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

Use the convention that the inner product is linear in its first argument. The precise polarization identity isIf (1) holds, apply this formula before and after $U$ to get preservation of all inner products, proving (2). If (2) holds, the identity $⟨Ux,Uy⟩=⟨x,U^{\ast }Uy⟩$ gives $U^{\ast }U=I$. For $z=Ux$, surjectivity then gives $U{U}^{\ast }z=U{U}^{\ast }Ux=Ux=z$, proving (3).

From (3), $\Vert Ux{\Vert }^2=⟨x,U^{\ast }Ux⟩=\Vert x{\Vert }^2$ and likewise $\Vert U^{\ast }x\Vert =\Vert x\Vert$, proving (4). Finally, if (4) holds, apply the same polarization identity to $U^{\ast }$ to obtain $U{U}^{\ast }=I$. This makes $U$ onto; the other norm identity supplies the remaining requirement of (1). ThusThese four statements are conditions in a single equivalence proof, rather than four independent source questions.

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

If the real part vanishes identically, the norm-difference identity becomes equality, soThus the same substitution as above gives $\Vert Sz\Vert =\Vert z\Vert$ for every $z$. To prove surjectivity as well, observe that both $T$ and $-T$ satisfy the nonnegative-real-part hypothesis. The preceding part makes both $I+T$ and $I-T$ invertible. Their product $S=(I-T)(I+T{)}^{-1}$ is therefore invertible and onto. It is a surjective norm-preserving linear operator, so the equivalences in (a) prove