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Hilbert-Schmidt operators on a Hilbert space $\mathcal H$ are linear operators $T\in B(\mathcal H)$ such that $$\|T\|_2=\Big(\sum_{\alpha\in I}\|Te_\alpha\|^2\Big)^{\frac{1}{2}}<\infty$$ Trace-class operators are defined as $T\in B(\mathcal H)$ such $$\|T\|_1=\sum_{\alpha\in I}\left<|T|e_\alpha,e_\alpha\right> <\infty$$ Both $B_2(\mathcal H)$ and $B_1(\mathcal H)$ form Banach $*$-algebras with respect to the $\|\cdot\|_2$ and $\|\cdot\|_1$ norms, respectively. I was wondering why they don't form a $C^*$-algebra? Is the $C^*$-identity not satisfied? If so, then please give an explicit example.

What about when $\mathcal H$ is finite-dimensional? I guess in this cases both $B_1(\mathcal H)$ and $B_2(\mathcal H)$ will form a $C^*$-algebra, but can't prove it.

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  • $\begingroup$ Is $I$ the set of indices for the orthonormal basis of $\mathcal H$? $\endgroup$ Commented Sep 6, 2019 at 10:13
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    $\begingroup$ The norm doesn't satisfy $\|T^* T\| = \|T\|^2$ for example when $T$ is normal compact $T \sum_j u_j = \sum_j c_j u_j$ for some orthonormal basis $(u_j)$ then $\|T\|_r = (\sum_j |c_j|^r)^{1/r}, \|T^*T\|_r = (\sum_j |c_j|^{2r})^{1/r}$ $\endgroup$
    – reuns
    Commented Sep 6, 2019 at 10:44
  • $\begingroup$ @niki di giano yes $\endgroup$
    – NewB
    Commented Sep 6, 2019 at 11:21

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In both cases you are looking for a positive operator $T$ such that $\operatorname{Tr}(T^2)\ne\operatorname{Tr}(T)^2$. Those are plenty. For instance take $$ T=\begin{bmatrix} 1&0\\0&2\end{bmatrix}. $$ Then $\operatorname{Tr}(T^2)=5$, while $\operatorname{Tr}(T)^2=9$.

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  • $\begingroup$ Thank you so much. This works! $\endgroup$
    – NewB
    Commented Sep 6, 2019 at 16:50

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