and $$\mathbb R\ni\lambda\mapsto\pi_\lambda\tag2$$ is right-continuous with respect to the strong operator topology. Moreover, assume that $$\pi_\lambda\xrightarrow{\lambda\to-\infty}0\tag3$$ and $$\pi_\lambda\xrightarrow{\lambda\to\infty}\operatorname{id}_H\tag4.$$
We know that a function $\mathbb R\to\mathbb R$ is of bounded variation if and only if it is the difference of two bounded nondecreasing functions $\mathbb R\to\mathbb R$. Now we may notice that $$\varrho_x(\lambda):=\langle\pi_\lambda x,x\rangle_H=\left\|\pi_\lambda x\right\|_H^2\;\;\;\text{for }\lambda\in\mathbb R$$ is bounded, nondecreasing and right-continuous for all $x\in H$. Thus, $$\varrho_{x,\:y}(\lambda):=\langle\pi_\lambda x,y\rangle_H=\frac{\varrho_{x+y}(\lambda)-\varrho_x(\lambda)-\varrho_y(\lambda)}2\;\;\;\text{for }\lambda\in\mathbb R$$ is right-continuous and of bounded variation for all $x,y\in H$. Now, with the notion of nondecreasingness given by $(1)$, I wonder whether we are able to show that $(2)$ is of bounded variation as well.