Applying Ascoli's theorem to the space of continuous cumulative distribution functions

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I want to show that the space of continuous,strictly increasing commutative distribution functions are relatively compact. Though, I am not sure if I can do this.

Firstly, I consider the space of continuous, strictly increasing cumulative distribution functions on a compact interval $E_1$. Then, by Ascoli's theorem, this space is relatively compact. Now, I apply Ascoli's theorem again on a compact interval $E_2supset E_1$, and so on, $E_nsupset E_n-1$.

The space of strictly increasing, continuous cumulative distribution functions on $E_n$ is relatively compact for all $n$. Hence this space is relatively compact on $mathbbR$.

Can I make such an argument?

functional-analysis

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edited Jul 28 at 9:00

Bernard

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asked Jul 28 at 4:21

user1292919

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up vote
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I want to show that the space of continuous,strictly increasing commutative distribution functions are relatively compact. Though, I am not sure if I can do this.

Firstly, I consider the space of continuous, strictly increasing cumulative distribution functions on a compact interval $E_1$. Then, by Ascoli's theorem, this space is relatively compact. Now, I apply Ascoli's theorem again on a compact interval $E_2supset E_1$, and so on, $E_nsupset E_n-1$.

The space of strictly increasing, continuous cumulative distribution functions on $E_n$ is relatively compact for all $n$. Hence this space is relatively compact on $mathbbR$.

Can I make such an argument?

functional-analysis

share|cite|improve this question

edited Jul 28 at 9:00

Bernard

110k635102

asked Jul 28 at 4:21

user1292919

650512

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I want to show that the space of continuous,strictly increasing commutative distribution functions are relatively compact. Though, I am not sure if I can do this.

Firstly, I consider the space of continuous, strictly increasing cumulative distribution functions on a compact interval $E_1$. Then, by Ascoli's theorem, this space is relatively compact. Now, I apply Ascoli's theorem again on a compact interval $E_2supset E_1$, and so on, $E_nsupset E_n-1$.

The space of strictly increasing, continuous cumulative distribution functions on $E_n$ is relatively compact for all $n$. Hence this space is relatively compact on $mathbbR$.

Can I make such an argument?

functional-analysis

share|cite|improve this question

edited Jul 28 at 9:00

Bernard

110k635102

asked Jul 28 at 4:21

user1292919

650512

I want to show that the space of continuous,strictly increasing commutative distribution functions are relatively compact. Though, I am not sure if I can do this.

Firstly, I consider the space of continuous, strictly increasing cumulative distribution functions on a compact interval $E_1$. Then, by Ascoli's theorem, this space is relatively compact. Now, I apply Ascoli's theorem again on a compact interval $E_2supset E_1$, and so on, $E_nsupset E_n-1$.

The space of strictly increasing, continuous cumulative distribution functions on $E_n$ is relatively compact for all $n$. Hence this space is relatively compact on $mathbbR$.

Can I make such an argument?

functional-analysis

share|cite|improve this question

edited Jul 28 at 9:00

Bernard

110k635102

asked Jul 28 at 4:21

user1292919

650512

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 9:00

Bernard

110k635102

edited Jul 28 at 9:00

Bernard

110k635102

edited Jul 28 at 9:00

Bernard

110k635102

110k635102

asked Jul 28 at 4:21

user1292919

650512

asked Jul 28 at 4:21

user1292919

650512

asked Jul 28 at 4:21

user1292919

650512

650512

add a comment | 

add a comment | 

2 Answers
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No. First Arzela's theorem requires the compactness; localizing as you suggest results in uniform convergence on compact subsets which is not really what you want. More to the point, the local result fails too because equicontinuity fails under these assumptions. Consider $f_n(x)=x^n$ on $[0,1]$ and extended to be $0$ on the left and $1$ on the right.

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answered Jul 28 at 5:35

Ian

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Let $F_n$ be normal with mean $0$ and variance $n$. Then there is no convergent subsequence.

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answered Jul 28 at 12:20

Kavi Rama Murthy

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

No. First Arzela's theorem requires the compactness; localizing as you suggest results in uniform convergence on compact subsets which is not really what you want. More to the point, the local result fails too because equicontinuity fails under these assumptions. Consider $f_n(x)=x^n$ on $[0,1]$ and extended to be $0$ on the left and $1$ on the right.

share|cite|improve this answer

answered Jul 28 at 5:35

Ian

65k24681

add a comment | 

up vote
1
down vote

No. First Arzela's theorem requires the compactness; localizing as you suggest results in uniform convergence on compact subsets which is not really what you want. More to the point, the local result fails too because equicontinuity fails under these assumptions. Consider $f_n(x)=x^n$ on $[0,1]$ and extended to be $0$ on the left and $1$ on the right.

share|cite|improve this answer

answered Jul 28 at 5:35

Ian

65k24681

add a comment | 

up vote
1
down vote

up vote
1
down vote

No. First Arzela's theorem requires the compactness; localizing as you suggest results in uniform convergence on compact subsets which is not really what you want. More to the point, the local result fails too because equicontinuity fails under these assumptions. Consider $f_n(x)=x^n$ on $[0,1]$ and extended to be $0$ on the left and $1$ on the right.

share|cite|improve this answer

answered Jul 28 at 5:35

Ian

65k24681

No. First Arzela's theorem requires the compactness; localizing as you suggest results in uniform convergence on compact subsets which is not really what you want. More to the point, the local result fails too because equicontinuity fails under these assumptions. Consider $f_n(x)=x^n$ on $[0,1]$ and extended to be $0$ on the left and $1$ on the right.

share|cite|improve this answer

answered Jul 28 at 5:35

Ian

65k24681

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 5:35

Ian

65k24681

answered Jul 28 at 5:35

Ian

65k24681

answered Jul 28 at 5:35

Ian

65k24681

65k24681

add a comment | 

add a comment | 

up vote
0
down vote

Let $F_n$ be normal with mean $0$ and variance $n$. Then there is no convergent subsequence.

share|cite|improve this answer

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

add a comment | 

up vote
0
down vote

Let $F_n$ be normal with mean $0$ and variance $n$. Then there is no convergent subsequence.

share|cite|improve this answer

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let $F_n$ be normal with mean $0$ and variance $n$. Then there is no convergent subsequence.

share|cite|improve this answer

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

Let $F_n$ be normal with mean $0$ and variance $n$. Then there is no convergent subsequence.

share|cite|improve this answer

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

answered Jul 28 at 12:20

Kavi Rama Murthy

19.9k2829

19.9k2829

add a comment | 

add a comment | 

 

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