Return to Answer

• Let U be thet set of functions f(x)∈C^∞(ℝ) which vanish on x≤0 and are positive on x>0.
• Let V be the set of functions f:ℝ⁺→ℝ such that x^-nf(x)→0 as x→0, for each positive integer n.

Lemma 1: For any f∈V, there is a g∈U such that f(x)/g(x)→0 as x→0.

Proof: Choose any smooth function r:ℝ⁺→ℝ⁺ with r(0) = 1 and r(x)=0 for x≥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 α_k→0 satisfying ∑_kα_k < ∞, and set

for x>0 and g(x)=0 for x≤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. Clearly, θ(x)→∞ and, therefore, x^-ng(x)→0 as x→0. It needs to be shown that all the derivatives of g vanish at 0 so that g∈U. As r and all its derivatives are bounded with compact support, r⁽ⁿ⁾(x) ≤K_nx^-n-1 for some constants K_n. The n^th 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. 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 as x→0.

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

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

Proof: The idea is to apply Lemma 1 to f(x)=Σ_kλ_k|f_k(x)| for positive reals λ_k. This works as long as f∈V, which is the case if Σ_kλsup_x≤kkmin(x,1)^-k|f_k(x)| is finite, and this condition is easy to ensure. █

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

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

The result follows from applying lemma 3 to the triply indexed sequence f_i,j,k(x)=max{|(d^i+j/dxⁱdy^j)g(x,y)|:|y|≤k}∈V. Then, there is an a∈U such that f_ijk(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 are bounded by a sum of terms, each of which is a product of f_ijk(x,y)/a(x)ⁿ with derivatives of a(x), so this vanishes as x→0. Therefore, G∈C^∞(ℝ²). █

• Let U be thet set of functions f(x) ∈ C^∞(ℝ) which vanish on x ≤ 0 and are positive on x > 0.
• Let V be the set of functions f:  ℝ⁺→ ℝ such that x^-n  f(x) → 0 as x → 0, for each positive integer n.

Lemma 1: For any f ∈ V, there is a g ∈ U such that f(x)/g(x) → 0 as x → 0.

Proof: Choose any smooth function r:  ℝ⁺→ ℝ⁺ with r(0) = 1 and r(x) = 0 for x ≥ 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 α_k → 0 satisfying ∑_k  α_k < ∞, and set

for x > 0 and g(x) = 0 for x ≤ 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. Clearly, θ(x) → ∞ and, therefore, x^-n  g(x) → 0 as x → 0. It needs to be shown that all the derivatives of g vanish at 0 so that g ∈ U. As r and all its derivatives are bounded with compact support, r⁽ⁿ⁾(x) ≤ K_nx^-n-1 for some constants K_n. The n^th 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. 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 as x → 0.

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

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

Proof: The idea is to apply Lemma 1 to f(x) = Σ_k  λ_k|f_k(x)| for positive reals λ_k. This works as long as f ∈ V, which is the case if Σ_k λ_ksup_x≤kmin(x,1)^-k|f_k(x)| is finite, and this condition is easy to ensure. █

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

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

The result follows from applying lemma 3 to the triply indexed sequence f_i,j,k(x) = max{|(d^i+j/dxⁱdy^j)g(x,y)|: |y| ≤ k} ∈ V. Then, there is an a ∈ U such that f_ijk(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 are bounded by a sum of terms, each of which is a product of f_ijk(x,y)/a(x)ⁿ with derivatives of a(x), so this vanishes as x → 0. Therefore, G ∈ C^∞(ℝ²). █

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

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 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 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?

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

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

Sketch Proof: The idea is to apply Lemma 1 to f(x)=Σ_kλ_k|f_k(x)| for positive reals λ_k. This works as long as f∈V, which is the case if Σ_kλ_kmin(x,1)^-ksup_x≤k|f_k(x)| is finite, and this condition is easy to ensure.

The result follows from applying lemma 3 to the triply indexed sequence f_i,j,k(x)=max{|(d^i+j/dxⁱdy^j)g(x,y)|:|y|≤k}∈V. Then, there is an a∈U such that f_ijk(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 a sum of terms, each of which is a product of f_ijk(x,y)/a(x)ⁿ with derivatives of a(x), so this vanishes as x→0. Therefore, G∈C^∞(ℝ²).

Proof: Choose any smooth function r:ℝ⁺→ℝ⁺ with r(0) = 1 and r(x)=0 for x≥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 α_k→0 satisfying ∑_kα_k < ∞, and set

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

Proof: The idea is to apply Lemma 1 to f(x)=Σ_kλ_k|f_k(x)| for positive reals λ_k. This works as long as f∈V, which is the case if Σ_kλsup_x≤kkmin(x,1)^-k|f_k(x)| is finite, and this condition is easy to ensure. █

The result follows from applying lemma 3 to the triply indexed sequence f_i,j,k(x)=max{|(d^i+j/dxⁱdy^j)g(x,y)|:|y|≤k}∈V. Then, there is an a∈U such that f_ijk(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 are bounded by a sum of terms, each of which is a product of f_ijk(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...