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• Let U be thet set of functions f(x)∈C ∈ C(ℝ) which vanish on x≤0  ≤ 0 and are positive on x>0. > 0.
• Let V be the set of functions f: ℝ+→ℝ → ℝ such that x-nf(x)→0  f(x) → 0 as x→0,  → 0, for each positive integer n.
• Lemma 1: For any f∈V,  ∈ V, there is a g∈U  ∈ U such that f(x)/g(x)→0  → 0 as x→0. → 0.

Proof: Choose any smooth function r: ℝ+→ℝ→ ℝ+ with r(0) = 1 and r(x)=0 r(x) = 0 for x≥1.  ≥ 1. For example, we can use r(x) = exp(1-1/(1-x)) for x < 1. Then, the idea is to choose a sequence of positive reals αk→0  → 0 satisfying ∑kα αk < ∞, and set

for x>0  > 0 and g(x)=0 g(x) = 0 for x≤0.  ≤ 0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x>0.  > 0. Clearly, θ(x)→∞ θ(x) → ∞ and, therefore, x-ng(x)→0  g(x) → 0 as x→0.  → 0. It needs to be shown that all the derivatives of g vanish at 0 so that g∈U.  ∈ U. As r and all its derivatives are bounded with compact support, r(n)(x) ≤K ≤ Knx-n-1 for some constants Kn. The nth derivative of θ is

which also has polynomially bounded growth in 1/x. However, the derivative on the left hand side is g(n)(x)/g(x) plus a polynomial in g(i)(x)/g(x) for i<n.  < n. So, induction gives that g(n)(x)/g(x) has polynomially bounded growth in 1/x and, multiplying by g(x), g(n)(x)→0  → 0 as x→0. → 0.

By definition of f∈V,  ∈ V, there is a decreasing sequence of positive reals εk such that f(x)≤x ≤ xn for x≤ε ≤ εn. We just need to make sure that αk≤ε ≤ εn+1 for k≥n  ≥ n to ensure that g(x)≥x ≥ xn-1 for εn+1≤x≤min(ε ≤ x ≤ min(εn,1). Then f(x)/g(x) goes to zero at rate x as x→0.

Lemma 2: For any sequence f1,f2,...∈V  ∈ V there is a g∈U  ∈ U such that fk(x)/g(x)→0  → 0 as x→0  → 0 for all k.

Proof: The idea is to apply Lemma 1 to f(x)=f(x) = Σkλ λk|fk(x)| for positive reals λk. This works as long as f∈V,  ∈ V, which is the case if Σkλsup λksupx≤kkmin(x,1)-k|fk(x)| is finite, and this condition is easy to ensure.

Lemma 3: For any sequence f1,f2,...∈V  ∈ V there is a g∈U  ∈ U such that fk(x)/g(x)n→0  → 0 as x→0  → 0 for all positive integers k,n.

Proof: Apply Lemma 2 to the doubly indexed sequence fk,n=|f = |fk|1/n.

The result follows from applying lemma 3 to the triply indexed sequence fi,j,k(x)=max{|(dx) = max{|(di+j/dxidyj)g(x,y)|:|y|≤k}∈V. g(x,y)|: |y| ≤ k} ∈ V. Then, there is an a∈U  ∈ U such that fijk(x)/a(x)n→0  → 0 as x→0.  → 0. Set G(x,y) = f(x,y)/a(x) for x>0  > 0 and G(x,y) = 0 for x ≤ 0. On any bounded region for x>0,  > 0, the derivatives of G(x,y) to any order are bounded by a sum of terms, each of which is a product of fijk(x,y)/a(x)n with derivatives of a(x), so this vanishes as x→0.  → 0. Therefore, G∈C ∈ C(ℝ2).

Very Rough Sketch:If S ⊂ N is the open set {h≠0} then g and all its derivatives vanish on p-1(S).The idea is to choose a smooth parameter u:N-S →ℝ+ which vanishes linearly with the distance to S. This can be done locally and then extended to the whole of this N (I'm assuming manifolds satisfy the second countability property). As all the derivatives of g vanish on p-1(S), u-ng tends to zero at the boundary of S. This uses the fact that p is a submersion, so that u also goes to zero linearly with the distance from p-1(S) in M.

Then, following a moment..similar argument as above, a can be expressed a function of u so that g/a and all its derivatives tend to zero at the boundary of S. Finally, G=0 on the closure of p-1(S) and G=g/a elsewhere.

I suppose the next question is: does proving the special case above of a single g and h reduce the proof of flatness to algebraic manipulation?

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Proof: Choose any smooth function r:ℝ+→ℝ+ with 1≥r≥0, r(xr(0)>0 for small x,  = 1 and r(x)=0 for x≥1. For example, we can use r(x) = exp(-1/(1-x) exp(1-1/(1-x)) for x≤1.  < 1. Then, the idea is to choose a sequence of positive reals αk→0 satisfying ∑kαk < ∞, and set

Sketch

Proof: The idea is to apply Lemma 1 to f(x)=Σkλk|fk(x)| for positive reals λk. This works as long as f∈V, which is the case if Σkλλsupx≤kkmin(x,1)-ksupx≤k|fk(x)| is finite, and this condition is easy to ensure.

The result follows from applying lemma 3 to the triply indexed sequence fi,j,k(x)=max{|(di+j/dxidyj)g(x,y)|:|y|≤k}∈V. Then, there is an a∈U such that fijk(x)/a(x)n→0 as x→0. Set G(x,y) = f(x,y)/a(x) for x>0 and G(x,y) = 0 for x ≤ 0. On any bounded region for x>0, the derivatives of G(x,y) to any order is are bounded by a sum of terms, each of which is a product of fijk(x,y)/a(x)n with derivatives of a(x), so this vanishes as x→0. Therefore, G∈C(ℝ2).

In fact, using a similar method, the simple case can be generalized to arbitrary submersions.

Let p: M →N be a submersion. If h ∈ C(N) and g ∈ C(M) satisfy hg = 0 then, g = aG for some G ∈ C(M) and a ∈ C(N) satisfying ha = 0.

I'll add more explanation of this in a moment...

9 more to proof of Lemma 1; [made Community Wiki]
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