Proving that a family of functions is compact
Clash Royale CLAN TAG#URR8PPP
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- Suppose that I'm dealing with a family of complex functions analytic in the right-half plane and that each $f$ has a representation:
$$f(z) = int^1_-1 frac2z(1+ z^2) + t(1 - z^2) , dmu(t),$$
where $dmu(t)$ is some probability measure. How do I show that such a family is compact?
- Similarly, let each $f(z_1, z_2)$ be analytic in $mathbbC_+ times mathbbC_+$.
$$f(z_1, z_2) = int^1_-1int^1_-1 frac2z_1(1+ z_1^2) + t_1(1 - z_1^2) frac2z_2(1+ z_2^2) + t_2(1 - z_2^2) , dmu_1(t), dmu_2(t).$$
Do those functions form a compact family?
complex-analysis convex-analysis several-complex-variables
asked Sep 3 at 13:36
skids
33
add a comment |Â
up vote
0
down vote
favorite
- Suppose that I'm dealing with a family of complex functions analytic in the right-half plane and that each $f$ has a representation:
$$f(z) = int^1_-1 frac2z(1+ z^2) + t(1 - z^2) , dmu(t),$$
where $dmu(t)$ is some probability measure. How do I show that such a family is compact?
- Similarly, let each $f(z_1, z_2)$ be analytic in $mathbbC_+ times mathbbC_+$.
$$f(z_1, z_2) = int^1_-1int^1_-1 frac2z_1(1+ z_1^2) + t_1(1 - z_1^2) frac2z_2(1+ z_2^2) + t_2(1 - z_2^2) , dmu_1(t), dmu_2(t).$$
Do those functions form a compact family?
complex-analysis convex-analysis several-complex-variables
asked Sep 3 at 13:36
skids
33
1
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
- Suppose that I'm dealing with a family of complex functions analytic in the right-half plane and that each $f$ has a representation:
$$f(z) = int^1_-1 frac2z(1+ z^2) + t(1 - z^2) , dmu(t),$$
where $dmu(t)$ is some probability measure. How do I show that such a family is compact?
- Similarly, let each $f(z_1, z_2)$ be analytic in $mathbbC_+ times mathbbC_+$.
$$f(z_1, z_2) = int^1_-1int^1_-1 frac2z_1(1+ z_1^2) + t_1(1 - z_1^2) frac2z_2(1+ z_2^2) + t_2(1 - z_2^2) , dmu_1(t), dmu_2(t).$$
Do those functions form a compact family?
complex-analysis convex-analysis several-complex-variables
asked Sep 3 at 13:36
skids
33
- Suppose that I'm dealing with a family of complex functions analytic in the right-half plane and that each $f$ has a representation:
$$f(z) = int^1_-1 frac2z(1+ z^2) + t(1 - z^2) , dmu(t),$$
where $dmu(t)$ is some probability measure. How do I show that such a family is compact?
- Similarly, let each $f(z_1, z_2)$ be analytic in $mathbbC_+ times mathbbC_+$.
$$f(z_1, z_2) = int^1_-1int^1_-1 frac2z_1(1+ z_1^2) + t_1(1 - z_1^2) frac2z_2(1+ z_2^2) + t_2(1 - z_2^2) , dmu_1(t), dmu_2(t).$$
Do those functions form a compact family?
complex-analysis convex-analysis several-complex-variables
complex-analysis convex-analysis several-complex-variables
asked Sep 3 at 13:36
skids
33
asked Sep 3 at 13:36
skids
33
asked Sep 3 at 13:36
skids
33
asked Sep 3 at 13:36
skids
33
asked Sep 3 at 13:36
skids
33
33
1
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26
add a comment |Â
1
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26
1
1
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26
add a comment |Â
1 Answer
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up vote
1
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These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $tin [-1,1]$, define $$K(z,t) = frac2z(1+z^2)+t(1-z^2).$$ It is a routine exercise that the map $mumapsto f(z)=int_[-1,1] K(z,t) mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $mu_ntomu$ weak* then the corresponding $f_nto f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $Ctimes[-1,1]$. Hence the collection of restrictions $K_t: zmapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $delta$-balls, and bound $|f_n-f|_C=sup$ with a finite sum of differences of integrals, one per ball center, plus a term $omega(delta)$. The weak* convergence of the $mu_n$ to $mu$ shows the differences of the integrals converges to $0$; choice of $delta$ small enough then forces $|f_n-f|_C<epsilon$ for any desired $epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact.
Alternatively, if $f_n$ are elements of your set, corresponding to measure $mu_n$, there is a weak* convergent subsequence of the $mu_n$ with limit $mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.
answered Sep 3 at 17:25
kimchi lover
8,91031128
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $tin [-1,1]$, define $$K(z,t) = frac2z(1+z^2)+t(1-z^2).$$ It is a routine exercise that the map $mumapsto f(z)=int_[-1,1] K(z,t) mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $mu_ntomu$ weak* then the corresponding $f_nto f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $Ctimes[-1,1]$. Hence the collection of restrictions $K_t: zmapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $delta$-balls, and bound $|f_n-f|_C=sup$ with a finite sum of differences of integrals, one per ball center, plus a term $omega(delta)$. The weak* convergence of the $mu_n$ to $mu$ shows the differences of the integrals converges to $0$; choice of $delta$ small enough then forces $|f_n-f|_C<epsilon$ for any desired $epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact.
Alternatively, if $f_n$ are elements of your set, corresponding to measure $mu_n$, there is a weak* convergent subsequence of the $mu_n$ with limit $mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.
answered Sep 3 at 17:25
kimchi lover
8,91031128
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
add a comment |Â
up vote
1
down vote
accepted
These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $tin [-1,1]$, define $$K(z,t) = frac2z(1+z^2)+t(1-z^2).$$ It is a routine exercise that the map $mumapsto f(z)=int_[-1,1] K(z,t) mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $mu_ntomu$ weak* then the corresponding $f_nto f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $Ctimes[-1,1]$. Hence the collection of restrictions $K_t: zmapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $delta$-balls, and bound $|f_n-f|_C=sup$ with a finite sum of differences of integrals, one per ball center, plus a term $omega(delta)$. The weak* convergence of the $mu_n$ to $mu$ shows the differences of the integrals converges to $0$; choice of $delta$ small enough then forces $|f_n-f|_C<epsilon$ for any desired $epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact.
Alternatively, if $f_n$ are elements of your set, corresponding to measure $mu_n$, there is a weak* convergent subsequence of the $mu_n$ with limit $mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.
answered Sep 3 at 17:25
kimchi lover
8,91031128
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $tin [-1,1]$, define $$K(z,t) = frac2z(1+z^2)+t(1-z^2).$$ It is a routine exercise that the map $mumapsto f(z)=int_[-1,1] K(z,t) mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $mu_ntomu$ weak* then the corresponding $f_nto f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $Ctimes[-1,1]$. Hence the collection of restrictions $K_t: zmapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $delta$-balls, and bound $|f_n-f|_C=sup$ with a finite sum of differences of integrals, one per ball center, plus a term $omega(delta)$. The weak* convergence of the $mu_n$ to $mu$ shows the differences of the integrals converges to $0$; choice of $delta$ small enough then forces $|f_n-f|_C<epsilon$ for any desired $epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact.
Alternatively, if $f_n$ are elements of your set, corresponding to measure $mu_n$, there is a weak* convergent subsequence of the $mu_n$ with limit $mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.
answered Sep 3 at 17:25
kimchi lover
8,91031128
These are suggestions, not proofs. (I'm not very familiar with spaces of analytic functions, and am away from reference books now.)
For $tin [-1,1]$, define $$K(z,t) = frac2z(1+z^2)+t(1-z^2).$$ It is a routine exercise that the map $mumapsto f(z)=int_[-1,1] K(z,t) mu(dt)$ is a continuous map from the probability measures on $[-1,1]$ (with the weak* topology) into the analytic functions on the right half plane.
At the risk of boring you, here is one way to argue.
One has to check that if $mu_ntomu$ weak* then the corresponding $f_nto f$ uniformly on compacts. Let $C$ be a compact subset of the half plane. Then $K$ is uniformly continuous on $Ctimes[-1,1]$. Hence the collection of restrictions $K_t: zmapsto K(z,t)$ of $K$ obtained by holding $t$ fixed is uniformly continuous: the 2-dimensional modulus of continuity $omega$ of $K$ serves as a common 1-dimensional modulus of continuity for the $K_t$. Cover $C$ with finitely many $delta$-balls, and bound $|f_n-f|_C=sup$ with a finite sum of differences of integrals, one per ball center, plus a term $omega(delta)$. The weak* convergence of the $mu_n$ to $mu$ shows the differences of the integrals converges to $0$; choice of $delta$ small enough then forces $|f_n-f|_C<epsilon$ for any desired $epsilon$.
Then your set is the continuous image of the set of probability measures on $[-1,1]$, which is compact.
Alternatively, if $f_n$ are elements of your set, corresponding to measure $mu_n$, there is a weak* convergent subsequence of the $mu_n$ with limit $mu$. Then the foregoing argument shows that the corresponding subsequence of $f_n$ converges as desired.
Similarly for 2. Or more directly, probably. Surely the cartesian product of a compact set of functions of $z_1$ with a compact set of functions of $z_2$ is compact in the space of functions of $(z_1,z_2)$?
In some sense, by using integrals against probability measures $mu$ you are automatically closing your sets, in a way you wouldn't if you used finite weighted sums.
answered Sep 3 at 17:25
kimchi lover
8,91031128
edited Sep 3 at 23:31
answered Sep 3 at 17:25
kimchi lover
8,91031128
answered Sep 3 at 17:25
kimchi lover
8,91031128
answered Sep 3 at 17:25
kimchi lover
8,91031128
8,91031128
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
add a comment |Â
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
Great, thank you! (I'd welcome any potential reference suggestions, especially concerning $mu mapsto f(z)$ being continuous.)
– skids
Sep 3 at 17:42
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
I have added detail. The spaces involved are all separable metric spaces, so sequences suffice for compactness checking. For a general reference on convergence of probability measures, Billingsley's Convergence of Probability Measures is hard to beat.
– kimchi lover
Sep 3 at 23:38
add a comment |Â
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1
Where does convex analysis come in to this? What topology are you using on the functions $f$?
– copper.hat
Sep 3 at 14:08
Ad topology - the one given by uniform convergence on compact subsets. Ad convexity - probably not much, but since I'm ultimately interesting in applying some results concerning convex functionals on convex compact families of analytic functions, I hoped this might of relevance (Will remove the tag if that's mislabeling.)
– skids
Sep 3 at 14:28
I would start by trying to show that the collection of functions $2z/((1+z^2)+t(1-z^2)): tin [-1,1]$ is compact. Your set is the convex hull of this collection, so the end game should be easy.
– kimchi lover
Sep 3 at 15:07
@kimchilover Is it obvious that my set is the convex hull, as opposed to: contained in the convex hull?
– skids
Sep 3 at 15:27
Also, is it straightforward to see that $2z/((1+ z^2) + t(1 - z^2)): t in [-1, 1]$ is compact? It would be compact if it was closed and locally uniformly bounded; and it would be l.c.b. if if it was normal, but not sure if this gets me anywhere so far...
– skids
Sep 3 at 17:26