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

44.7k10119272




44.7k10119272










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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1 Answer
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active

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active

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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

3,601517




3,601517












  • $begingroup$
    +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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Ellipse (mathématiques)

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Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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