Which convergence when discussing density in $L^p$?
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For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?
measure-theory lp-spaces
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
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add a comment |
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For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?
measure-theory lp-spaces
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
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2
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$L^p$ convergence
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– mathworker21
Dec 28 '18 at 21:13
2
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When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
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– Jakobian
Dec 28 '18 at 21:16
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All right, thanks a lot to both.
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– Bermudes
Dec 28 '18 at 21:27
add a comment |
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For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?
measure-theory lp-spaces
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
$endgroup$
For an exercise in my measure theory course, I need to prove some density results in the $L^p(mathbb R^n)$ space. But density requires convergence and this gets me confused. So when discussing density of a subset (e.g. the measurable simple functions) in $L^p$, what convergence are we talking about? Is it pointwise convergence or $L^p$ convergence?
measure-theory lp-spaces
measure-theory lp-spaces
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
asked Dec 28 '18 at 21:11
BermudesBermudes
309214
309214
2
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$L^p$ convergence
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– mathworker21
Dec 28 '18 at 21:13
2
$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16
$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27
add a comment |
2
$begingroup$
$L^p$ convergence
$endgroup$
– mathworker21
Dec 28 '18 at 21:13
2
$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16
$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27
2
2
$begingroup$
$L^p$ convergence
$endgroup$
– mathworker21
Dec 28 '18 at 21:13
$begingroup$
$L^p$ convergence
$endgroup$
– mathworker21
Dec 28 '18 at 21:13
2
2
$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16
$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16
$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27
$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27
add a comment |
1 Answer
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A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
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1 Answer
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$begingroup$
A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
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add a comment |
$begingroup$
A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
A subset $S$ of $mathbb L^pleft(X,mathcal F,muright)$ is dense in $mathbb L^pleft(X,mathcal F,muright)$ if for each element $f$ of $mathbb L^pleft(X,mathcal F,muright)$ and each positive $varepsilon$, there exists an element $g$ of $S$ such that $leftlVert f-grightrVert_p:=left(int_Xleftlvert f(x)-g(x)rightrvert^pmathrm dmu(x)right)^{1/p}lt varepsilon$, or equivalently, there exists a sequence $left(g_nright)_{ngeqslant 1}$ of elements of $S$ such that $lim_{nto +infty}leftlVert f-g_nrightrVert_p=0$.
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 21 at 10:07
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
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2
$begingroup$
$L^p$ convergence
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– mathworker21
Dec 28 '18 at 21:13
2
$begingroup$
When we say $L^p$ space, we mean $L^p$ set with $L^p$ norm.
$endgroup$
– Jakobian
Dec 28 '18 at 21:16
$begingroup$
All right, thanks a lot to both.
$endgroup$
– Bermudes
Dec 28 '18 at 21:27