Mixing
Separation of closed sets with distance $>0$ by function $f in C_b^k(mathbb{R}^d)$
$begingroup$
I'm interested in the following problem on the separation of closed sets:
Let $A,B subseteq mathbb{R}^d$ be closed sets such that $$d(A,B) := inf{|x-y|; x in A, y in B}>0.$$ Question: Does there exist a function $f in C_b^k(mathbb{R}^d)$ such that $$f^{-1}({0}) = A quad text{and} quad f^{-1}({1})=B tag{1}$$ ...?
Here, $C_b^k(mathbb{R}^d)$ denotes the space of functions $f: mathbb{R}^d to mathbb{R}$ with bounded derivatives up to order $k$, and $k in mathbb{N}$ is some fixed number.
It is a classical result that there exists a continuous function $f$ satisfying $(1)$. With a bit more effort it can be shown that there exists a smooth function $f in C^{infty}(mathbb{R}^d)$ which satisfies $(1)$; this is e.g. discussed in this question.
I strongly believe that the answer to my question is "yes" but I couldn't find any results concerning the boundedness of the derivates of Urysohn function $f$. The condition $d(A,B)>0$ means intuitively that the function $f$ does not need to be arbitrarily steep, and hence it would be naturally that $f$ can be chosen in such a way that its derivates are bounded. However, it is not obvious for me how to prove this rigorously; the construction discussed in the question, which I linked above, does not seem to be helpful.
I would be happy about references and/or your thoughts on the question.
real-analysis general-topology
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
$endgroup$
add a comment |
$begingroup$
I'm interested in the following problem on the separation of closed sets:
Let $A,B subseteq mathbb{R}^d$ be closed sets such that $$d(A,B) := inf{|x-y|; x in A, y in B}>0.$$ Question: Does there exist a function $f in C_b^k(mathbb{R}^d)$ such that $$f^{-1}({0}) = A quad text{and} quad f^{-1}({1})=B tag{1}$$ ...?
Here, $C_b^k(mathbb{R}^d)$ denotes the space of functions $f: mathbb{R}^d to mathbb{R}$ with bounded derivatives up to order $k$, and $k in mathbb{N}$ is some fixed number.
It is a classical result that there exists a continuous function $f$ satisfying $(1)$. With a bit more effort it can be shown that there exists a smooth function $f in C^{infty}(mathbb{R}^d)$ which satisfies $(1)$; this is e.g. discussed in this question.
I strongly believe that the answer to my question is "yes" but I couldn't find any results concerning the boundedness of the derivates of Urysohn function $f$. The condition $d(A,B)>0$ means intuitively that the function $f$ does not need to be arbitrarily steep, and hence it would be naturally that $f$ can be chosen in such a way that its derivates are bounded. However, it is not obvious for me how to prove this rigorously; the construction discussed in the question, which I linked above, does not seem to be helpful.
I would be happy about references and/or your thoughts on the question.
real-analysis general-topology
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
$endgroup$
add a comment |
$begingroup$
I'm interested in the following problem on the separation of closed sets:
Let $A,B subseteq mathbb{R}^d$ be closed sets such that $$d(A,B) := inf{|x-y|; x in A, y in B}>0.$$ Question: Does there exist a function $f in C_b^k(mathbb{R}^d)$ such that $$f^{-1}({0}) = A quad text{and} quad f^{-1}({1})=B tag{1}$$ ...?
Here, $C_b^k(mathbb{R}^d)$ denotes the space of functions $f: mathbb{R}^d to mathbb{R}$ with bounded derivatives up to order $k$, and $k in mathbb{N}$ is some fixed number.
It is a classical result that there exists a continuous function $f$ satisfying $(1)$. With a bit more effort it can be shown that there exists a smooth function $f in C^{infty}(mathbb{R}^d)$ which satisfies $(1)$; this is e.g. discussed in this question.
I strongly believe that the answer to my question is "yes" but I couldn't find any results concerning the boundedness of the derivates of Urysohn function $f$. The condition $d(A,B)>0$ means intuitively that the function $f$ does not need to be arbitrarily steep, and hence it would be naturally that $f$ can be chosen in such a way that its derivates are bounded. However, it is not obvious for me how to prove this rigorously; the construction discussed in the question, which I linked above, does not seem to be helpful.
I would be happy about references and/or your thoughts on the question.
real-analysis general-topology
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
$endgroup$
I'm interested in the following problem on the separation of closed sets:
Let $A,B subseteq mathbb{R}^d$ be closed sets such that $$d(A,B) := inf{|x-y|; x in A, y in B}>0.$$ Question: Does there exist a function $f in C_b^k(mathbb{R}^d)$ such that $$f^{-1}({0}) = A quad text{and} quad f^{-1}({1})=B tag{1}$$ ...?
Here, $C_b^k(mathbb{R}^d)$ denotes the space of functions $f: mathbb{R}^d to mathbb{R}$ with bounded derivatives up to order $k$, and $k in mathbb{N}$ is some fixed number.
It is a classical result that there exists a continuous function $f$ satisfying $(1)$. With a bit more effort it can be shown that there exists a smooth function $f in C^{infty}(mathbb{R}^d)$ which satisfies $(1)$; this is e.g. discussed in this question.
I strongly believe that the answer to my question is "yes" but I couldn't find any results concerning the boundedness of the derivates of Urysohn function $f$. The condition $d(A,B)>0$ means intuitively that the function $f$ does not need to be arbitrarily steep, and hence it would be naturally that $f$ can be chosen in such a way that its derivates are bounded. However, it is not obvious for me how to prove this rigorously; the construction discussed in the question, which I linked above, does not seem to be helpful.
I would be happy about references and/or your thoughts on the question.
real-analysis general-topology
real-analysis general-topology
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
edited Dec 2 '18 at 14:58
saz
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
asked Nov 4 '18 at 19:10
sazsaz
81.4k861127
81.4k861127
add a comment |
add a comment |
1 Answer
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I think I managed to prove the assertion; it turns out that it is possible to construct $f in C_b^{infty}(mathbb{R}^d)$ satisfying $(1)$.
As $d(A,B)>0$ we can choose $epsilon>0$ such that the sets $$A_{epsilon} := A+overline{B(0,epsilon)} qquad B_{epsilon} := B+overline{B(0,epsilon)}$$ are disjoint. According to this result, there exists $h in C^{infty}(mathbb{R}^d)$, $0 leq h leq 1$, such that $$h^{-1}({0}) = A_{epsilon} quad text{and} quad h^{-1}({1})=B_{epsilon}.$$Now let $varphi in C_c^{infty}(mathbb{R}^d)$ be such that $text{spt} , varphi = overline{B(0,epsilon)}$, $int varphi(y) , dy=1$ and $varphi geq 0$. Define $$f(x) := (h ast varphi)(x) = int_{mathbb{R}^d} h(y) varphi(x-y) , dy, qquad x in mathbb{R}^d.$$ Since $f$ is the convolution of a bounded continuous function with a smooth function with compact support, it is well known that $f$ is smooth and its derivatives are given by
$$partial_x^{alpha} f(x) = int_{mathbb{R}^d} h(y) partial_x^{alpha} varphi(x-y) , dy$$
for any multi-index $alpha in mathbb{N}_0^d$. This implies, in particular, $|partial^{alpha} f|_{infty} leq |partial^{alpha} varphi|_{L^1}< infty$, and so $f in C_b^{infty}(mathbb{R}^d)$. Moreover, as $text{spt} , varphi subseteq overline{B(0,epsilon)}$, it is obvious that $f(x)=0$ for any $x in A$ and $f(x)=1$ for $x in B$. It remains to check that $0<f(x)<1$ for any $x in (A cup B)^c$. To this end, we consider several cases separately:
Case 1: $x in mathbb{R}^d backslash (A_{epsilon} cup B_{epsilon})$. Then $0 < h(x)<1$, and therefore we can choose $r in (0,epsilon)$ such that
$$0 < inf_{|y-x| leq r} h(y) leq sup_{|y-x| leq r} h(y)<1;$$ this implies $$begin{align*} f(x) &leq int_{overline{B(x,r)}^c} varphi(x-y) , dy + underbrace{sup_{|y-x| leq r} h(y)}_{<1} int_{overline{B(x,r)}}varphi(x-y) , dy \ &< int_{mathbb{R}^d} varphi(x-y) , dy =1; end{align*}$$ here we have used that $text{spt} , varphi = overline{B(0,epsilon)} supseteq overline{B(x,r)}$. A very similar estimate shows $f(x)>0$.
Case 2: $x in A_{epsilon} backslash A$. We have $overline{B(x,epsilon)} cap A^c neq emptyset$, and therefore there exist $y in mathbb{R}^d$ and $r>0$ such that $$overline{B(y,r)} subseteq A^c cap overline{B(x,epsilon)}.$$ In particular $$0 < inf_{z in overline{B(y,r)}} h(z) leq sup_{z in overline{B(y,r)}} h(z) < 1.$$Since $text{spt} , varphi = overline{B(0,epsilon)}$, it follows very similar as in the first case that $0<f(x)<1$.
Case 3: $x in B_{epsilon} backslash B$. This works as case 2.
Consequently, we have shown that $f$ has all the desired properties.
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
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add a comment |
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I think I managed to prove the assertion; it turns out that it is possible to construct $f in C_b^{infty}(mathbb{R}^d)$ satisfying $(1)$.
As $d(A,B)>0$ we can choose $epsilon>0$ such that the sets $$A_{epsilon} := A+overline{B(0,epsilon)} qquad B_{epsilon} := B+overline{B(0,epsilon)}$$ are disjoint. According to this result, there exists $h in C^{infty}(mathbb{R}^d)$, $0 leq h leq 1$, such that $$h^{-1}({0}) = A_{epsilon} quad text{and} quad h^{-1}({1})=B_{epsilon}.$$Now let $varphi in C_c^{infty}(mathbb{R}^d)$ be such that $text{spt} , varphi = overline{B(0,epsilon)}$, $int varphi(y) , dy=1$ and $varphi geq 0$. Define $$f(x) := (h ast varphi)(x) = int_{mathbb{R}^d} h(y) varphi(x-y) , dy, qquad x in mathbb{R}^d.$$ Since $f$ is the convolution of a bounded continuous function with a smooth function with compact support, it is well known that $f$ is smooth and its derivatives are given by
$$partial_x^{alpha} f(x) = int_{mathbb{R}^d} h(y) partial_x^{alpha} varphi(x-y) , dy$$
for any multi-index $alpha in mathbb{N}_0^d$. This implies, in particular, $|partial^{alpha} f|_{infty} leq |partial^{alpha} varphi|_{L^1}< infty$, and so $f in C_b^{infty}(mathbb{R}^d)$. Moreover, as $text{spt} , varphi subseteq overline{B(0,epsilon)}$, it is obvious that $f(x)=0$ for any $x in A$ and $f(x)=1$ for $x in B$. It remains to check that $0<f(x)<1$ for any $x in (A cup B)^c$. To this end, we consider several cases separately:
Case 1: $x in mathbb{R}^d backslash (A_{epsilon} cup B_{epsilon})$. Then $0 < h(x)<1$, and therefore we can choose $r in (0,epsilon)$ such that
$$0 < inf_{|y-x| leq r} h(y) leq sup_{|y-x| leq r} h(y)<1;$$ this implies $$begin{align*} f(x) &leq int_{overline{B(x,r)}^c} varphi(x-y) , dy + underbrace{sup_{|y-x| leq r} h(y)}_{<1} int_{overline{B(x,r)}}varphi(x-y) , dy \ &< int_{mathbb{R}^d} varphi(x-y) , dy =1; end{align*}$$ here we have used that $text{spt} , varphi = overline{B(0,epsilon)} supseteq overline{B(x,r)}$. A very similar estimate shows $f(x)>0$.
Case 2: $x in A_{epsilon} backslash A$. We have $overline{B(x,epsilon)} cap A^c neq emptyset$, and therefore there exist $y in mathbb{R}^d$ and $r>0$ such that $$overline{B(y,r)} subseteq A^c cap overline{B(x,epsilon)}.$$ In particular $$0 < inf_{z in overline{B(y,r)}} h(z) leq sup_{z in overline{B(y,r)}} h(z) < 1.$$Since $text{spt} , varphi = overline{B(0,epsilon)}$, it follows very similar as in the first case that $0<f(x)<1$.
Case 3: $x in B_{epsilon} backslash B$. This works as case 2.
Consequently, we have shown that $f$ has all the desired properties.
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
$endgroup$
add a comment |
$begingroup$
I think I managed to prove the assertion; it turns out that it is possible to construct $f in C_b^{infty}(mathbb{R}^d)$ satisfying $(1)$.
As $d(A,B)>0$ we can choose $epsilon>0$ such that the sets $$A_{epsilon} := A+overline{B(0,epsilon)} qquad B_{epsilon} := B+overline{B(0,epsilon)}$$ are disjoint. According to this result, there exists $h in C^{infty}(mathbb{R}^d)$, $0 leq h leq 1$, such that $$h^{-1}({0}) = A_{epsilon} quad text{and} quad h^{-1}({1})=B_{epsilon}.$$Now let $varphi in C_c^{infty}(mathbb{R}^d)$ be such that $text{spt} , varphi = overline{B(0,epsilon)}$, $int varphi(y) , dy=1$ and $varphi geq 0$. Define $$f(x) := (h ast varphi)(x) = int_{mathbb{R}^d} h(y) varphi(x-y) , dy, qquad x in mathbb{R}^d.$$ Since $f$ is the convolution of a bounded continuous function with a smooth function with compact support, it is well known that $f$ is smooth and its derivatives are given by
$$partial_x^{alpha} f(x) = int_{mathbb{R}^d} h(y) partial_x^{alpha} varphi(x-y) , dy$$
for any multi-index $alpha in mathbb{N}_0^d$. This implies, in particular, $|partial^{alpha} f|_{infty} leq |partial^{alpha} varphi|_{L^1}< infty$, and so $f in C_b^{infty}(mathbb{R}^d)$. Moreover, as $text{spt} , varphi subseteq overline{B(0,epsilon)}$, it is obvious that $f(x)=0$ for any $x in A$ and $f(x)=1$ for $x in B$. It remains to check that $0<f(x)<1$ for any $x in (A cup B)^c$. To this end, we consider several cases separately:
Case 1: $x in mathbb{R}^d backslash (A_{epsilon} cup B_{epsilon})$. Then $0 < h(x)<1$, and therefore we can choose $r in (0,epsilon)$ such that
$$0 < inf_{|y-x| leq r} h(y) leq sup_{|y-x| leq r} h(y)<1;$$ this implies $$begin{align*} f(x) &leq int_{overline{B(x,r)}^c} varphi(x-y) , dy + underbrace{sup_{|y-x| leq r} h(y)}_{<1} int_{overline{B(x,r)}}varphi(x-y) , dy \ &< int_{mathbb{R}^d} varphi(x-y) , dy =1; end{align*}$$ here we have used that $text{spt} , varphi = overline{B(0,epsilon)} supseteq overline{B(x,r)}$. A very similar estimate shows $f(x)>0$.
Case 2: $x in A_{epsilon} backslash A$. We have $overline{B(x,epsilon)} cap A^c neq emptyset$, and therefore there exist $y in mathbb{R}^d$ and $r>0$ such that $$overline{B(y,r)} subseteq A^c cap overline{B(x,epsilon)}.$$ In particular $$0 < inf_{z in overline{B(y,r)}} h(z) leq sup_{z in overline{B(y,r)}} h(z) < 1.$$Since $text{spt} , varphi = overline{B(0,epsilon)}$, it follows very similar as in the first case that $0<f(x)<1$.
Case 3: $x in B_{epsilon} backslash B$. This works as case 2.
Consequently, we have shown that $f$ has all the desired properties.
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
$endgroup$
add a comment |
$begingroup$
I think I managed to prove the assertion; it turns out that it is possible to construct $f in C_b^{infty}(mathbb{R}^d)$ satisfying $(1)$.
As $d(A,B)>0$ we can choose $epsilon>0$ such that the sets $$A_{epsilon} := A+overline{B(0,epsilon)} qquad B_{epsilon} := B+overline{B(0,epsilon)}$$ are disjoint. According to this result, there exists $h in C^{infty}(mathbb{R}^d)$, $0 leq h leq 1$, such that $$h^{-1}({0}) = A_{epsilon} quad text{and} quad h^{-1}({1})=B_{epsilon}.$$Now let $varphi in C_c^{infty}(mathbb{R}^d)$ be such that $text{spt} , varphi = overline{B(0,epsilon)}$, $int varphi(y) , dy=1$ and $varphi geq 0$. Define $$f(x) := (h ast varphi)(x) = int_{mathbb{R}^d} h(y) varphi(x-y) , dy, qquad x in mathbb{R}^d.$$ Since $f$ is the convolution of a bounded continuous function with a smooth function with compact support, it is well known that $f$ is smooth and its derivatives are given by
$$partial_x^{alpha} f(x) = int_{mathbb{R}^d} h(y) partial_x^{alpha} varphi(x-y) , dy$$
for any multi-index $alpha in mathbb{N}_0^d$. This implies, in particular, $|partial^{alpha} f|_{infty} leq |partial^{alpha} varphi|_{L^1}< infty$, and so $f in C_b^{infty}(mathbb{R}^d)$. Moreover, as $text{spt} , varphi subseteq overline{B(0,epsilon)}$, it is obvious that $f(x)=0$ for any $x in A$ and $f(x)=1$ for $x in B$. It remains to check that $0<f(x)<1$ for any $x in (A cup B)^c$. To this end, we consider several cases separately:
Case 1: $x in mathbb{R}^d backslash (A_{epsilon} cup B_{epsilon})$. Then $0 < h(x)<1$, and therefore we can choose $r in (0,epsilon)$ such that
$$0 < inf_{|y-x| leq r} h(y) leq sup_{|y-x| leq r} h(y)<1;$$ this implies $$begin{align*} f(x) &leq int_{overline{B(x,r)}^c} varphi(x-y) , dy + underbrace{sup_{|y-x| leq r} h(y)}_{<1} int_{overline{B(x,r)}}varphi(x-y) , dy \ &< int_{mathbb{R}^d} varphi(x-y) , dy =1; end{align*}$$ here we have used that $text{spt} , varphi = overline{B(0,epsilon)} supseteq overline{B(x,r)}$. A very similar estimate shows $f(x)>0$.
Case 2: $x in A_{epsilon} backslash A$. We have $overline{B(x,epsilon)} cap A^c neq emptyset$, and therefore there exist $y in mathbb{R}^d$ and $r>0$ such that $$overline{B(y,r)} subseteq A^c cap overline{B(x,epsilon)}.$$ In particular $$0 < inf_{z in overline{B(y,r)}} h(z) leq sup_{z in overline{B(y,r)}} h(z) < 1.$$Since $text{spt} , varphi = overline{B(0,epsilon)}$, it follows very similar as in the first case that $0<f(x)<1$.
Case 3: $x in B_{epsilon} backslash B$. This works as case 2.
Consequently, we have shown that $f$ has all the desired properties.
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
$endgroup$
I think I managed to prove the assertion; it turns out that it is possible to construct $f in C_b^{infty}(mathbb{R}^d)$ satisfying $(1)$.
As $d(A,B)>0$ we can choose $epsilon>0$ such that the sets $$A_{epsilon} := A+overline{B(0,epsilon)} qquad B_{epsilon} := B+overline{B(0,epsilon)}$$ are disjoint. According to this result, there exists $h in C^{infty}(mathbb{R}^d)$, $0 leq h leq 1$, such that $$h^{-1}({0}) = A_{epsilon} quad text{and} quad h^{-1}({1})=B_{epsilon}.$$Now let $varphi in C_c^{infty}(mathbb{R}^d)$ be such that $text{spt} , varphi = overline{B(0,epsilon)}$, $int varphi(y) , dy=1$ and $varphi geq 0$. Define $$f(x) := (h ast varphi)(x) = int_{mathbb{R}^d} h(y) varphi(x-y) , dy, qquad x in mathbb{R}^d.$$ Since $f$ is the convolution of a bounded continuous function with a smooth function with compact support, it is well known that $f$ is smooth and its derivatives are given by
$$partial_x^{alpha} f(x) = int_{mathbb{R}^d} h(y) partial_x^{alpha} varphi(x-y) , dy$$
for any multi-index $alpha in mathbb{N}_0^d$. This implies, in particular, $|partial^{alpha} f|_{infty} leq |partial^{alpha} varphi|_{L^1}< infty$, and so $f in C_b^{infty}(mathbb{R}^d)$. Moreover, as $text{spt} , varphi subseteq overline{B(0,epsilon)}$, it is obvious that $f(x)=0$ for any $x in A$ and $f(x)=1$ for $x in B$. It remains to check that $0<f(x)<1$ for any $x in (A cup B)^c$. To this end, we consider several cases separately:
Case 1: $x in mathbb{R}^d backslash (A_{epsilon} cup B_{epsilon})$. Then $0 < h(x)<1$, and therefore we can choose $r in (0,epsilon)$ such that
$$0 < inf_{|y-x| leq r} h(y) leq sup_{|y-x| leq r} h(y)<1;$$ this implies $$begin{align*} f(x) &leq int_{overline{B(x,r)}^c} varphi(x-y) , dy + underbrace{sup_{|y-x| leq r} h(y)}_{<1} int_{overline{B(x,r)}}varphi(x-y) , dy \ &< int_{mathbb{R}^d} varphi(x-y) , dy =1; end{align*}$$ here we have used that $text{spt} , varphi = overline{B(0,epsilon)} supseteq overline{B(x,r)}$. A very similar estimate shows $f(x)>0$.
Case 2: $x in A_{epsilon} backslash A$. We have $overline{B(x,epsilon)} cap A^c neq emptyset$, and therefore there exist $y in mathbb{R}^d$ and $r>0$ such that $$overline{B(y,r)} subseteq A^c cap overline{B(x,epsilon)}.$$ In particular $$0 < inf_{z in overline{B(y,r)}} h(z) leq sup_{z in overline{B(y,r)}} h(z) < 1.$$Since $text{spt} , varphi = overline{B(0,epsilon)}$, it follows very similar as in the first case that $0<f(x)<1$.
Case 3: $x in B_{epsilon} backslash B$. This works as case 2.
Consequently, we have shown that $f$ has all the desired properties.
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
edited Jan 20 at 20:51
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
answered Nov 5 '18 at 14:20
sazsaz
81.4k861127
81.4k861127
add a comment |
add a comment |
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