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](https://mathoverflow.net/questions/166149/approximation-of-smooth-compactly-supported-functions-on-mathbbr2-using-su)


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?



Approximation of a $C^{\infty}_c$ function by tensor products
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 $f\_y$.


