Past exam of the analysis course of the University of Oxford / 2011 / 6 / b

The four claims below show that $I+T$ has a bounded inverse and that the Cayley transform of a bounded accretive operator is contractive. For an arbitrary $z\in X$, write $z=(I+T)x$. ThenTaking the supremum over the unit ball gives

Expanding both squared norms givesBoth norms are nonnegative, so taking square roots proves $\boxed{\Vert (I+T)x\Vert \ge \Vert (I-T)x\Vert }$.

A separate expansion givesHence $\boxed{\Vert (I+T)x\Vert \ge \Vert x\Vert }$. In particular $I+T$ is injective and bounded below by one.

By the definition of the adjoint operator and conjugate symmetry,Taking real parts proves

The lower bound in (ii) makes the image of $I+T$ closed: if $(I+T)x_n$ converges, the bound makes $(x_n)$ Cauchy; completeness gives its limit and hence a image preimage. The same bound applies to $I+T^{\ast }$ by (iii), so its kernel is zero. The orthogonal complement identityshows that the image is dense, using the orthogonal decomposition by a closed subspace for its closure. Being both closed and dense, it equals $X$. Thus $I+T$ is onto and bounded below, and by part 3(a)(iii) it has a bounded inverse. Moreover