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Hilbert Schmidt Norm Beispiel Essay

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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 ($s \in \mathcal>$)?

Thank you in advance!

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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Hilbert–Schmidt operator

Hilbert–Schmidt operator

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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.


Hilbert schmidt norm beispiel essay

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