Can a real continuous bounded function on $ \Bbb{R}^{2} $ be expressed as a finite sum of products of real continuous functions on $ \Bbb{R} $? Can a real-valued continuous bounded function on $ \Bbb{R}^{2} $ always be expressed as a finite sum of products of real-valued continuous functions on $ \Bbb{R} $?
 A: No: $f(x,y)=\sin xy$, for example, cannot be so written, even if we allow the constituent functions to be discontinuous. Suppose that there existed $n\in\Bbb N$ and functions $p_1,\dots,p_n,q_1,\dots,q_n$ such that $\sin xy = \sum_{k=1}^n p_k(x)q_k(y)$. Then in particular, for every fixed $y_0$ the function $\sin xy_0$ would be an $\Bbb R$-linear combination of $\{p_1(x),\dots,p_n(x)\}$. In particular, the vector space spanned by $\{\sin xy_0\colon y_0\in\Bbb R\}$ would have dimension at most $n$, which is easily refuted (take $n+1$ algebraically independent values for $y_0$, for example).
A: Just to add something to Greg’s beautiful answer. If $ f \in C(\Bbb{R}^{2};\Bbb{R}) $ satisfies the property that
$$
S_{f,A} \stackrel{\text{df}}{=}
\{ f(\bullet,y) \in C(\Bbb{R};\Bbb{R}) \mid y \in A \}
$$
is an infinite $ \Bbb{R} $-linearly independent subset of $ C(\Bbb{R};\Bbb{R}) $ for some infinite subset $ A $ of $ \Bbb{R} $, then $ f $ cannot be so expressed.
Greg’s example consists of $ f = \left\{ \begin{matrix} \Bbb{R}^{2} & \to & \Bbb{R} \\ (x,y) & \mapsto & \sin(x y) \end{matrix} \right\} $ and $ A = \Bbb{N} $. We have
$$
S_{f,A} = \{ \sin(n \bullet) \in C(\Bbb{R};\Bbb{R}) \mid n \in \Bbb{N} \},
$$
which is easily shown to be an infinite $ \Bbb{R} $-linearly independent subset of $ C(\Bbb{R};\Bbb{R}) $. The argument goes as follows: If $ N \in \Bbb{N} $ and $ a: \{ 1,\ldots,N \} \to \Bbb{R} $ satisfies
$$
\sum_{n = 1}^{N} a_{n} \cdot \sin(n \bullet) \equiv 0,
$$
then for each $ k \in \{ 1,\ldots,N \} $, multiplying both sides of the equation by $ \sin(k \bullet) $ and integrating over $ [0,2 \pi] $ will yield $ a_{k} = 0 $.
Another example is $ f = \left\{ \begin{matrix} \Bbb{R}^{2} & \to & \Bbb{R} \\ (x,y) & \mapsto & e^{x y} \end{matrix} \right\} $ and $ A = \Bbb{N}_{0} $. If $ N \in \Bbb{N}_{0} $ and $ a: \{ 0,\ldots,N \} \to \Bbb{R} $ satisfies
$$
\sum_{n = 0}^{N} a_{n} e^{n \bullet} \equiv 0,
$$
then dividing both sides of the equation by $ e^{N \bullet} $ and taking the limit as $ x \to \infty $ yield $ a_{N} = 0 $. An induction argument will furnish the final step to establish that $ a_{n} = 0 $ for all $ n \in \{ 0,\ldots,N \} $.
