Example of non-locally integrable $f$ that $int_Omega fvarphi=0$.
$begingroup$
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
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
$endgroup$
|
show 7 more comments
$begingroup$
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
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
$endgroup$
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
|
show 7 more comments
$begingroup$
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
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
$endgroup$
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
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
edited Dec 25 '18 at 6:45
Martin Sleziak
44.8k10118272
44.8k10118272
asked Dec 24 '18 at 23:48
P.RP.R
184
asked Dec 24 '18 at 23:48
P.RP.R
184
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
|
show 7 more comments
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
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
+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
add a comment |
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$begingroup$
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.
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
+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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
$begingroup$
+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
add a comment |
$begingroup$
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.
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
$endgroup$
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.
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
edited Dec 25 '18 at 2:03
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
answered Dec 25 '18 at 1:30
SmileyCraftSmileyCraft
3,591517
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
add a comment |
$begingroup$
+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
add a comment |
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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
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@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 $fvarphi in L^1(Omega)$?
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– mathworker21
Dec 25 '18 at 0:11
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@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.
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– mathworker21
Dec 25 '18 at 1:43
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@mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/…
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– SmileyCraft
Dec 25 '18 at 2:31