For example, if X=RX=\mathbb{R} with the usual Lebesuge measure, then there is a natural map Φ:L2(X)⊗L2(X)∋f⊗g↦((x,y)↦f(x)g(y))∈L2(X2)\Phi:L^2(X)\otimes L^2(X)\ni f\otimes g\mapsto((x,y)\mapsto f(x)g(y))\in L^2(X^2) which is also unitary map between Hilbert spaces, by Fubinis theorem (which requires XX and YY to be complete and σ\sigma-finite).

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I am not an expert but I imagine that you will have to take some sort of completed tensor product, and that once you do that this will be true for all XX. (The reason for the completion is that your map will only hit functions that are finite sums of functions on XX times functions on YY, but you can write functions which aren’t of this form.)

– hunter

2 days ago

Thats true and I have implicitly assumed that when saying that Φ\Phi is a unitary map between Hilbert spaces, so L2(X)⊗L2(X)L^2(X)\otimes L^2(X) is to be taken as the tensor product between Hilbert spaces, which is exactly what you said: the completion of the algebraic tensor product with respect to the canonical inner product. But nevertheless I could have pointed this out more clearly, thanks!

– Robert Rauch

2 days ago

oh I see, sorry!

– hunter

2 days ago

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