proving a quadratic form is closed

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I'm trying to show that, given a spectral measure $d\mu_\psi(\lambda)$ for a self-adjont operator $A$, for the following quadratic form

$$q_\lambda(\psi)=\int_{\mathbb R}\chi_{(-\infty,\lambda]}(\tau) d\mu_\psi (\tau)$$

there exists an operator $P(\lambda)$ such that:

$$<\psi|P(\lambda)\psi>= q_\lambda(\psi)$$

In particular, $q_\lambda(\psi)$ is bounded from below and so I have just to prove that it is closed.

Well, I know that in order to show closedness I should prove that the domain of the quadratic form $Q$ is complete with respect to the norm form

$$||\psi||_q= q_\lambda(\psi)+||\psi||_H$$ where $H$ is our generic Hilbert space.

I have some troubles proving this, since chosen a Cauchy sequence on $H$, I can't understand how $q_\lambda(\psi_n-\psi_m)$ behaves.

Any help would be greatly apppreciated!

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This is a bounded form: $0 \le q_{\lambda}(f) \le \|f\|^{2}$, regardless of $\lambda$, and it is defined everywhere on the Hilbert space. So the form norm $q_{\lambda}(f)+\|f\|^{2}$ and the usual norm are equivalent norms.