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### Are Hilbert-Schmidt operators on separable Hilbert spaces - Hilbert Schmidt - on the space of Hilbert Schmidt Operators?

Let's consider a separable Hilbert space $(\mathcal H, \langle\cdot, \cdot\rangle_<\mathcal H>)$ with Norm $||\cdot||_<\mathcal H> := \langle\cdot, \cdot\rangle^<1/2>_<\mathcal H>,$ orthonomal basis $(e_j)$ of $\mathcal H$ and let $s\colon \mathcal H \rightarrow \mathcal H$ be a Hilbert Schmidt operator, denoted by $s \in \mathcal.$ It's known that $(\mathcal, \langle\cdot, \cdot\rangle_<\mathcal>)$ is a (separable. ) Hilbert space and for $s_1, s_2 \in \mathcal\colon$ $\langle s_1, s_2\rangle_<\mathcal>=\sum_^\infty\langle s_1(e_j), s_2(e_j)\rangle_<\mathcal H>,$ such that $||s_1||^2_<\mathcal> = \sum_^\infty||s_1(e_j)||^2_<\mathcal H>.$ Now, assuming that $\mathcal< S_H>$ indeed is separable, let $(\phi_j)$ be an orthonormal basis of $\mathcal S_<\mathcal H>.$ It easily can be shown, since $||\cdot||_<\mathcal>$ is sub-multiplicative, that for $s \in \mathcal\colon s\colon \mathcal \rightarrow \mathcal$ is a bounded operator as well, denoted by $s \in \mathcal>.$

Is $s\colon\mathcal \rightarrow \mathcal$ "Hilbert-Schmidt" as well ($s \in \mathcal>$)?

asked Mar 23 '16 at 8:58

The answer is No, assuming of course that for you $s$ acts on $\mathcal_\mathcal$ by left-multiplication. There is no need to assume that $\mathcal_\mathcal$ is separable, since it actually is, with an explicit countable orthonormal basis given by the elementary operators $e_$ that act as $e_(e_k) = e_i \delta_$ on the orthonormal basis of $\mathcal$.

The Hilbert-Schmidt norm of $s$ on $\mathcal_\mathcal$ is $\operatorname(s' s)$, where $s'$ is the adjoint of $s$ on $\mathcal_\mathcal$. It is not hard to tell that $s' = s^*$, that is, left-multiplication by $s^*$, where $s^*$ is the adjoint of $s$ on $\mathcal$. Let $r = s^* s$ on $\mathcal$ and denote by $r$ also the left-multiplication by $r$ on $\mathcal_\mathcal$. Its trace is $\operatorname(r) = \sum_ (e_, r e_)_<\mathcal_\mathcal>$. A quick calculation gives $(e_, r e_)_<\mathcal_\mathcal> = (e_i, r e_i)_\mathcal$. So, each of the diagonal elements $(e_i, r e_i)_\mathcal$ occurs infinitely many times in the summation for $\operatorname(r)$, which hence cannot be finite unless all diagonal elements of $r$ vanish. But we know that that's impossible from $r = s^* s$.

This answer is essentially the same as the one by André Henriques, which appeared as I was typing, but goes into more explicit detail.

answered Mar 23 '16 at 9:47

If $s:H\to H$ is a rank one projection, then $s\otimes\mathrm:H\otimes \overline H\to H\otimes \overline H$ is a projection of rank $\dim(H)$.

In particular, if $H$ is infinite dimensional, then $s\otimes\mathrm:H\otimes \overline H\to H\otimes \overline H$ is a projection of infinite rank, and thus not Hilbert-Schmidt.

Here, I've identified the space of Hilbert-Schmidt operators with the Hilbert space tensor product $H\otimes \overline H$.

answered Mar 23 '16 at 9:31

Thank you very much André! Just to make sure: I work on hilbert spaces on $\mathbb R.$ I've got some questions: 1. I'm not familiar with the tensor product - I googled it, but why is it important? 2. What do you mean with $\bar$. 3. Projections of infinite rank automatically are'nt Hilbert-Schmidt? I have an idea, but I am not 100% sure! – Obriareos Mar 24 '16 at 9:59

The bar indicates that I take the complex conjugate. When working over the real numbers, $\overline H = H$. – André Henriques Mar 24 '16 at 16:48

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### Measuring Statistical Dependence with Hilbert-Schmidt Norms

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### Encyclopedia of Mathematics

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### Hilbert–Schmidt operator

Hilbert–Schmidt operator

The product of two Hilbert–Schmidt operators has finite trace class norm ; therefore, if A and B are two Hilbert–Schmidt operators, the Hilbert–Schmidt inner product can be defined as

The Hilbert–Schmidt operators form a two-sided *-ideal in the Banach algebra of bounded operators on H. They also form a Hilbert space, which can be shown to be naturally isometrically isomorphic to the tensor product of Hilbert spaces

The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, H is finite-dimensional.

An important class of examples is provided by Hilbert–Schmidt integral operators .

Hilbert–Schmidt operators are nuclear operators of order 2, and are therefore compact.

References

### Hilbert schmidt norm beispiel essay

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Hilbert-Schmidt-Klasse — In der Mathematik ist ein Hilbert Schmidt Operator (nach David Hilbert und Erhard Schmidt) ein stetiger linearer Operator auf einem Hilbertraum, für den eine gewisse Zahl, die sogenannte Hilbert Schmidt Norm, endlich ist. Die Hilbert Schmidt… … Deutsch Wikipedia

Hilbert-Schmidt-Operator — In der Mathematik ist ein Hilbert Schmidt Operator (nach David Hilbert und Erhard Schmidt) ein stetiger linearer Operator auf einem Hilbertraum, für den eine gewisse Zahl, die sogenannte Hilbert Schmidt Norm, endlich ist. Die Hilbert Schmidt… … Deutsch Wikipedia

Hilbert-Schmidt Operator — In der Mathematik ist ein Hilbert Schmidt Operator (nach David Hilbert und Erhard Schmidt) ein stetiger linearer Operator auf einem Hilbertraum, für den eine gewisse Zahl, die sogenannte Hilbert Schmidt Norm, endlich ist. Die Hilbert Schmidt… … Deutsch Wikipedia

Hilbert–Schmidt operator — In mathematics, a Hilbert–Schmidt operator is a bounded operator A on a Hilbert space H with finite Hilbert–Schmidt norm, meaning that there exists an orthonormal basis of H with the property:sum |Ae i|^2

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