Identification of tensor product spaces

Question

Let $P_1, P_2$ be (probability) measures, $\Omega_1, \Omega_2 \subset \mathbb{R}^n$ . Prove that $L_{P_1 \otimes P_2}^2(\Omega_1 \times \Omega_2)$ and $L_{P_1}^2(\Omega_1) \otimes L_{P_2}^2(\Omega_2)$ are isometric isomorph.

So basically I would need to find the isomorphism and use two exercises which I have solved before. But how could I use them?

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In two preceeding exercises we have proved:

1) If $\{\phi_i\}_i$, $\{\psi_j\}_j$ are orthonormal bases of real Hilbert spaces $H_1$, $H_2$, then $\{\phi_i \otimes \psi_j\}_{ij}$ is an orthonormal basis of $H_1 \otimes H_2$.

2) If $\{\phi_i\}_i$, $\{\psi_j\}_j$ are orthonormal bases of (the seperable Hilbert spaces) $L_{P_1}^2(\Omega_1)$, $L_{P_2}^2(\Omega_2)$, then $\{\phi_i \psi_j\}_{ij}$ is an orthonormal basis of $L_{P_1 \otimes P_2}^2(\Omega_1 \times \Omega_2)$.

Answer 1

Here you seem to have two basis, and these should be sent one on another via a good isomorphism.

All you have to do is:

1. to define the application on the elements of the basis to the other, and check that the inner products are preserved;
2. then you define the morphism on the vector spaces generated by these basis (there is only one way if you want to keep the morphism property);
3. now define it on the closure of the first space to the second (use the fact that this is an isometry, so it is uniformly continuous)

and you are done, because the isometry property passes to the limit.