# Hilbert-Schmidt operators and Trace-class operators

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.

• Is $I$ the set of indices for the orthonormal basis of $\mathcal H$? Commented Sep 6, 2019 at 10:13
• 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 $\|T\|_r = (\sum_j |c_j|^r)^{1/r}, \|T^*T\|_r = (\sum_j |c_j|^{2r})^{1/r}$ Commented Sep 6, 2019 at 10:44
• @niki di giano yes
– NewB
Commented Sep 6, 2019 at 11:21

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 $$\operatorname{Tr}(T)^2=9$$.

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