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Exercise in Conway's Functional Analysis book:

Let $T$ be a trace class operator on a Hilbert space ${\cal H}$.

Prove: $$\sup\{|\mbox{tr}(CT)|:\ C\ \mbox{is compact}, ||C||\leq 1\}=||T||_1.$$

Here, $||T||_1=\mbox{tr}[(T^*T)^{\frac{1}{2}}]$ is the trace norm.

I can prove that $\leq$ holds. I can prove the equality in the finite dimensional case using polar decomposition. This led me to believe that a polar decomposition argument should also work for the infinite dimensional case. However, I am not sure how to use the compactness assumption. Any hints for the $\geq $ inequality?

Khashayar Baghizadeh
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chhro
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1 Answers1

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Let $|T|=(T^*T)^{1/2}$. Then $\|\,|T|\,\|_1=\|T\|_1$. As $|T|$ is compact and positive, we can write via the Spectral Theorem $$ |T|=\sum_{j=1}^\infty\lambda_jP_j, $$ where $\{P_j\}$ are rank-one and $\lambda_j\geq0$ for all $j$. Note that $\operatorname{Tr}(|T|)=\sum_j\lambda_j$. Let $Q_k=\sum_{j=1}^kP_j$. Then $$ \operatorname{Tr}(Q_k|T|)=\sum_{j=1}^k\lambda_j. $$ So, given $\varepsilon>0$, we can choose $k$ big enough such that $\sum_{j>k}\lambda_j<\varepsilon$. Then $$ \operatorname{Tr}(Q_k|T|)>\operatorname{Tr}(|T|)-\varepsilon=\|T\|_1-\varepsilon. $$ Now using the Polar Decomposition, $T=V|T|$ for a partial isometry $V$, and $|T|=V^*T$. Then $$ \operatorname{Tr}(Q_kV^*T)>\|T\|_1-\varepsilon. $$ The operator $Q_kV^*$ is compact because $Q_k$ is finite-rank, and $\|Q_kV^*\|\leq\|Q_k\|\,\|V^*\|=1$.

Martin Argerami
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  • Sorry, but partial isometry does not necessarily mean $V^V=I$, correct? So why is $|T|=V^T$? – chhro Jul 17 '20 at 22:19
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    It's a basic feature of the polar decomposition. The polar decomposition gives you $T=V|T|$ with $V$ a partial isometry with initial space the closure of the range of $T^$ and final space the closure of the range of $T$. In particular, $V^V|T|=|T|$. – Martin Argerami Jul 17 '20 at 22:40