Example of non-locally integrable $f$ that $int_Omega fvarphi=0$.












0












$begingroup$


In real analysis we know that if $fin L_{loc}^1(Omega)$ and $int_Omega fvarphi=0$ for any $varphiin C_0^infty(Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $fin L_{loc}^1(Omega)$ is necessary.






real-analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 23:58










  • $begingroup$
    I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
    $endgroup$
    – Math1000
    Dec 25 '18 at 0:03








  • 1




    $begingroup$
    @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 0:11






  • 1




    $begingroup$
    @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:43








  • 1




    $begingroup$
    @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:31
















0












$begingroup$


In real analysis we know that if $fin L_{loc}^1(Omega)$ and $int_Omega fvarphi=0$ for any $varphiin C_0^infty(Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $fin L_{loc}^1(Omega)$ is necessary.






real-analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 23:58










  • $begingroup$
    I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
    $endgroup$
    – Math1000
    Dec 25 '18 at 0:03








  • 1




    $begingroup$
    @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 0:11






  • 1




    $begingroup$
    @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:43








  • 1




    $begingroup$
    @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:31














0












0








0





$begingroup$


In real analysis we know that if $fin L_{loc}^1(Omega)$ and $int_Omega fvarphi=0$ for any $varphiin C_0^infty(Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $fin L_{loc}^1(Omega)$ is necessary.






real-analysis







share|cite|improve this question











$endgroup$




In real analysis we know that if $fin L_{loc}^1(Omega)$ and $int_Omega fvarphi=0$ for any $varphiin C_0^infty(Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $fin L_{loc}^1(Omega)$ is necessary.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 6:45









Martin Sleziak

44.8k10118272




44.8k10118272










asked Dec 24 '18 at 23:48









P.RP.R

184




184








  • 4




    $begingroup$
    I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 23:58










  • $begingroup$
    I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
    $endgroup$
    – Math1000
    Dec 25 '18 at 0:03








  • 1




    $begingroup$
    @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 0:11






  • 1




    $begingroup$
    @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:43








  • 1




    $begingroup$
    @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:31














  • 4




    $begingroup$
    I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
    $endgroup$
    – SmileyCraft
    Dec 24 '18 at 23:58










  • $begingroup$
    I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
    $endgroup$
    – Math1000
    Dec 25 '18 at 0:03








  • 1




    $begingroup$
    @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
    $endgroup$
    – mathworker21
    Dec 25 '18 at 0:11






  • 1




    $begingroup$
    @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:43








  • 1




    $begingroup$
    @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:31








4




4




$begingroup$
I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
$endgroup$
– SmileyCraft
Dec 24 '18 at 23:58




$begingroup$
I believe the $fin L_{loc}^1(Omega)$ is just necessary such that $int_Omega fvarphi$ is well-defined.
$endgroup$
– SmileyCraft
Dec 24 '18 at 23:58












$begingroup$
I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
$endgroup$
– Math1000
Dec 25 '18 at 0:03






$begingroup$
I suppose here we mean $int_Omega f mathsf dvarphi$ where $varphi$ is some measure?
$endgroup$
– Math1000
Dec 25 '18 at 0:03






1




1




$begingroup$
@SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
$endgroup$
– mathworker21
Dec 25 '18 at 0:11




$begingroup$
@SmileyCraft why is that? I mean $f$ is measurable (presumably) and $varphi in C_0^infty(Omega)$. So $fvarphi$ is measurable. So we just need to know whether $fvarphi in L^1(Omega)$. And it's not clear to me that it isn't. Consider, for example, $Omega = {|z| < 1} subseteq mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $varphi$ vanishing at the boundary of $Omega$, that $fvarphi in L^1(Omega)$?
$endgroup$
– mathworker21
Dec 25 '18 at 0:11




1




1




$begingroup$
@SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
$endgroup$
– mathworker21
Dec 25 '18 at 1:43






$begingroup$
@SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want.
$endgroup$
– mathworker21
Dec 25 '18 at 1:43






1




1




$begingroup$
@mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
$endgroup$
– SmileyCraft
Dec 25 '18 at 2:31




$begingroup$
@mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
$endgroup$
– SmileyCraft
Dec 25 '18 at 2:31










1 Answer
1






active

oldest

votes


















1












$begingroup$

The condition $fin L_{loc}^1(Omega)$ is necessary for the statement to even make sense. If $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$, then $fin L_{loc}^1(Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $Omegasubseteqmathbb{R}^n$.



Proof of claim: Let $Omegasubseteqmathbb{R}^n$ open and $f:Omegatomathbb{R}$ such that $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$. Let $KsubseteqOmega$ compact. We can find (*) $UsubseteqOmega$ open and $LsubseteqOmega$ compact such that $Ksubseteq Usubseteq L$. By smooth Urysohn we find $varphiin C^infty$ such that $varphi|_{K}equiv1$ and $mbox{supp}(varphi)subseteq U$. Because $Usubseteq LsubseteqOmega$ we get $varphiin C_0^infty(Omega)$. By hypothesis, $fvarphi$ is integrable on $Omega$, and hence also on $K$. Since $(fvarphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $fin L_{loc}^1(Omega)$.



(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(Omegacap B_R(0))^c$ is closed and disjoint from $K$. Because $mathbb{R}^2$ is normal, we find disjoint open sets $U,Vsubseteqmathbb{R}^2$ such that $Ksubseteq U$ and $Ssubseteq V$. Choosing $L=V^c$ we get the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:40












  • $begingroup$
    Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:04











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1 Answer
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active

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1












$begingroup$

The condition $fin L_{loc}^1(Omega)$ is necessary for the statement to even make sense. If $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$, then $fin L_{loc}^1(Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $Omegasubseteqmathbb{R}^n$.



Proof of claim: Let $Omegasubseteqmathbb{R}^n$ open and $f:Omegatomathbb{R}$ such that $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$. Let $KsubseteqOmega$ compact. We can find (*) $UsubseteqOmega$ open and $LsubseteqOmega$ compact such that $Ksubseteq Usubseteq L$. By smooth Urysohn we find $varphiin C^infty$ such that $varphi|_{K}equiv1$ and $mbox{supp}(varphi)subseteq U$. Because $Usubseteq LsubseteqOmega$ we get $varphiin C_0^infty(Omega)$. By hypothesis, $fvarphi$ is integrable on $Omega$, and hence also on $K$. Since $(fvarphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $fin L_{loc}^1(Omega)$.



(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(Omegacap B_R(0))^c$ is closed and disjoint from $K$. Because $mathbb{R}^2$ is normal, we find disjoint open sets $U,Vsubseteqmathbb{R}^2$ such that $Ksubseteq U$ and $Ssubseteq V$. Choosing $L=V^c$ we get the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:40












  • $begingroup$
    Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:04
















1












$begingroup$

The condition $fin L_{loc}^1(Omega)$ is necessary for the statement to even make sense. If $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$, then $fin L_{loc}^1(Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $Omegasubseteqmathbb{R}^n$.



Proof of claim: Let $Omegasubseteqmathbb{R}^n$ open and $f:Omegatomathbb{R}$ such that $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$. Let $KsubseteqOmega$ compact. We can find (*) $UsubseteqOmega$ open and $LsubseteqOmega$ compact such that $Ksubseteq Usubseteq L$. By smooth Urysohn we find $varphiin C^infty$ such that $varphi|_{K}equiv1$ and $mbox{supp}(varphi)subseteq U$. Because $Usubseteq LsubseteqOmega$ we get $varphiin C_0^infty(Omega)$. By hypothesis, $fvarphi$ is integrable on $Omega$, and hence also on $K$. Since $(fvarphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $fin L_{loc}^1(Omega)$.



(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(Omegacap B_R(0))^c$ is closed and disjoint from $K$. Because $mathbb{R}^2$ is normal, we find disjoint open sets $U,Vsubseteqmathbb{R}^2$ such that $Ksubseteq U$ and $Ssubseteq V$. Choosing $L=V^c$ we get the desired properties.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    +1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:40












  • $begingroup$
    Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:04














1












1








1





$begingroup$

The condition $fin L_{loc}^1(Omega)$ is necessary for the statement to even make sense. If $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$, then $fin L_{loc}^1(Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $Omegasubseteqmathbb{R}^n$.



Proof of claim: Let $Omegasubseteqmathbb{R}^n$ open and $f:Omegatomathbb{R}$ such that $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$. Let $KsubseteqOmega$ compact. We can find (*) $UsubseteqOmega$ open and $LsubseteqOmega$ compact such that $Ksubseteq Usubseteq L$. By smooth Urysohn we find $varphiin C^infty$ such that $varphi|_{K}equiv1$ and $mbox{supp}(varphi)subseteq U$. Because $Usubseteq LsubseteqOmega$ we get $varphiin C_0^infty(Omega)$. By hypothesis, $fvarphi$ is integrable on $Omega$, and hence also on $K$. Since $(fvarphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $fin L_{loc}^1(Omega)$.



(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(Omegacap B_R(0))^c$ is closed and disjoint from $K$. Because $mathbb{R}^2$ is normal, we find disjoint open sets $U,Vsubseteqmathbb{R}^2$ such that $Ksubseteq U$ and $Ssubseteq V$. Choosing $L=V^c$ we get the desired properties.






share|cite|improve this answer











$endgroup$



The condition $fin L_{loc}^1(Omega)$ is necessary for the statement to even make sense. If $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$, then $fin L_{loc}^1(Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $Omegasubseteqmathbb{R}^n$.



Proof of claim: Let $Omegasubseteqmathbb{R}^n$ open and $f:Omegatomathbb{R}$ such that $fvarphi$ is integrable for all $varphiin C_0^infty(Omega)$. Let $KsubseteqOmega$ compact. We can find (*) $UsubseteqOmega$ open and $LsubseteqOmega$ compact such that $Ksubseteq Usubseteq L$. By smooth Urysohn we find $varphiin C^infty$ such that $varphi|_{K}equiv1$ and $mbox{supp}(varphi)subseteq U$. Because $Usubseteq LsubseteqOmega$ we get $varphiin C_0^infty(Omega)$. By hypothesis, $fvarphi$ is integrable on $Omega$, and hence also on $K$. Since $(fvarphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $fin L_{loc}^1(Omega)$.



(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(Omegacap B_R(0))^c$ is closed and disjoint from $K$. Because $mathbb{R}^2$ is normal, we find disjoint open sets $U,Vsubseteqmathbb{R}^2$ such that $Ksubseteq U$ and $Ssubseteq V$. Choosing $L=V^c$ we get the desired properties.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 25 '18 at 2:03

























answered Dec 25 '18 at 1:30









SmileyCraftSmileyCraft

3,591517




3,591517












  • $begingroup$
    +1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:40












  • $begingroup$
    Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:04


















  • $begingroup$
    +1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
    $endgroup$
    – mathworker21
    Dec 25 '18 at 1:40












  • $begingroup$
    Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
    $endgroup$
    – SmileyCraft
    Dec 25 '18 at 2:04
















$begingroup$
+1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
$endgroup$
– mathworker21
Dec 25 '18 at 1:40






$begingroup$
+1 nice! Just one (most likely stupid) question. Is it obvious that [$forall x$ there exists compact $K ni x$ s.t. $int_K |f(y)|dy < infty$] implies [for all compact $K, int_K |f(y)|dy < infty$]. The latter is what I thought the definition of $L^1_{loc}$ is.
$endgroup$
– mathworker21
Dec 25 '18 at 1:40














$begingroup$
Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
$endgroup$
– SmileyCraft
Dec 25 '18 at 2:04




$begingroup$
Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer.
$endgroup$
– SmileyCraft
Dec 25 '18 at 2:04


















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Cabo Verde

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  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
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Gyllenstierna

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