$\int_{\lambda+\varepsilon}^{+\infty}{d||E_{t}v||^{2}}=0$ implies $v\in \text{Im}1_{\{\lambda\}}(H)$ Let $H$ be a non-negative definite self-adjoint operator in $L^{2}$ and $\{E_{t}\}_{t\in \mathbb{R}}$ be its spectral resolution, that is $E_{t}=1_{(-\infty, t)}(H)$. Suppose that we have shown that for some $\lambda\geq 0$ and $v\in L^{2}$ that
$$\lambda\int_{\lambda}^{+\infty}{d||E_{t}v||^{2}}=\int_{\lambda}^{+\infty}{td||E_{t}v||^{2}}$$
Then why does it follow that $\forall \varepsilon>0$ $$\int_{\lambda+\varepsilon}^{+\infty}{d||E_{t}v||^{2}}=0$$ and why does this imply that $v\in \text{Im}1_{\{\lambda\}}(H)$?
Thanks in advance!
 A: Let me set $\mu_v(A) = \|E(A)v\|^2$ for a Borel set $A$. You have
$$
0=\int_{\lambda}^\infty (1-t)\,d\mu_v(t) = \int_{-\infty}^\infty\chi_{[\lambda,\infty)}(t)(1-t)\,d\mu_v(t) = \big(E([\lambda,\infty))(I-H)v,v\big).
$$
This always holds when $v$ has its support in $(-\infty,\lambda)$, i.e., $E((-\infty,\lambda))x = x$, showing that the second implication $v\in\operatorname{Im} E(\{\lambda\})$ is false. The first implication is true if $\lambda\ge 1$. But it is false if $\lambda < 1$. Let us go through the cases. But before, define $w = E([\lambda,\infty))v$. Then we have $((I-H)w,w) = 0$. Also set $X_\lambda = \operatorname{Im} E([\lambda,\infty))$.
$\lambda > 1$: For $u\in X_\lambda$ we have $(Hu,u)\ge\lambda\|u\|^2$. Since $w\in X_\lambda$, it follows that $0 = ((H-I)w,w) = (Hw,w)-\|w\|^2\ge(\lambda-1)\|w\|^2$, hence $w=0$. The first implication now easily follows.
$\lambda=1$: Then $((H-I)u,u)\ge 0$ for all $u\in X_\lambda$. If $T$ denotes the restriction of $H-I$ to $X_\lambda$, we have $(Tu,u)\ge 0$ for $u\in X_\lambda$, hence $T\ge 0$. Thus, $0 = ((H-I)w,w) = (Tw,w) = \|T^{1/2}w\|^2$, which implies $T^{1/2}w = 0$ and therefore $Tw = 0$, which means $Hw=w = \lambda w$. So, $E(\{\lambda\})w = w$ and whence $E([\lambda+\epsilon,\infty))v = E([\lambda+\epsilon,\infty))E([\lambda,\infty))v = E([\lambda+\epsilon,\infty))w = 0$.
$\lambda < 1$: A counterexample is $H|_{X_\lambda} = I|_{X_\lambda}$. Then $((I-H)w,w) = 0$, so your condition is satisfied, but $\int_{\lambda+\epsilon}^\infty\,d\mu_v(t) = \mu_v([\lambda+\epsilon,\infty)) = \mu_w([\lambda+\epsilon,\infty)) = \|w\|^2$.
