This is true for your "simple" case. In fact, assuming that the details of my proof below goes through correctly, the following is true.
If g(x,y) ∈ C∞(ℝ2) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C∞(ℝ2).
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∞(ℝ) which vanish on x≤0.
- V be the set of functions f:ℝ+→ℝ+ such that x-nf(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 f1,f2,...∈V there is a g∈U such that fk(x)/g(x)→0 as x→0 for all k.
Lemma 3: For any sequence f1,f2,...∈V there is a g∈U such that fk(x)/g(x)n→0 as x→0 for all positive integers k,n.
The result follows from applying lemma 3 to fk(x)=max{|g(x,y)|:|y|≤k}∈V. details to follow...