This question is from section 10.6 in Kreyszig's Introductory Functional Analysis text. Let $H$ be a Hilbert space and let $D$ be a dense subspace of $H$. Let $T:D\rightarrow H$ be a self-adjoint operator with Cayley transform $U$. Then $1\in \rho (U) $ if and only if $T$ is bounded.
It was easy enough to show that if $1\in \rho (U) $ then $T$ is bounded because in that situation, $$T=i(I+U)(I-U)^{-1} $$ is a composition of bounded operators on $H$.
The converse has proved to be a little bit trickier. Kreyszig explicitly proves that 1 cannot be an eigenvalue of $U$ and because $D$ is dense and the above equation, it follows that $(I-U)^{-1} $ is densely defined. However, I have not been able to show that $(I-U)^{-1} $ is bounded and hence have not ruled out the possibility that 1 is in the continuous spectrum for $U$.
Suppose $T$ is bounded. Check that $\tfrac 12 (iT-I)$ is an inverse to $U-I$. Hence $1 \in \rho(U)$.