Closed Subset of ${ L }^{ p }$












2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















2












2








2





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 22:20









Davide Giraudo

126k16150261




126k16150261










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
















  • 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












2 Answers
2






active

oldest

votes


















2












$begingroup$

$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ 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.






share|cite|improve this answer











$endgroup$



$d_p(f,g)=|f-g|_{L^p}=left(int_X |f-g|^p,dmuright)^{1/p}$ 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 '18 at 19:57

























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


















  • $begingroup$
    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











2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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






$begingroup$
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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