What is the natural inner product for space $H^{1}\otimes L^{2}$? What is the natural inner product for space $H^{1}\otimes L^{2}$ in one dimension? I've read that this would given by $$(f,g)=(df/dx,dg/dx)_{L^{2}}+(f,g)_{L^{2}},$$
where $f\in H^{1}$ and $g\in L^{2}$. Let I remember that $H^{1}$ is the Sobolev Space $W^{1,2}$. Ok, I know that this norm is just the norm of $H^{1}$. I'm confused with those differences. Most generally, what is the inner product of a space which is the tensor product of two different spaces?
 A: I'll formulate my comment into an answer.
If you have two hilbert spaces $V,W$ with inner products $<,>_V$ and $<,>_W$, the inner product of $V\otimes W$ is given by
$$<f\otimes g,h\otimes i>_{V\otimes w}=<f,h>_V <g,i>_W,$$
and extending linearly.
A: Another point of view:
An inner product on finite dimensional $V$ is a map $\psi_V: V \otimes V \to \mathbb{C}$, satisfying certain properties (symmetry, non-degeneracy, positive definite). If we have two (finite dimensional) Hilbert spaces, their inner products together give us maps $\psi_V \otimes \psi_W : V \otimes V \otimes W \otimes W \to \mathbb{C} \otimes \mathbb{C}$.
If we want a map from $\mathbb{C} \otimes \mathbb{C} \to \mathbb{C}$, then there is really only one choice (because of dimension), and the natural choice is multiplication. Composition of $\psi_V \otimes \psi_W$ with multiplication then gives us a map $V \otimes V \otimes W \otimes W \to \mathbb{C}$, which is symmetric under the action of $(\mathbb{Z}/2\mathbb{Z})^2$ that interchanges the copies of $V$ or the copies of $W$. Re-arranging, we get a map $(V \otimes W) \otimes (V \otimes W) \to \mathbb{C}$, which is symmetric.
This suggests that the inner product should be the product of the inner products. Non-degeneracy can be quickly checked by remembering it is equivalent to the map $V \to V^*$ being an isomorphism, and using that the tensor product of two isomorphisms is an isomorphism. For positive definiteness, you have to argue by hand after picking an orthonormal basis for $V$ and $W$, and writing  vectors of $V\otimes W$ as linear combinations $e_i \otimes f_j$.
There is an additional subtlety in infinite dimensions, since the linear algebraically defined tensor product is not complete. Suppose we consider the space $L^2(S^1) \otimes L^2(S^1)$. This space is not complete, since there are Cauchy sequence obtained by summing more and more independent pure tensors: if $\phi_m$ is the usual orthonormal basis for $L^2(S^1)$, then $s_n \Sigma_{m = 0}^{n} (1 / m^2) \phi_m \otimes \phi_m$ is a Cauchy sequence without a limit in the tensor product $L^2(S^1) \otimes L^2(S^1)$. (... if it converged to some tensor $s$, that tensor would be a sum of tensors finitely many $\phi_m$, meaning that eventually we would be converging to $s$ by adding tensors to $s_n$ that were orthogonal to $s$ and $s_n$, but since adding an orthogonal tensor to $s- s_n$ increased the norm, this is not possible...)
This is why one usually takes the completion after tensoring two infinite dimensional Hilbert spaces: https://en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces
It's useful to point out that a complete orthonromal basis can be obtained in the obvious way, namely $\phi_m \otimes \psi_n$, where those are complete orthormal basis for the factors.
There is another point of view on the definition, which you can read about more on the above wikipedia link. Namely, the incompleted space $V \otimes W \cong V^* \otimes W$ is identified with the space of finite rank maps $V \to W$. Finite rank maps have an inner product $Tr(T_1^* T_2)$, and the space of bounded operators on which the corresponding norm is finite gives the completion: https://en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator . I'm not sure how to make sense of this trace in a canonicaly way in the infinite dimensional case (wikipedia gives a computational formula using a basis), so I'll leave this forsomeone else to (hopefully) explain.
