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free and bound variables

free variable, bound variable, free occurrence, bound occurrence, occurs free, occurs bound
occur free, occur bound, closed, open
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03C07 no label found03B10 no label found


If F and G are hilbert spaces, is the banach space L(F,G) of bounded linear operators from F to G necessarily a hilbert space?

Either it's way too hot, or, if we equip L(F,G) with the operator norm (|A|=sup_{|x|=1} |Ax|), F=G=R^2 will be a counterexample.
It cannot be a Hilbert space because A=((1,0),(0,0)) and B=((0,0),(0,1)) don't satisfy the parallelogram identity |A+B|^2+|A-B|^2=2(|A|^2+|B|^2) valid in any Hilbert space.
Of course, in this particular example, you can use the Frechet norm (identifying the 2x2-Matrices with R^4) and get the same topology on L(F,G) from a scalar product...

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.
PS (think what you can do when F, G are not separable)

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