In relation to various problems in understanding entanglement and non-locality,

Question

In relation to various problems in understanding entanglement and non-locality, I have come across the following mathematical problem. It is most concise by far to state in its most mathematical form and not go into the background much. However, I hope people interested in entanglement theory might be able to see how the problem is interesting/useful.

Here goes. I have two finite dimension vector spaces A and B and each is equipped with a norm (Banach spaces) such that ||...||:A?R and ||...||:B?R. Both the vector spaces and norms are isomorphic to each other. My question concerns norms on the tensor product of these spaces (for simplicity, lets say just the algebraic tensor product) A?B and the dual norms. First let me state something I know to be true.

Lem 1:
If a norm ||...|| on A?B satisfies:
||a?b||?||a||.||b|| (sub-multiplicativity)
then the dual norm satisfies
||a?b||D?||a||D.||b||D (super-multiplicativity)

where we define the dual of a norm in the usual way as
||a||D=sup{|b

Explanation / Answer

The converse is obviously not true. The asymmetry between the super-multiplicativity and sub-multiplicativity arises because the dual norm is always defined as a supremum and never as an infimum.

To see a counterexample, choose a direction in A?B, for example a direction of vectors that are of the form a?b, and in a very small "ray" vicinity of this direction, define the norm on the tensor product space as
||v||=1000||a||?||b||
Super-multiplicativity will still obviously hold because we have increased the norm somewhere on the tensor product space while kept it constant on the rest of it.

However, the dual norm skyrockets by this tiny change because it's a supremum over all c with ||c||?1 which includes c?a?b where the norm was amplified. Correspondingly, the dual norm for certain dual vectors has been essentially increased to 1,000 times what it was before and is no-longer sub-multiplicative.

Warning: the argument above is wrong. I have misinterpreted |b

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