Is every function a linear combination of separable functions? Let $\mathcal{F}(\mathbb{R}^n\longrightarrow\mathbb{R})$ be the space of functions from $\Bbb R^n$ to $\Bbb R$. If I'm not mistaken, this space is canonically isomorphic to the tensor product of $n$ times $\mathcal{F}(\mathbb{R}\longrightarrow\mathbb{R})$ with the isomorphism defined so that
$$(f_1\otimes\cdots\otimes f_n)(x^1,\dots,x^n)\equiv f_1(x^1)\cdots f_n(x^n).$$
Pure tensors in $\mathcal{F}(\mathbb{R}\longrightarrow\mathbb{R})\otimes\cdots\otimes\mathcal{F}(\mathbb{R}\longrightarrow\mathbb{R})$ correspond to separable functions in $\mathcal{F}(\mathbb{R}^n\longrightarrow\mathbb{R})$, and every element in a tensor product can be written as a linear combination of pure tensors, so every function from $\mathbb{R}^n$ to $\mathbb{R}$ must be a linear combination of separable functions, shouldn't  it?
I know tensor products and linear algebra in general get tricky when dealing with uncountable-dimension spaces, that's the main reason I'm asking this here.
 A: The answer to your question is NO. Consider the Kronecker delta defined by $\delta(x,y)=1$ if $x=y$ and $0$ otherwise. This is a map ${\mathbb R}^2 \to {\mathbb R}$.
Suppose by contradiction that $\delta$ is a linear combination of separable functions,
$$
\delta(x,y)=\sum_{k=1}^{m} f_k(x)g_k(y) \tag{1}
$$
Let $I_{m+1}$ be the $(m+1)\times(m+1)$ identity matrix, let $F$ be the $(m+1)\times m$ matrix defined by $F=(f_j(i))_{ 1 \leq i \leq m+1, 1\leq j \leq m}$, and let $G$ be the $m\times (m+1)$ matrix defined by $G=(g_i(j))_{ 1 \leq i \leq m, 1\leq j \leq m+1}$. Then (1) can be rewritten as
$$
I_{m+1}=FG \tag{2}
$$
Since $G$ has only $m$ lines, its rank cannot exceed $m$, so
 ${\mathsf{rank}}(I_{m+1})={\mathsf{rank}}(FG) \leq {\mathsf{rank}}(G) \leq m$, so $I_{m+1}$ is a singular matrix which is absurd.
Remark 1: this argument also works on ${\cal F}({\mathbb N}^2,{\mathbb Q})$ which is a much smaller space than ${\cal F}({\mathbb R}^2,{\mathbb R})$.
Remark 2: The Kronecker delta is not smooth, but with a little more work using convolutions one can transform it into a smooth function while keeping its value on ${\mathbb N}^2$, leaving the argument intact.
