If one wants to define a continuous functional calculus for unbounded self-adjoint operators, it seems like an obvious approach to use is to use the functional calculus for bounded normal operators and apply it to the resolvent. Concretely, if $A$ is an unbounded self-adjoint operator on a Hilbert space, then for any function $f$ which is continuous on the compactified real line one can define $f(A)=g(R(\lambda))$ for an appropriate continuous function $g$, where $R(\lambda)=(A-\lambda)^{-1}$ is a resolvent for $A$ associated to some $\lambda$ in the resolvent set. My question is if there is a way to generalize this approach to allow for bounded functions which are continuous on the real line but which do not have a limit at infinity. This limit seems to be important because without it, $g$ will not be continuous. And, plenty of important examples are of this class, such as $e^{-iAt}$.
Two ideas with which I am familiar: first, one can use the functional calculus for bounded operators and functions which are bounded but not necessarily continuous. I assume that this immediately solves the problem (I would appreciate it if someone would confirm this), but it seems funny to need to invoke non-continuous functions in order to solve this problem. Second, I read Lax's proof of the Hille-Yosida theorem in his Functional Analysis book, and there he uses resolvents to define exponentials of unbounded self-adjoint operators. So, at least in this special case it is possible to directly define the operator by using resolvents, but it would be nice to have an approach that works for a general $f$. (Perhaps the approach Lax describes does generalize to a general $f$? Namely, can you define $f(A)v = \lim_{n\to\infty} f(-inAR(in))v$ for all vectors $v$?)
Finally, I am aware that spectral measures allow one to easily define general functional calculi. But, it seems to me that the technology of measure theory should not be necessary to solve this problem, since the approaches above using resolvents should suffice.
Thanks!