Closed Subset of ${ L }^{ p }$
$begingroup$
I'd like to show that the set
$$
S:=left{ f:Xrightarrow R:quad fquad mu -mathrm{measurable},quad int_{ X } { f }^{ 2 }dmu le 1 right}
$$
is closed in ${ L }^{ p }(dmu )$, where $X$ is an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
First of all is it correct that this given set is the closed unit ball in ${ L }^{ 2 }$? I tried going with the standard definition and tried to show that our set contains all its limits points but I am having struggels to define arbitrary sequences in this space. Could someone offer a hint.
real-analysis general-topology measure-theory lp-spaces
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
$endgroup$
add a comment |
$begingroup$
I'd like to show that the set
$$
S:=left{ f:Xrightarrow R:quad fquad mu -mathrm{measurable},quad int_{ X } { f }^{ 2 }dmu le 1 right}
$$
is closed in ${ L }^{ p }(dmu )$, where $X$ is an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
First of all is it correct that this given set is the closed unit ball in ${ L }^{ 2 }$? I tried going with the standard definition and tried to show that our set contains all its limits points but I am having struggels to define arbitrary sequences in this space. Could someone offer a hint.
real-analysis general-topology measure-theory lp-spaces
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
$endgroup$
1
$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
$endgroup$
– mathworker21
Nov 29 '18 at 19:09
$begingroup$
Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
$endgroup$
– Giuseppe Negro
Nov 29 '18 at 19:32
$begingroup$
Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 19:50
add a comment |
$begingroup$
I'd like to show that the set
$$
S:=left{ f:Xrightarrow R:quad fquad mu -mathrm{measurable},quad int_{ X } { f }^{ 2 }dmu le 1 right}
$$
is closed in ${ L }^{ p }(dmu )$, where $X$ is an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
First of all is it correct that this given set is the closed unit ball in ${ L }^{ 2 }$? I tried going with the standard definition and tried to show that our set contains all its limits points but I am having struggels to define arbitrary sequences in this space. Could someone offer a hint.
real-analysis general-topology measure-theory lp-spaces
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
$endgroup$
I'd like to show that the set
$$
S:=left{ f:Xrightarrow R:quad fquad mu -mathrm{measurable},quad int_{ X } { f }^{ 2 }dmu le 1 right}
$$
is closed in ${ L }^{ p }(dmu )$, where $X$ is an arbitrary set and $mu$ a finite measure on the sigma algebra $Sigma subset { 2 }^{ X }$.
First of all is it correct that this given set is the closed unit ball in ${ L }^{ 2 }$? I tried going with the standard definition and tried to show that our set contains all its limits points but I am having struggels to define arbitrary sequences in this space. Could someone offer a hint.
real-analysis general-topology measure-theory lp-spaces
real-analysis general-topology measure-theory lp-spaces
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
edited Nov 29 '18 at 22:20
Davide Giraudo
126k16150261
126k16150261
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
asked Nov 29 '18 at 19:07
MasterPIMasterPI
21818
21818
1
$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
$endgroup$
– mathworker21
Nov 29 '18 at 19:09
$begingroup$
Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
$endgroup$
– Giuseppe Negro
Nov 29 '18 at 19:32
$begingroup$
Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 19:50
add a comment |
1
$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
$endgroup$
– mathworker21
Nov 29 '18 at 19:09
$begingroup$
Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
$endgroup$
– Giuseppe Negro
Nov 29 '18 at 19:32
$begingroup$
Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 19:50
1
1
$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
$endgroup$
– mathworker21
Nov 29 '18 at 19:09
$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
$endgroup$
– mathworker21
Nov 29 '18 at 19:09
$begingroup$
Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
$endgroup$
– Giuseppe Negro
Nov 29 '18 at 19:32
$begingroup$
Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
$endgroup$
– Giuseppe Negro
Nov 29 '18 at 19:32
$begingroup$
Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 19:50
$begingroup$
Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 19:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ is a distance on $L^p(X,mu)$; this follows from the Minkowski inequality.
Your set is $B(0,1)={fin L^p: d_p(f,0)leq 1}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.
Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)={xin X:d(x,x_0)leq r}$ is closed in the induced topology.
Proof. Take $xnotin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<delta<d(x,x_0)-r$. Then I claim that $B(x,delta)subset D(x_0,r)^c$. Indeed, if $yin B(x,delta)$, then
$$
d(y,x_0)geq d(x,x_0)-d(y,x)> d(x,x_0)-delta > r.
$$
This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
$endgroup$
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
|
show 5 more comments
$begingroup$
Let $left(f_nright)_{ngeqslant 1}$ be a sequence in the involved set with converges in $mathbb L^p$ to some $f$. Extract a subsequence $left(f_{n_k}right)_{kgeqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain
$$
int_Xf^2mathrm dmu=int_Xliminf_{kto +infty} f_{n_k}^2mathrm dmuleqslant
liminf_{kto +infty} int_X f_{n_k}^2mathrm dmu leqslant 1$$
hence $f$ is also an element of the consider set.
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ is a distance on $L^p(X,mu)$; this follows from the Minkowski inequality.
Your set is $B(0,1)={fin L^p: d_p(f,0)leq 1}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.
Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)={xin X:d(x,x_0)leq r}$ is closed in the induced topology.
Proof. Take $xnotin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<delta<d(x,x_0)-r$. Then I claim that $B(x,delta)subset D(x_0,r)^c$. Indeed, if $yin B(x,delta)$, then
$$
d(y,x_0)geq d(x,x_0)-d(y,x)> d(x,x_0)-delta > r.
$$
This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
$endgroup$
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
|
show 5 more comments
$begingroup$
$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ is a distance on $L^p(X,mu)$; this follows from the Minkowski inequality.
Your set is $B(0,1)={fin L^p: d_p(f,0)leq 1}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.
Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)={xin X:d(x,x_0)leq r}$ is closed in the induced topology.
Proof. Take $xnotin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<delta<d(x,x_0)-r$. Then I claim that $B(x,delta)subset D(x_0,r)^c$. Indeed, if $yin B(x,delta)$, then
$$
d(y,x_0)geq d(x,x_0)-d(y,x)> d(x,x_0)-delta > r.
$$
This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
$endgroup$
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
|
show 5 more comments
$begingroup$
$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ is a distance on $L^p(X,mu)$; this follows from the Minkowski inequality.
Your set is $B(0,1)={fin L^p: d_p(f,0)leq 1}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.
Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)={xin X:d(x,x_0)leq r}$ is closed in the induced topology.
Proof. Take $xnotin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<delta<d(x,x_0)-r$. Then I claim that $B(x,delta)subset D(x_0,r)^c$. Indeed, if $yin B(x,delta)$, then
$$
d(y,x_0)geq d(x,x_0)-d(y,x)> d(x,x_0)-delta > r.
$$
This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
$endgroup$
$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ is a distance on $L^p(X,mu)$; this follows from the Minkowski inequality.
Your set is $B(0,1)={fin L^p: d_p(f,0)leq 1}$ and it's a general fact that closed balls are, indeed, closed. This is just an immediate consequence of the definition of a topology induced by a metric.
Fact. In a metric space $(X,d)$, any "closed ball" $D(x_0,r)={xin X:d(x,x_0)leq r}$ is closed in the induced topology.
Proof. Take $xnotin D(x_0,r)$. Then $d(x,x_0)>r$. Take $0<delta<d(x,x_0)-r$. Then I claim that $B(x,delta)subset D(x_0,r)^c$. Indeed, if $yin B(x,delta)$, then
$$
d(y,x_0)geq d(x,x_0)-d(y,x)> d(x,x_0)-delta > r.
$$
This means that $D(x_0,r)^c$ is open, which is equivalent to the thesis.
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
edited Nov 29 '18 at 19:57
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
answered Nov 29 '18 at 19:26
FedericoFederico
5,034514
5,034514
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
|
show 5 more comments
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
I don't think it is just by definition. You have to prove it (but it is very easy to prove it).
$endgroup$
– mathworker21
Nov 29 '18 at 19:34
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
@mathworker21 Well, yeah... but still... I rephrased it however
$endgroup$
– Federico
Nov 29 '18 at 19:35
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
$begingroup$
It doesn't make a difference to people like me and you, who know the subject, but it might make a difference to a beginner.
$endgroup$
– mathworker21
Nov 29 '18 at 19:43
1
1
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
@mathworker21 happy now? :)
$endgroup$
– Federico
Nov 29 '18 at 19:57
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
$begingroup$
Is it even possible to use the Minkowski inequality when going from ${L}^2$ to ${L}^p$?? (sorry maybe a dump question but im not very familiar with this topic)
$endgroup$
– MasterPI
Nov 29 '18 at 19:58
|
show 5 more comments
$begingroup$
Let $left(f_nright)_{ngeqslant 1}$ be a sequence in the involved set with converges in $mathbb L^p$ to some $f$. Extract a subsequence $left(f_{n_k}right)_{kgeqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain
$$
int_Xf^2mathrm dmu=int_Xliminf_{kto +infty} f_{n_k}^2mathrm dmuleqslant
liminf_{kto +infty} int_X f_{n_k}^2mathrm dmu leqslant 1$$
hence $f$ is also an element of the consider set.
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
add a comment |
$begingroup$
Let $left(f_nright)_{ngeqslant 1}$ be a sequence in the involved set with converges in $mathbb L^p$ to some $f$. Extract a subsequence $left(f_{n_k}right)_{kgeqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain
$$
int_Xf^2mathrm dmu=int_Xliminf_{kto +infty} f_{n_k}^2mathrm dmuleqslant
liminf_{kto +infty} int_X f_{n_k}^2mathrm dmu leqslant 1$$
hence $f$ is also an element of the consider set.
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
add a comment |
$begingroup$
Let $left(f_nright)_{ngeqslant 1}$ be a sequence in the involved set with converges in $mathbb L^p$ to some $f$. Extract a subsequence $left(f_{n_k}right)_{kgeqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain
$$
int_Xf^2mathrm dmu=int_Xliminf_{kto +infty} f_{n_k}^2mathrm dmuleqslant
liminf_{kto +infty} int_X f_{n_k}^2mathrm dmu leqslant 1$$
hence $f$ is also an element of the consider set.
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
Let $left(f_nright)_{ngeqslant 1}$ be a sequence in the involved set with converges in $mathbb L^p$ to some $f$. Extract a subsequence $left(f_{n_k}right)_{kgeqslant 1}$ which converges almost everywhere to $f$. Using Fatou's lemma, we obtain
$$
int_Xf^2mathrm dmu=int_Xliminf_{kto +infty} f_{n_k}^2mathrm dmuleqslant
liminf_{kto +infty} int_X f_{n_k}^2mathrm dmu leqslant 1$$
hence $f$ is also an element of the consider set.
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
answered Nov 29 '18 at 21:08
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
add a comment |
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
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– MasterPI
Nov 29 '18 at 22:44
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
$begingroup$
it almost seems to easy to be true, a question that came in to my mind was if it is possible for a sequence to converge in our set or generally ${L}^2$ but not in ${L}^p$
$endgroup$
– MasterPI
Nov 29 '18 at 22:44
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$begingroup$
you have $2$ instead of $p$. Also, it's obviously true by the sequential version of closed-ness.
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– mathworker21
Nov 29 '18 at 19:09
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Are you sure about the 2? If it's $p$, then the answer is trivial. If it is $2$ it is interesting.
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– Giuseppe Negro
Nov 29 '18 at 19:32
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Thank for all your answers. Yes i'm sure about the squared Integral. So the statement is that the Closed unit ball of ${L}^2$ is closed in ${L}^p$
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– MasterPI
Nov 29 '18 at 19:50