Why $L_{p}(mathbb{R})$ is separable? [duplicate]
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This question already has an answer here:
Separability of $l^{p}$ spaces
1 answer
It's easy to prove that $L_{p}(K)$ is separable , using Stone theorem. But how can we show the same result for real line ?
I thought about considering : $ mathbb{R} = cup (a_i , b_i] $ and $mathbb{1}((a_i , b_i])$.
But my teacher said that they are not in $L_{p}$.
real-analysis functional-analysis
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
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marked as duplicate by Cesareo, user91500, mrtaurho, José Carlos Santos real-analysis
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Dec 26 '18 at 11:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Separability of $l^{p}$ spaces
1 answer
It's easy to prove that $L_{p}(K)$ is separable , using Stone theorem. But how can we show the same result for real line ?
I thought about considering : $ mathbb{R} = cup (a_i , b_i] $ and $mathbb{1}((a_i , b_i])$.
But my teacher said that they are not in $L_{p}$.
real-analysis functional-analysis
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
$endgroup$
marked as duplicate by Cesareo, user91500, mrtaurho, José Carlos Santos real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Dec 26 '18 at 11:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53
add a comment |
$begingroup$
This question already has an answer here:
Separability of $l^{p}$ spaces
1 answer
It's easy to prove that $L_{p}(K)$ is separable , using Stone theorem. But how can we show the same result for real line ?
I thought about considering : $ mathbb{R} = cup (a_i , b_i] $ and $mathbb{1}((a_i , b_i])$.
But my teacher said that they are not in $L_{p}$.
real-analysis functional-analysis
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
$endgroup$
This question already has an answer here:
Separability of $l^{p}$ spaces
1 answer
It's easy to prove that $L_{p}(K)$ is separable , using Stone theorem. But how can we show the same result for real line ?
I thought about considering : $ mathbb{R} = cup (a_i , b_i] $ and $mathbb{1}((a_i , b_i])$.
But my teacher said that they are not in $L_{p}$.
This question already has an answer here:
Separability of $l^{p}$ spaces
1 answer
real-analysis functional-analysis
real-analysis functional-analysis
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
edited Dec 25 '18 at 16:49
Lord Shark the Unknown
104k1160132
104k1160132
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
asked Dec 25 '18 at 16:41
openspaceopenspace
3,4502822
3,4502822
marked as duplicate by Cesareo, user91500, mrtaurho, José Carlos Santos real-analysis
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Dec 26 '18 at 11:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Cesareo, user91500, mrtaurho, José Carlos Santos real-analysis
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Dec 26 '18 at 11:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53
add a comment |
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have more or less the right idea, but the problem is that $1_{mathbb R}$ does not lie in $L_p(mathbb R)$ because $mathbb R$ has infinite Lebesgue measure. Instead consider indicators of finite intervals $1_{[a,b]}$ with rational endpoints, to $a,b in mathbb Q.$ If $mathcal{Q} subset L_p(mathbb R)$ is the set of all such indicator functions, we claim that,
$$ V := overline{operatorname{span}} mathcal{Q} = L_p(mathbb R).$$
Indeed if this holds, we can show the set of rational linear combinations in dense, which in turn proves separability.
The key step lies in showing that $1_A in V$ for any $A subset mathbb R$ (Lebesgue) measurable with finite measure. The added condition is important, because $1_A notin L_p(mathbb R)$ is $A$ has infinite measure. To show this, for each $k in mathbb Z$ consider,
$$ mathcal{D}_k = left{ A subset [k,k+1] : A text{ measurable and} 1_A in overline{operatorname{span}}mathcal{Q_k}right},$$
where $mathcal{Q_k} subset mathcal{Q}$ contains all functions in $mathcal{Q}$ supported in $[k,k+1].$ One can show using the Dynkin $lambda-pi$ lemma that each $mathcal{D}_k$ contains all measurable subsets of $[k,k+1].$ Hence by monotone convergence we get $1_A in V$ for all $A subset mathbb R$ measurable with finite measure.
To conclude the argument, pick $f in L_p(mathbb R)$ and write $f = f_+ - f_-,$ where $f_+ = max{f,0}$ and $f_- = max{-f,0}.$ It suffices to show we can approximate both in $L_p$ by elements in $V,$ so wlog assume $f geq 0$ almost everywhere. Then we can find a sequence of simple functions $f_n$ such that $f_n rightarrow f$ pointwise monotonically. By monotone convergence, $f_n rightarrow f$ in $L_p(mathbb R)$ and hence $f in V.$
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
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add a comment |
$begingroup$
Hint: Show the set
$$bigcup_{n=1}^infty{Pchi_{[-n,n]}: P text { is a polynomial with rational coefficients}}$$
is dense in $L^p(mathbb R).$
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
$endgroup$
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have more or less the right idea, but the problem is that $1_{mathbb R}$ does not lie in $L_p(mathbb R)$ because $mathbb R$ has infinite Lebesgue measure. Instead consider indicators of finite intervals $1_{[a,b]}$ with rational endpoints, to $a,b in mathbb Q.$ If $mathcal{Q} subset L_p(mathbb R)$ is the set of all such indicator functions, we claim that,
$$ V := overline{operatorname{span}} mathcal{Q} = L_p(mathbb R).$$
Indeed if this holds, we can show the set of rational linear combinations in dense, which in turn proves separability.
The key step lies in showing that $1_A in V$ for any $A subset mathbb R$ (Lebesgue) measurable with finite measure. The added condition is important, because $1_A notin L_p(mathbb R)$ is $A$ has infinite measure. To show this, for each $k in mathbb Z$ consider,
$$ mathcal{D}_k = left{ A subset [k,k+1] : A text{ measurable and} 1_A in overline{operatorname{span}}mathcal{Q_k}right},$$
where $mathcal{Q_k} subset mathcal{Q}$ contains all functions in $mathcal{Q}$ supported in $[k,k+1].$ One can show using the Dynkin $lambda-pi$ lemma that each $mathcal{D}_k$ contains all measurable subsets of $[k,k+1].$ Hence by monotone convergence we get $1_A in V$ for all $A subset mathbb R$ measurable with finite measure.
To conclude the argument, pick $f in L_p(mathbb R)$ and write $f = f_+ - f_-,$ where $f_+ = max{f,0}$ and $f_- = max{-f,0}.$ It suffices to show we can approximate both in $L_p$ by elements in $V,$ so wlog assume $f geq 0$ almost everywhere. Then we can find a sequence of simple functions $f_n$ such that $f_n rightarrow f$ pointwise monotonically. By monotone convergence, $f_n rightarrow f$ in $L_p(mathbb R)$ and hence $f in V.$
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
You have more or less the right idea, but the problem is that $1_{mathbb R}$ does not lie in $L_p(mathbb R)$ because $mathbb R$ has infinite Lebesgue measure. Instead consider indicators of finite intervals $1_{[a,b]}$ with rational endpoints, to $a,b in mathbb Q.$ If $mathcal{Q} subset L_p(mathbb R)$ is the set of all such indicator functions, we claim that,
$$ V := overline{operatorname{span}} mathcal{Q} = L_p(mathbb R).$$
Indeed if this holds, we can show the set of rational linear combinations in dense, which in turn proves separability.
The key step lies in showing that $1_A in V$ for any $A subset mathbb R$ (Lebesgue) measurable with finite measure. The added condition is important, because $1_A notin L_p(mathbb R)$ is $A$ has infinite measure. To show this, for each $k in mathbb Z$ consider,
$$ mathcal{D}_k = left{ A subset [k,k+1] : A text{ measurable and} 1_A in overline{operatorname{span}}mathcal{Q_k}right},$$
where $mathcal{Q_k} subset mathcal{Q}$ contains all functions in $mathcal{Q}$ supported in $[k,k+1].$ One can show using the Dynkin $lambda-pi$ lemma that each $mathcal{D}_k$ contains all measurable subsets of $[k,k+1].$ Hence by monotone convergence we get $1_A in V$ for all $A subset mathbb R$ measurable with finite measure.
To conclude the argument, pick $f in L_p(mathbb R)$ and write $f = f_+ - f_-,$ where $f_+ = max{f,0}$ and $f_- = max{-f,0}.$ It suffices to show we can approximate both in $L_p$ by elements in $V,$ so wlog assume $f geq 0$ almost everywhere. Then we can find a sequence of simple functions $f_n$ such that $f_n rightarrow f$ pointwise monotonically. By monotone convergence, $f_n rightarrow f$ in $L_p(mathbb R)$ and hence $f in V.$
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
You have more or less the right idea, but the problem is that $1_{mathbb R}$ does not lie in $L_p(mathbb R)$ because $mathbb R$ has infinite Lebesgue measure. Instead consider indicators of finite intervals $1_{[a,b]}$ with rational endpoints, to $a,b in mathbb Q.$ If $mathcal{Q} subset L_p(mathbb R)$ is the set of all such indicator functions, we claim that,
$$ V := overline{operatorname{span}} mathcal{Q} = L_p(mathbb R).$$
Indeed if this holds, we can show the set of rational linear combinations in dense, which in turn proves separability.
The key step lies in showing that $1_A in V$ for any $A subset mathbb R$ (Lebesgue) measurable with finite measure. The added condition is important, because $1_A notin L_p(mathbb R)$ is $A$ has infinite measure. To show this, for each $k in mathbb Z$ consider,
$$ mathcal{D}_k = left{ A subset [k,k+1] : A text{ measurable and} 1_A in overline{operatorname{span}}mathcal{Q_k}right},$$
where $mathcal{Q_k} subset mathcal{Q}$ contains all functions in $mathcal{Q}$ supported in $[k,k+1].$ One can show using the Dynkin $lambda-pi$ lemma that each $mathcal{D}_k$ contains all measurable subsets of $[k,k+1].$ Hence by monotone convergence we get $1_A in V$ for all $A subset mathbb R$ measurable with finite measure.
To conclude the argument, pick $f in L_p(mathbb R)$ and write $f = f_+ - f_-,$ where $f_+ = max{f,0}$ and $f_- = max{-f,0}.$ It suffices to show we can approximate both in $L_p$ by elements in $V,$ so wlog assume $f geq 0$ almost everywhere. Then we can find a sequence of simple functions $f_n$ such that $f_n rightarrow f$ pointwise monotonically. By monotone convergence, $f_n rightarrow f$ in $L_p(mathbb R)$ and hence $f in V.$
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
$endgroup$
You have more or less the right idea, but the problem is that $1_{mathbb R}$ does not lie in $L_p(mathbb R)$ because $mathbb R$ has infinite Lebesgue measure. Instead consider indicators of finite intervals $1_{[a,b]}$ with rational endpoints, to $a,b in mathbb Q.$ If $mathcal{Q} subset L_p(mathbb R)$ is the set of all such indicator functions, we claim that,
$$ V := overline{operatorname{span}} mathcal{Q} = L_p(mathbb R).$$
Indeed if this holds, we can show the set of rational linear combinations in dense, which in turn proves separability.
The key step lies in showing that $1_A in V$ for any $A subset mathbb R$ (Lebesgue) measurable with finite measure. The added condition is important, because $1_A notin L_p(mathbb R)$ is $A$ has infinite measure. To show this, for each $k in mathbb Z$ consider,
$$ mathcal{D}_k = left{ A subset [k,k+1] : A text{ measurable and} 1_A in overline{operatorname{span}}mathcal{Q_k}right},$$
where $mathcal{Q_k} subset mathcal{Q}$ contains all functions in $mathcal{Q}$ supported in $[k,k+1].$ One can show using the Dynkin $lambda-pi$ lemma that each $mathcal{D}_k$ contains all measurable subsets of $[k,k+1].$ Hence by monotone convergence we get $1_A in V$ for all $A subset mathbb R$ measurable with finite measure.
To conclude the argument, pick $f in L_p(mathbb R)$ and write $f = f_+ - f_-,$ where $f_+ = max{f,0}$ and $f_- = max{-f,0}.$ It suffices to show we can approximate both in $L_p$ by elements in $V,$ so wlog assume $f geq 0$ almost everywhere. Then we can find a sequence of simple functions $f_n$ such that $f_n rightarrow f$ pointwise monotonically. By monotone convergence, $f_n rightarrow f$ in $L_p(mathbb R)$ and hence $f in V.$
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
answered Dec 25 '18 at 17:00
ktoiktoi
2,4061617
2,4061617
add a comment |
add a comment |
$begingroup$
Hint: Show the set
$$bigcup_{n=1}^infty{Pchi_{[-n,n]}: P text { is a polynomial with rational coefficients}}$$
is dense in $L^p(mathbb R).$
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
$endgroup$
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
add a comment |
$begingroup$
Hint: Show the set
$$bigcup_{n=1}^infty{Pchi_{[-n,n]}: P text { is a polynomial with rational coefficients}}$$
is dense in $L^p(mathbb R).$
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
$endgroup$
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
add a comment |
$begingroup$
Hint: Show the set
$$bigcup_{n=1}^infty{Pchi_{[-n,n]}: P text { is a polynomial with rational coefficients}}$$
is dense in $L^p(mathbb R).$
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
$endgroup$
Hint: Show the set
$$bigcup_{n=1}^infty{Pchi_{[-n,n]}: P text { is a polynomial with rational coefficients}}$$
is dense in $L^p(mathbb R).$
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
answered Dec 25 '18 at 17:40
zhw.zhw.
72.9k43175
72.9k43175
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
add a comment |
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
$P chi [-n,n]$ is polynomial with support on $[-n,n]$?
$endgroup$
– openspace
Dec 25 '18 at 18:35
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
$begingroup$
It equals $P$ in $[-n,n]$ and $0$ elsewhere.
$endgroup$
– zhw.
Dec 25 '18 at 18:36
add a comment |
$begingroup$
This is not a duplicate of the problem at the given link, which is about $l^p.$
$endgroup$
– zhw.
Dec 26 '18 at 17:53