Let $A$ be an unbounded self-adjoint operator on a Hilbert space $H$, I was wondering if there was an 'elementary proof', i.e. that doesn't use the full functional calculus, of the strong limit :
$$\displaystyle s-\lim_{\varepsilon \to 0} (I+i\varepsilon A)^{-1}=I$$
where $I$ is the identity map of $H$.
I was hoping that something like this might work: let $u\in \mathcal{D}(A)$, set $y_\varepsilon=(I+i\varepsilon A)^{-1}u$, $y_\varepsilon$ is uniformly bounded and: $(I+i\varepsilon A)y_\varepsilon=u$ for any $\varepsilon>0$. If one can show that $i\varepsilon Ay_\varepsilon\to 0$ when $\varepsilon \to 0$ then the result follows, but showing this seems to be pretty much the same as the original problem.
Edit: I mean by $s-\lim$ a limit in the strong operator topology.
You can use the following result:
If $\|T-I\|<1$, then $T^{-1}=\sum_{k=0}^\infty(I-T)^k$, which implies that $\|T^{-1}\|\leq \dfrac{1}{1-\|I-T\|}$.
For small $\varepsilon$, let $T=I+i\varepsilon A$. Then one has \begin{eqnarray} &&\|(I+i\varepsilon A)^{-1}-I\|\\ &=&\|(I-(I+i\varepsilon A))(I+i\varepsilon A)^{-1}\|\\ &=&|\varepsilon|\|A(I+i\varepsilon A)^{-1}\|\\ &\le&|\varepsilon|\|A\|\|(I+i\varepsilon A)^{-1}\|\\ &\le&\varepsilon\|A\|\frac{1}{1-\|I-(I+i\varepsilon A)\|}\\ &=&\varepsilon\|A\|\frac{1}{1-|\varepsilon| \|A\|\|}. \end{eqnarray} Now it is easy to see $$ \lim_{\varepsilon\to0}\|(I+i\varepsilon A)^{-1}-I\}\|=0 $$ or equivalently $$\displaystyle s-\lim_{\varepsilon \to 0} (I+i\varepsilon A)^{-1}=I.$$