# Show that the Cayley transform of a bounded self-adjoint operator on a Hilbert space has the number 1 in its resolvent.

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 $$1 \in \rho(U)$$.