# Approximation of smooth compactly supported functions on $\mathbb{R}^2$ using sums of products of one variable functions

Let $f \in C^{\infty}(\mathbb{R}^2)$ be smooth and compactly supported. Can we approximate $f(x,y)$ by sums of the form $\sum_{i=1}^m g_i(x) h_i (y)$ where $g_i, h_i \in C^{\infty}(\mathbb{R})$ are smooth with compact support.

Exact formulation: Suppose $f \in C^{\infty}(\mathbb{R}^2)$ with $supp(f)\subseteq [a,b] \times [c,d]$, and $\varepsilon > 0$. Can we find a sum $\sum_{i=1}^m g_i(x) h_i (x)$ such that $supp(g_i) \subseteq [a,b]$ and $supp(h_i) \subseteq [c,d]$, with $\| f(x,y)- \sum_{i=1}^m g_i(x) h_i (y)\|_\infty < \varepsilon$ ?

• Does not Fourier sine expansion work for this? You may be also interested in ACA (adaptive cross approximation) and continuous LU factorization (by Nick Trefthen). May 14 '14 at 15:24
• Stone and Weierstrass say yes, even if $f$ is only continuous. May 14 '14 at 15:50
• I don't see how to use Stone and Weierstrass theorem to deduce approximation with only compactly supported functions.
– ebg
May 14 '14 at 15:52
• The most obvious way to use Stone-Weierstrass may get you smooth functions that don't vanish on the boundary on your rectangle, but cutoff functions will take care of that. Remember that, sufficiently close to the boundary, we have $|f| < \epsilon$. May 14 '14 at 16:01

The situation is even much better, because $C^\infty([a,b]\times [c,d])=C^\infty([a,b])\tilde{\otimes}_\pi C^\infty([c,d])$ (the completed projective tensor product) and due to a celebrated result of Grothendieck every Element of $X\tilde{\otimes}_\pi Y$ for two Frechet spaces is even a series $\sum\limits_{n=0}^\infty x_n\otimes y_n$ which converges in the topology of the projective tensor product. In the concerte situation this implies uniform convergence not only of the functions but of all partial derivatives.

• Thank you ! can you please provide a reference for this fact?
– ebg
May 16 '14 at 22:43

As everyone says, yes it is possible. Here is an explicit way to do it though. Take a smooth, orthonormal basis of $L^2([a,b])$. Then project the function $f$ onto each of these basis elements.

More precisely, let $\left\{a_i(x)\right\}$ denote one such basis. Let $$m_i(y) = \frac{\int_a^b f(x,y) a_i(x) dx}{(\int_a^b a_i(x)dx)^2}$$ Then $$f(x,y) = \sum_i m_i(y) a_i(x)$$ on $[a,b]$. Take a partition of unity of the interval $[a,b]$ called $\left\{ p_i(x) \right\}$ so that for each $n\in \mathbb{N}$ we have $$\sum_{i=1}^n p_i(x) = \left\{ \begin{array}{cl} 1 & \textrm{ for } x\in [a + \frac{b-a}{2^n},b- \frac{b-a}{2^n}] \\ 0 & \textrm{ otherwise} \end{array}\right.$$

Then I believe $$\sum_{i=1}^n m_i(y) \left[ a_i(x) \sum_{j=1}^n p_j(x) \right]$$ is the approximation you are looking for.

• Can you please explain the last equality ?
– ebg
May 14 '14 at 17:19
• I was being stupid before. You probobably don't want each of the $a_i$'s to depend on $n$. This answer seems incomplete. I'll keep thinking about it though. May 15 '14 at 15:04