Suppose that $f \in C^{\infty}_c ( \mathbb{R}^2 )$, i.e. $f$ is a $C^{\infty}$ function with compact support defined on $\mathbb{R}^2$. The following link

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

tells us that $f$ can be approximated by a sequence of functions $(f_n)$ in $C^{\infty}_c ( \mathbb{R}) \otimes C^{\infty}_c ( \mathbb{R})$ in the infinity norm, where each $f_n$ is of the form

$$ f_n( x,y) = \sum_{i=1}^{k_n} a^{(n)}_i (x) b^{(n)}_i (y), $$

for some functions $a^{(n)}_i, b^{(n)}_i \in C^{\infty}_c ( \mathbb{R})$, $1 \leq i \leq k_n$.

I wonder whether it is possible to construct such a sequence of functions $(f_n)$ such that $$ \sup_{n \in \mathbb{N}, 1 \leq i \leq k_n} \big\| a^{(n)}_i \big\|_{\infty} < + \infty, \quad \quad \sup_{n \in \mathbb{N}, 1 \leq i \leq k_n} \big\| b^{(n)}_i \big\|_{\infty} < + \infty, $$ and $$ \sup_{n \in \mathbb{N}, 1 \leq i \leq k_n} \big\| \big( a^{(n)}_i \big)' \big\|_{\infty} < + \infty, \quad \quad \sup_{n \in \mathbb{N}, 1 \leq i \leq k_n} \big\| \big( b^{(n)}_i \big)' \big\|_{\infty} < + \infty,$$ where as usual, $h'$ denotes the derivative to the function $h$.

This proposition would be very useful for my proof. But is this statement true? I am unable to prove this by the standard construction of the sequence $(f_n)$ using the Stone-Weierstrass theorem. Any ideas?

  • 1
    $\begingroup$ Am I missing something? As you put it, it is not forbidden to break the $a_i^{(n)}$ and $b_i^{(n)}$ as a sum of a large number $N=N(i,n)$ of repeated $a_i^{(n)}/N$ resp. $b_i^{(n)}/N$, gaining the properties you want. $\endgroup$ Nov 25, 2016 at 0:20
  • $\begingroup$ @PietroMajer You're right that we can break the $a^{(n)}_i$ and $b^{(n)}_i$ into repeated smaller parts respectively. But if their infinity norms explode to infinity, then this still doesn't solve the problem, regardless of how large the $N$ is chosen to be. Am I right? $\endgroup$
    – Richard
    Nov 25, 2016 at 0:55
  • 2
    $\begingroup$ Why, $N$ is not fixed for all $n$. You can write each single $f_n$ as a sum of several $a(x)b(y)$, each function $a$ and $b$ having $C^1$ norm less than $1$. $\endgroup$ Nov 25, 2016 at 8:01
  • $\begingroup$ If you look at the accepted answer of the question mathoverflow.net/questions/166149/… you mention you get an answer to your version, too. $\endgroup$ Nov 25, 2016 at 9:07
  • $\begingroup$ @PietroMajer Sorry, of course you're right. But actually what I really want in my proof is for the sequence $(k_n)_{n \in \mathbb{N}}$ to bounded as well... Do you think that this is possible? $\endgroup$
    – Richard
    Nov 26, 2016 at 2:19

1 Answer 1


Let $Q$ be a square with side $T$ which contains the support. Extend your function to a doubly periodic one with periods $(0,T)$ and $(T,0)$. Then expand into the double Fourier series: $$f(x,y)=\sum a_{m,n}e^{2\pi inx/T}e^{2\pi im/T}.$$ This series can be differentiated term-by-term. As $f\in C^\infty$ this converges uniformly, so you have the approximation. Fourier coefficients are evidently bounded. Differentiating in $x$ you obtain a Fourier series of $f_x$ whose coefficients are also bounded. Same for $f_y$.


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