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

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

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

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

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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.

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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.

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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.

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