Past exam of the analysis course of the University of Oxford / 2010 / B / 5

A bounded linear operator on a complex Hilbert space is self-adjoint if $T=T^{\ast }$, equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y$.

If $Tx=0$, then $⟨x,Ty⟩=⟨Tx,y⟩=0$ for all $y$, so $x\in (\mathrm{Ran}T{)}^{\perp }$. Conversely, membership in that orthogonal complement gives $⟨Tx,y⟩=0$ for every $y$, and choosing $y=Tx$ yields $Tx=0$. Therefore

Self-adjointness and conjugate symmetry of the inner product giveA complex number equal to its conjugate is real, so $⟨Tx,x⟩\in \mathbb{R}$ for every $x$.

For a positive operator, $⟨Tx,x⟩\ge 0$. We show directly that every nonreal or negative real $\lambda$ lies in the resolvent set.

If $\mathrm{Im}\lambda \ne 0$, the Cauchy-Schwarz inequality and reality of $⟨Tx,x⟩$ giveThus $T-\lambda I$ is bounded below, hence injective with closed range: a convergent sequence of outputs has Cauchy preimages, whose limit maps to that output. Its adjoint is $T-\bar{\lambda }I$, which has trivial kernel by the same estimate. The identity $\mathrm{Ran}(A{)}^{\perp }=\ker A^{\ast }$, proved by the same inner-product calculation as in (i), makes its range dense. Being closed and dense, it is all of $X$, and the lower bound makes the inverse bounded.

If $\lambda <0$ is real, positivity gives $⟨(T-\lambda I)x,x⟩\ge (-\lambda )\Vert x{\Vert }^2$, so the same argument applies with lower bound $-\lambda$. Here the operator is self-adjoint, so its range is dense because its own kernel is zero. Hence

Set $B(x,y)=⟨Tx,y⟩$. It is a positive semidefinite Hermitian form. Write $A=B(x,x)$, $C=B(y,y)$ and $c=B(x,y)$. For every complex $t$,If $C>0$, choosing $t=-c/C$ gives $A-|c{|}^2/C\ge 0$. If $C=0$ and $c\ne 0$, choosing $t=-sc$ for large positive $s$ makes the expression negative, a contradiction. Thus $c=0$ in that case. In all cases,This is the Cauchy-Schwarz inequality for a positive operator.

For $x\in \ker (S+T)$,Both terms are nonnegative because $S,T$ are positive operators. Their sum is zero, so each vanishes. In particular,

By the Cauchy-Schwarz inequality for a positive operator, for every $y$,Taking $y=Tx$ gives $Tx=0$, and $(S+T)x=0$ then gives $Sx=0$. ThusThe reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.

Commutativity and equality of squares giveThus $S-T$ vanishes on $\mathrm{Ran}(S+T)$, and by continuity on its closure. The preceding kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Since $S+T$ is self-adjoint, part (i) and the Hilbert projection theorem give the orthogonal direct sumTherefore $S-T$ vanishes on all of $X$, provingThis proves commuting positive operators with equal squares are equal without assuming a square-root uniqueness theorem.