Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 3

An element $x$ of a unital C-star algebra is positive when $x=y^{\ast }y$ for some $y$, equivalently when $x=x^{\ast }$ and $\sigma (x)\subseteq [0,\infty )$. The continuous functional calculus for the nonnegative function $t\mapsto \sqrt{t}$ defines a positive element $x^{1/2}$ satisfying $(x^{1/2}{)}^2=x$. If a positive $z$ also satisfies $z^2=x$, functional calculus for $z$ gives $z=\sqrt{z^2}=\sqrt{x}$, proving uniqueness. For a positive operator $T=S^{\ast }S$ on ${\ell }^2$,

For arbitrary $T$, put $|T|=(T^{\ast }T{)}^{1/2}$. Thenso $\ker |T|=\ker T$. Defineon $\mathrm{im}|T|$. The kernel identity makes this well-defined, and the norm identity makes it an isometry. Extend it continuously toand set it equal to zero on $\ker T$. The resulting $U$ is a partial isometry, has $\ker U=\ker T$, and satisfies the polar decomposition of a bounded operator $T=U|T|$.

Finally $U^{\ast }U$ is the orthogonal projection onto $(\ker T{)}^{\perp }$, which contains the range of $|T|$. Therefore

Solved by gpt-5.6-sol high.

Twice applying Parseval identity for a Hilbertian basis to the matrix coefficients givesThus $\Vert T{\Vert }_2=\Vert T^{\ast }{\Vert }_2$.

Solved by gpt-5.6-sol high.

The operator norm givesUsing part i and the first inequality for the adjoints gives

Solved by gpt-5.6-sol high.

By the definition of the trace norm and positivity of $|T|$,

Solved by gpt-5.6-sol high.

Let $T=AB$ with $A,B$ Hilbert--Schmidt, and write $T=U|T|$. Since $|T|=U^{\ast }AB$, part iii and the Cauchy-Schwarz inequality for series giveThus $T$ is trace class with the required bound.

Solved by gpt-5.6-sol high.

For the polar decomposition $T=U|T|$, setThen $T=AB$. Part iii gives $\Vert B{\Vert }_2^2=\Vert T{\Vert }_1$. The range of $B$ lies in the initial space $(\ker T{)}^{\perp }$ on which $U$ is isometric, so $\Vert A{\Vert }_2=\Vert B{\Vert }_2$. Hence $A,B$ are Hilbert--Schmidt andproving the Hilbert-Schmidt factorization of a trace-class operator.

Solved by gpt-5.6-sol high.

Use the norm-attaining factorization $T=AB$ from part v. Then $ST=(SA)B$, parts ii and iv giveThus the trace-class operators form a left ideal. Applying the result to adjoints gives the corresponding right-ideal estimate as well.

Solved by gpt-5.6-sol high.

Factor $T=AB$ as in part b(v). Then Parseval identity for a Hilbertian basis and Cauchy-Schwarz inequality giveThe defining series for the operator trace is therefore absolutely convergent and satisfies $|\mathrm{tr}T|\le \Vert T{\Vert }_1$.

For unit vectors $x,y$, the adjoint operator of $R=x\otimes y$ is $R^{\ast }=y\otimes x$, andHence $|R|=P_{\mathbb{C}y}$ and part b(iii), followed by Parseval, givesScaling proves $\Vert x\otimes y{\Vert }_1=\Vert x\Vert \Vert y\Vert$. A second use of Parseval yieldswhich proves the asserted rank-one operator formulas.

Solved by gpt-5.6-sol high.

Part b(vi) and part c show that $\Vert T{\Vert }_1\le 1$ impliesConversely, choose a unit vector $x$ with $\Vert Sx\Vert$ arbitrarily close to $\Vert S\Vert$, put $y=Sx/\Vert Sx\Vert$, and take $T=x\otimes y$. Part c gives $\Vert T{\Vert }_1=1$ andTaking the supremum provesThus $S\mapsto [T\mapsto \mathrm{tr}(ST)]$ is an isometric embedding ${\mathcal{S}}_{\infty }\to {\mathcal{S}}_1^{\ast }$.

Suppose finite-rank operators are dense in ${\mathcal{S}}_1$ and let $L\in {\mathcal{S}}_1^{\ast }$. The sesquilinear formsatisfies $|b(x,y)|\le \Vert L\Vert \Vert x\Vert \Vert y\Vert$. The Riesz representation theorem gives $S\in {\mathcal{S}}_{\infty }$ with $b(x,y)=⟨Sx,y⟩$. Hence $L(x\otimes y)=\mathrm{tr}(S(x\otimes y))$, and linearity gives equality on every finite-rank operator. Density and continuity extend it to every $T\in {\mathcal{S}}_1$, proving surjectivity.

Conversely, if finite-rank operators were not dense, the Hahn-Banach theorem would give a nonzero $L\in {\mathcal{S}}_1^{\ast }$ vanishing on their closure. Surjectivity would represent it by some $S$, but thenfor all $x,y$, forcing $S=0$ and $L=0$, a contradiction. This proves the stated trace duality criterion.

Solved by gpt-5.6-sol high.