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Show that given a real-valued continuous function $f$ on $[0,1] \times [0,1]$ and an $\epsilon > 0,$ there exist real-valued continuous functions $g_1, \dots, g_n$ and $h_1 , \dots, h_n$ on $[0,1]$ such that for some finite $n\geq 1,$ $$ \bigg| f(x,y) - \sum_{i=1}^n g_i(x)h_i(y) \bigg| \leq \epsilon, \;\;\;\; 0 \leq x,y \leq 1. $$

I believe the Stone-Weierstrass approximation could be useful here, as I think this is just a 2D analogue, but I am not sure what the proof would consist of. Proving the Stone-Weierstrass theorem in 1D can consist of using convolutions with an approximate identity which is done in Rudin, I've seen proofs that use Bernstein polynomials, and I remember seeing a rather concise proof that used PDE's, but none of these seem like what would be expected on a qualifier.

Perhaps this problem doesn't call for the Weierstrass approximation?
Does anyone have any suggestions?

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You can indeed use Stone-Weierstrass, but if all you require of the $g_i$s and $h_i$s is that they're continuous, less than that can do.

Select some $m$, and for $0\le i\le m$, define $$ k_i : [0,1]\to\mathbb R : x \mapsto \max(0,1-|i-mx|)$$

 |      /\
_|_____/  \______
 0     i/m      1

Then, $$ \sum_{0\le i\le m} k_i(x) = 1 $$ for $x\in[0,1]$, and therefore $$ \sum_{\begin{matrix}0\le i\le m\\0\le j\le m\end{matrix}} k_i(x)k_j(y) = 1 $$ for $(x,y)\in[0,1]^2$, and $k_i(x)k_j(y)$ has its support bounded by $1/m$ away from $(i/m,j/m)$.

Now, since $f$ is continuous and its domain is compact, it is uniformly continuous, so there is an $m$ such that $f$ doesn't vary by more than $\varepsilon$ when $(x,y)$ varies by up to $1/m$ in each coordinate. But then $$ \tilde f(x,y) = \sum_{\begin{matrix}0\le i\le m\\0\le j\le m\end{matrix}} \underbrace{f(\tfrac im,\tfrac jm)k_i(x)}_{g_{ij}(x)} \cdot \underbrace{k_j(y)}_{h_{ij}(y)} $$ approximates $f(x,y)$, because its value is always some weighted average of values of $f(x,y)$ taken no more than $1/m$ away in each coordinate.

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  • $\begingroup$ Impressive! Very clear I see exactly what is going on... thanks! $\endgroup$ – Merkh Jun 27 '16 at 12:40

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