This is true for your "simple" case. In fact, assuming that the details of my proof below goes through correctly, the following is true.
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*If g(x,y) ∈ C<sup>∞</sup>(ℝ<sup>2</sup>) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C<sup>∞</sup>(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C<sup>∞</sup>(ℝ<sup>2</sup>).*
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This can be shown by proving the statements below. I'll state them without full proofs for now, so at least there is a partial argument, and will edit the post to extend the proofs (maybe there are references where these are already proven, but I don't know of any).
First, I'll refer to the following sets of functions.
- *U* be thet set of functions f(x)∈C<sup>∞</sup>(ℝ) which vanish on x≤0.
- *V* be the set of functions f:ℝ<sup>+</sup>→ℝ<sup>+</sup> such that x<sup>-n</sup>f(x)→0 as x→0, for each positive integer n.
Clearly, the restriction of any f∈U to the positive reals is in V.
The statements I require to show the main result are as follows.
**Lemma 1**: *For any f∈V, there is a g∈U such that f(x)/g(x)→0 as x→0.*
**Lemma 2**: For any sequence f<sub>1</sub>,f<sub>2</sub>,...∈V there is a g∈U such that f<sub>k</sub>(x)/g(x)→0 as x→0 for all k.
**Lemma 3**: For any sequence f<sub>1</sub>,f<sub>2</sub>,...∈V there is a g∈U such that f<sub>k</sub>(x)/g(x)<sup>n</sup>→0 as x→0 for all positive integers k,n.
The result follows from applying lemma 3 to f<sub>k</sub>(x)=max{|g(x,y)|:|y|≤k}∈V. details to follow...