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| ``Re: Space of hilbert space linear operators''
by vasco on 2004-06-28 14:49:32 |
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| In general L(F,G) won't be a Banach space. What you usually do is the following: take o.n. bases {f_i} and {g_j} (you have to suppose F and G separable). Then put
(A|B)=sum (A f_i|g_j)
where the sum is over all i,j. This is the Hilbert-Schmidt (H-S) product. Now call H-S(F,G) the subspace of all operators for which (A|A)<infinity (you have to prove it is a subspace). This is a Hilbert space under the H-S product, and in general it is smaller then L(F,G).
When you are dealing with R^n and R^m you just get the usual dot product in R^nm (identify matrices with long vectors).
The space H-S(F,G) has some nice properties and there is an alternative description when F=L^2 (X), G=L^2 (Y). In this case an operator A will belong to H-S iff it is of the following form:
(A f)(y) = int_X K(x,y)f(x)dx
where int is the integral and K belongs to L^2 (X*Y). Moreover you have (A|A)=|K|^2.
I hope I have somehow answered your question.
Bye
PS (think what you can do when F, G are not separable) |
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