To start with, let me apologize for my ignorance as I know next to nothing about partial differential equations. My question is about the tensor product of Banach spaces but actually I do not understand those either, so let me rephrase it as a Hilbert space question. Alright! Let us consider the Lebesgue Hilbert space $L^2(X)$ and why not also $L^2(Y)$ where $X$ and $Y$ are the domains of our respective functions.
Now the tensor product of $L^2(X) \otimes L^2(Y)$ is readily seen to be the closure of the space of functions that have the form $fg$ where $f$ is in $L^2(X)$ and $g$ is in $L^2(Y)$. Perfect. This is of course (Banach spaces so think isomorphic) also equal to $L^2(X \times Y)$. Nothing interesting here. (For an interesting related discussion check out my question Sequence of measurable functions and the excellent answer by Nate which gives a couple of very interesting tools.)
So let me start with a nice PDE. The Poisson equation. $\Delta u = -f$ where - if you have a inclination to thinking in practical terms - you could see $f$ as a normalized charge density function and then $u$ would be the potential this charge density gives us.
So, where does our solution live? As $u$ would be the convolution with the so called fundamental solution of the Laplace equation this would mean that the solution $u$ is at least as smooth as $f$ itself (that is why I took this equation...). Also, if we would look on the whole of $\mathbf R^3$ then we might have a problem with integrability: the maximum principle would state that if the function would be bounded (like if it would have compact support and $f$ continuous) then it would be a constant. That messes up the integrability. So as I know nothing about these things let us just consider the domain to be the unit cube $Q$. Integrable for sure!
So the solution would live in $L^2([0,1])$ tensored with itself three times. Actually, all its derivatives should be in this space as well, so we have the Sobolev space $W^{2, 2}(Q)$ if we take $f$ to be at least $C^2$. Also, as I have kinda shown here (need to restrict) here (for $L^2$ you can also use Plancherel) this is equal to the $L^2$ space intersected with the $L^2$ functions that have Laplacian as well in $L^2$.
Now the question...
As I have asked before in Non-separable linear PDE and about the excellent answer of Willy there I was wondering about solutions in these spaces.
If we solve for instance the heat equation by separation of variables in a 1D-heat bar we find two ODEs for both the position variable and the time variable. These belong to a set of simultaneous eigenvalues $\lambda_n$, and have corresponding solutions $s_n$ and $t_n$. Then we construct the solution as $$u = \sum_n \lambda u_n t_n.$$ Question: This is a solution in the product space. What makes it so that one cannot write a posteriori a solution of a non-separable equation in this way? The simultaneous $\lambda_n$ would not correspond to such an equation?
Any omissions and mistakes are solely due to my misunderstanding and/or misconception. I appreciate any constructive comment.
Oh blimey... For now, let me make it more abstract to find my source of misunderstanding. Let $H_1$ and $H_2$ be Hilbert spaces. A simple tensor (tensor of rank one) $x_1 \otimes x_2$ is the identification of $x_1$ with its dual element $x_1^*$. Kind of like $x_1 \times x_2$ where $x_1$ is in $H_1$ and $x_2$ is in $H_2$. Then the inner product defined on $H_1 \otimes H_2$ is given by $$\langle u_1 \otimes u_2, v_1 \otimes v_2 \rangle = \langle u_1, v_1 \rangle_{H_1} \langle u_2, v_2 \rangle_{H_2}.$$ Next, extend by linearity and complete under this inner product.
More abstractly, to each simple tensor $x_1 \otimes x_2$ associate the rank one operator $$ \begin{align} T:H_1^* &\to H_2\\ x^* &\mapsto x^*(x_1) x_2. \end{align} $$ We can use this to construct a linear mapping between $H_1 \otimes H_2$ and the space of finite rank operators from $H_1^*$ to $H_2$. These are a subspace of the Hilbert-Schmidt operators which has scalar product for an orthonormal basis $(e_n)$ of $H_1$, $$\langle \Lambda_1, \Lambda_2 \rangle = \sum_n \langle \Lambda_1 e_n^*, \Lambda_2 e_n^*\rangle.$$ Hence, we can identify the tensor product $H_1 \otimes H_2$ as the Hilbert Schmidt operators from $H_1^*$ to $H_2$.
Also to my knowledge, whenever the spaces are separable: $$L^2(X) \otimes L^2(Y) \cong L^2(X \times Y) \cong L^2(X; L^2(Y)).$$
So, if I have $f$ in $L^2$ and $g$ in $L^2$, then the set of $fg$ is in $L^2 \otimes L^2$. Furthermore, it is a dense set in $L^2(X \times Y)$.
The answer is simply that you made a logical fallacy. And nothing whatsoever can be said about the solution space.
You tried to take the converse of the statement $$ \text{solution writable as } f(x)\cdot g(y) \implies \text{solution is in a tensor product space}$$ which is false for any nontrivial vector spaces!
Let $V,W$ be vector spaces, an element of $V\otimes W$ is not always rank 1 (that is to say, expressible as $v\otimes w$ for $v\in V$ and $w\in W$). Given $v,v'\in V$ and $w,w'\in W$, in general $v\otimes w + v'\otimes w' \in V\otimes W$ cannot be "factored" as $v''\otimes w''$. (Just think about $V = W = \mathbb{R}$ and the space of two-by-two matrices....)
So just because a solution lives in $L^2(Q)$ does not imply that said solution can be written as $f(x)g(y)h(z)$ for $f,g,h\in L^2([0,1])$.
For the revised question:
Nothing prevents you from writing a solution that way. You've basically just described the Fourier series. (Any $L^2$ function $f$ on $Q$ can be written as the Fourier series $$ f(x,y,z) \approx \sum_{j,k,l = -\infty}^\infty a_{jkl} \exp[ 2\pi i(jx + ky + l z)]$$ Chain this with some bijection of $\mathbb{Z}^3 \to \mathbb{Z}$ you end up with a single sum like you described.)
What fails is for each $\exp[\ldots]$ or $u_nt_n$ to solve the equation. Because, let's face it: if every solution can be written as a linear combination of solutions which are pure tensors, then the linear equation is separable...