Talk:Hilbert–Schmidt operator

Functionals

The definition of Hilbert--Schmidt functional here is wrong. In particular, it fails to include the condition on page 126 of Kadison--Ringrose that the sum (3) there converges. Despite being a bit buried in this subsection of this article, it's quite an important point: as a consequence, the current definition of the tensor product of Hilbert spaces (in the section on universal property of the article `Tensor product of Hilbert spaces') is incorrect. (The notion of Hilbert--Schmidt functional written here leads to the projective, not the Hilbert, tensor product.) — Preceding unsigned comment added by 76.14.65.43 (talk) 21:58, 13 January 2014 (UTC)[reply]

Seperability

Is separability really required for an operator to be Hilbert-Schmidt? I understood from other pages that square integrability is sufficient, eg, the remark "regardless of separability" at the end of Mercer's theorem Biker333 (talk) 13:59, 18 June 2011 (UTC)[reply]

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

The article says currently: "The Hilbert–Schmidt operators are closed in the norm topology if, and only if, H is finite dimensional". Surely every Hilbert-Schmidt operator is closed since it is bounded. Should it be "The set of Hilbert–Schmidt operators is closed in the norm topology if, and only if, H is finite dimensional."? Jaan Vajakas (talk) 19:05, 16 August 2012 (UTC)[reply]

Yes. Mct mht (talk) 03:17, 17 August 2012 (UTC)[reply]

Now I improved the sentence in the article. Jaan Vajakas (talk) 15:00, 17 August 2012 (UTC)[reply]

Great. The closure of the Hilbert-Schimidt operators are the compact operators. :) Mct mht (talk) 23:29, 17 August 2012 (UTC)[reply]

Separable or not?

In Hilbert–Schmidt operator: Note that the index set need not be countable.

In trace class : Mimicking the definition for matrices, a bounded linear operator A over a separable Hilbert space H is said to be in the trace class if...

Does the Hilbert space have to be separable, or not, in the Hilbert–Schmidt operator article ?

Or in the trace class article : can separability be dropped, yet the trace Tr(A) still defined, if A is nonnegative self-adjoint? If so, a remark should be added to trace class to this effect. That would allow a reader to use the trace class article as the definition of Tr in the Hilbert-Schmidt operator article, for a fully general Hilbert space.

178.38.85.195 (talk) 18:33, 13 April 2015 (UTC)[reply]