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edited Feb 17 2010 at 22:49 |
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 → 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→ 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).
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edited Feb 17 2010 at 22:22 |
I'll add more explanation 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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edited Feb 17 2010 at 21:55 |
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...
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edited Feb 17 2010 at 21:35 |
I can show that this is true for your "simple" case.
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. They could possibly be standard results, but I've never seen them before.
The proofs are a bit sketchy at the moment, but I think they're good.
First, I'll refer to the following sets of functions.
- 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.
The statements I need 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.
Sketch proof
Proof: Choose any smooth function r:ℝ+→ℝ+ with 1≥r≥0, r(x)≥1 >0 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x) exp(-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0 satisfying ∑kαk < ∞, and set
$$g(x) = x^{\theta(x)},\ \ \ \theta(x)=\sum_{k=1}^\infty r(x/\alpha_k)$$
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) ≤Knx-n-1 for some constants Kn. The nth derivative of θ is
$$\theta^{(n)}(x)=\sum_k\alpha_k^{-n}r^{(n)}(x/\alpha_k)\le K_nx^{-n-1}\sum_k\alpha_k$$
which has polynomially bounded growth in 1/x. The derivatives of log(g) satisfy
$$\frac{d^n}{dx^n}\log(g(x))=\frac{d^n}{dx^n}\left(\log(x)\theta(x)\right)$$
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 of 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 εk such that f(x)≤xn for x≤εn. We just need to make sure that αk≤εn+1 for k≥n to ensure that g(x)≥xn-1 for εn+1≤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 there is a g∈U such that fk(x)/g(x)→0 as x→0 for all k.
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 Σkλkmin(x,1)-ksupx≤k|fk(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 fk(x)/g(x)n→0 as x→0 for all positive integers k,n.
Proof: Apply Lemma 2 to the doubly indexed sequence fk,n=|fk|1/n.
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 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).
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edited Feb 17 2010 at 21:27 |
Sketch proof: Choose any smooth function r:ℝ+→ℝ+ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0, |α|≤1 →0 satisfying ∑kαk < ∞, and set $$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$ $g(x) = x^{\theta(x)},\ \ \ \theta(x)=\sum_{k=1}^\infty r(x/\alpha_k)$$ for x>0 and g(x)=0 for x≤0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x>0. The nth derivative is bounded by KnClearly, θ(x)→∞ and, therefore, x-ng(x)Σk1{→0 as x<αk}αk-n→0. It needs to be shown that all the derivatives of g vanish at 0 so that g∈U. As r and all its derivatives are bounded with compact support, r(n)(x) ≤Knx-n-1 for positive some constants Kn. As long as αk go to zero fast enoughThe nth derivative of θ is $$\theta^{(n)}(x)=\sum_k\alpha_k^{-n}r^{(n)}(x/\alpha_k)\le K_nx^{-n-1}\sum_k\alpha_k$$ which has polynomially bounded growth in 1/x. The derivatives of log(g) satisfy $$\frac{d^n}{dx^n}\log(g(x))=\frac{d^n}{dx^n}\left(\log(x)\theta(x)\right)$$ which also has polynomially bounded growth in 1/x. However, this goes to zero as x→0, so the derivative on the left hand side is g∈C∞(ℝ).n)(x)/g(x) plus a polynomial of g(i)(x)/g(x) for i<n. So, induction gives that g(n)(x)/g(x) has polynomially bounded growth in 1/x and, multiplying by g(x), g(n)(x)→0 as x→0.
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edited Feb 17 2010 at 2:01 |
I can show that this is true for your "simple" case.
If g(x,y) ∈ C∞(ℝ2) vanishes on x ≤ 0 then it decomposes as g(x,y) = a(x)G(x,y) where a(x) ∈ C∞(ℝ) vanishes on x ≤ 0 and G(x,y) ∈ C∞(ℝ2).
This can be shown by proving the statements below. They could possibly be standard results, but I've never seen them before. The proofs are a bit sketchy at the moment, but I think they're good.
First, I'll refer to the following sets of functions.
- Let U be thet set of functions f(x)∈C∞(ℝ) which vanish on x≤0 and are positive on x>0.
- Let V be the set of functions f:ℝ+→ℝ such that x-nf(x)→0 as x→0, for each positive integer n.
The statements I need to show the main result are as follows.
Lemma 1: For any f∈V, there is a g∈U such that f(x)/g(x)→0 as x→0.
Sketch proof: Choose any smooth function r:ℝ+→ℝ+ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0 →0, |α|≤1 and set
$$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$
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. The nth derivative is bounded by Knx-ng(x)Σk1{x>α<αk}αk-n, for positive constants Kn. As long as αk go to zero fast enough, this will be of size O(xn), goes to zero as x→0, so g∈C∞(ℝ).
By definition of f∈V, there is a decreasing sequence of positive reals εk such that f(x)≤xn for x≤εn. We just need to make sure that αk≤εn+1 for k≥n to ensure that 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 fk(x)/g(x)→0 as x→0 for all k.
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 Σkλkmin(x,1)-ksupx≤k|fk(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 fk(x)/g(x)n→0 as x→0 for all positive integers k,n.
Proof: Apply Lemma 2 to the doubly indexed sequence fk,n=|fk|1/n.
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 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).
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edited Feb 17 2010 at 1:53 |
This I can show that this is true for your "simple" case. In fact, assuming that the details of my proof below goes through correctly, the following is true. This can be shown by proving the statements below. I'll state They could possibly be standard results, but I've never seen them without full before. The proofs for now, so at least there is are a partial argument, and will edit the post to extend bit sketchy at the proofs (maybe there are references where these are already provenmoment, but I don't know of any)think they're good. 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.The result follows from applying lemma 3 to fi,j,k(x)=max{|(di+j/dxidyj)g(x,y)|:|y|≤k}∈V. details 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 follow..any order is 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).
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edited Feb 17 2010 at 1:40 |
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 and are positive 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.
The statements I need 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.
Sketch proof: Choose any smooth function r:ℝ+→ℝ+ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0 and set
$$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$
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. The nth derivative is bounded by Knx-ng(x)Σk1{x>αk}αk-n, for positive constants Kn. As long as αk go to zero fast enough, this will be of size O(xn), so g∈C∞(ℝ).
By definition of f∈V, there is a decreasing sequence of positive reals εk such that f(x)≤xn for x≤εn. We just need to make sure that αk≤εn+1 for k≥n to ensure that 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 fk(x)/g(x)→0 as x→0 for all k.
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 Σkλkmin(x,1)-ksupx≤k|fk(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 fk(x)/g(x)n→0 as x→0 for all positive integers k,n.
Proof: Apply Lemma 2 to the doubly indexed sequence fk,n=|fk|1/n.
The result follows from applying lemma 3 to fi,j,k(x)=max{|(di+j/dxidyj)g(x,y)|:|y|≤k}∈V. details to follow...
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edited Feb 17 2010 at 1:29 |
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 and are positive 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.
The statements I need 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.
Sketch proof: Choose any smooth function r:ℝ+→ℝ+ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0 and set
$$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$
for x>0 and g(x)=0 for x≤0. Only finitely many terms in the summation will be nonzero outside any neighborhood of 0, so it is a well defined expression, and smooth on x>0. The nth derivative is bounded by Knx-ng(x)Σk1{x>αk}αk-n, for positive constants Kn. As long as αk go to zero fast enough, this will be of size O(xn), so g∈C∞(ℝ).
By definition of f∈V, there is a decreasing sequence of positive reals εk such that f(x)≤xn for x≤εn. We just need to make sure that αk≤εn+1 for k≥n to ensure that f(x)/g(x) goes to zero at rate x as x→0.
Lemma 2: For any sequence f1,f2,...∈V there is a g∈U such that fk(x)/g(x)→0 as x→0 for all k.
Lemma 3: For any sequence f1,f2,...∈V there is a g∈U such that fk(x)/g(x)n→0 as x→0 for all positive integers k,n.
The result follows from applying lemma 3 to fki,j,k(x)=max{|g(x,y)|:|y|x)=max{|(di+j/dxidyj)g(x,y)|:|y|≤k}∈V. details to follow...
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edited Feb 17 2010 at 1:24 |
U be thet set of functions f(x)∈C∞(ℝ) which vanish on x≤0.≤0 and are positive on x>0.V be the set of functions f:ℝ+→ℝ +such that f(x)>0 and x-nf(x)→0 as x→0, for each positive integer n.Lemma 1: For any f∈V, there is a g∈U such that f(x)/g(x)→0 as x→0. Sketch proof: Choose any smooth function r:ℝ+→ℝ+ with r(x)≥1 for small x, and r(x)=0 for x≥1. For example, we can use r(x) = exp(2-1/(1-x)) for x≤1. Then, the idea is to choose a sequence of positive reals αk→0 and set $$ g(x)=x^{\sum_kr_k(x/\alpha_k)}={\rm exp}\left({\rm log}(x)\sum_kr_k(x/\alpha_k)\right)$$ 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. The nth derivative is bounded by Knx-ng(x)Σk1{x>αk}αk-n, for positive constants Kn. As long as αk go to zero fast enough, this will be of size O(xn), so g∈C∞(ℝ). By definition of f∈V, there is a decreasing sequence of positive reals εk such that f(x)≤xn for x≤εn. We just need to make sure that αk≤εn+1 for k≥n to ensure that f(x)/g(x) goes to zero at rate x as x→0.
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edited Feb 17 2010 at 0:38 |
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 f(x)>0 and 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 need 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...
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answered Feb 17 2010 at 0:24 |
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...
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