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This is a question asking for reference.

I have a proof of the following.

Let $f=f(x,y)$ be a smooth function in $\mathbb R^2$ which vanishes at the origin. Then there exist smooth functions $f_1=f_1(xy,x)$ and $f_2=f_2(xy,y)$ such that $f=f_1+f_2$.

Its proof consists on: 1) deal with the formal problem, 2) deal with the flat terms.

However I do not think such a result is new, it must exist somewhere in the literature. Maybe in some more general context. Does anybody knows a reference for such a result?

Edit: Let me give a bit more detail. The formal series of $f$ can be written as

\begin{equation} \hat f=\sum_{i,j\geq 0}a_{ij}x^iy^j \end{equation}

and can be partitioned as

\begin{equation} \hat f=\hat f_1+\hat f_2=\sum_{i\geq j}a_{ij}(xy)^jx^{i-j}+\sum_{i<j}a_{ij}(xy)^iy^{j-i}. \end{equation}

Then, by Borel's lemma there exist smooth functions $f_1(xy,x)$ and $f_2(xy,y)$ such that $$ f=f_1+f_2+h(x,y) $$

where $h$ is flat. The next step is to "kill" the flat term. I won't go into the detail, but this should be enough to get the idea behind the proof.

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  • $\begingroup$ Did you not mean $f_2(x,xy)$ instead of $f_2(xy,y)$? $\endgroup$
    – GH from MO
    Commented Oct 30, 2014 at 16:40
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    $\begingroup$ @GHfromMO nope, I really meant what is written. Imagine in the formal series expansion of the form $x^iy^j$, $f_1$ contains monomials where $i\geq j$ and $f_2$ the rest. $\endgroup$
    – PepeToro
    Commented Oct 30, 2014 at 17:06
  • $\begingroup$ Well, the notation $f_1=f_1(xy,x)$ etc. is confusing in itself. Also, it is not clear to me what you mean by formal series expansion. In short, it is not clear to me what you are stating. $\endgroup$
    – GH from MO
    Commented Oct 30, 2014 at 17:32
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    $\begingroup$ @GHfromMO Formally, you are right. The statement means: For all $f\in C^\infty(\mathbb R^2)$ with $f(0,0)=0$ there are $g_1,g_2 \in C^\infty(\mathbb R^2)$ such that $f(x,y)=g_1(xy,x)+g_2(xy,y)$ for all $(x,y)\in\mathbb R^2$. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 31, 2014 at 9:39
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    $\begingroup$ Well there is not much difference from what I wrote. What @JochenWengenroth wrote above is correct. I hope things are clear now. $\endgroup$
    – PepeToro
    Commented Oct 31, 2014 at 10:09

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