Conditions for the differentiation of Fourier Transform of a function.












2












$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12
















2












$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12














2












2








2





$begingroup$


Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform







share|cite|improve this question











$endgroup$




Suppose $varphi(t)$ is smooth on $mathbb{R}$ and $f(x)=displaystyle int_{-infty}^{infty} varphi(t) e^{-2pi ixt}dt$ is the Fourier Transform of $varphi(t)$. Then Inverse Fourier Transform is given by:
begin{equation}
varphi(t)=displaystyle int_{-infty}^{infty} f(x) e^{-2pi ixt}dx
end{equation}
My question is what should be the conditions on $f(x)$ so that I can differentiate $varphi(t)$?
Can I differentiate to get:
begin{equation*}
varphi'(t)=(-2pi i)displaystyle int_{-infty}^{infty} xf(x) e^{-2pi ixt}dx
end{equation*}






real-analysis fourier-transform




real-analysis fourier-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 0:01







ersh

















asked Dec 18 '18 at 23:48









ershersh

389113




389113








  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12














  • 1




    $begingroup$
    Integrability of $xf(x)$ is a sufficient condition.
    $endgroup$
    – Kavi Rama Murthy
    Dec 18 '18 at 23:55










  • $begingroup$
    @KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
    $endgroup$
    – ersh
    Dec 19 '18 at 0:04






  • 2




    $begingroup$
    You are right and you get these by an application of DCT.
    $endgroup$
    – Kavi Rama Murthy
    Dec 19 '18 at 0:07






  • 1




    $begingroup$
    The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
    $endgroup$
    – reuns
    Dec 19 '18 at 8:12








1




1




$begingroup$
Integrability of $xf(x)$ is a sufficient condition.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:55




$begingroup$
Integrability of $xf(x)$ is a sufficient condition.
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:55












$begingroup$
@KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
$endgroup$
– ersh
Dec 19 '18 at 0:04




$begingroup$
@KaviRamaMurthy If I have to differentiate again then integrability of $x^2f(x)$ would be sufficient and son, right? What would be the good reference related to it?
$endgroup$
– ersh
Dec 19 '18 at 0:04




2




2




$begingroup$
You are right and you get these by an application of DCT.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 0:07




$begingroup$
You are right and you get these by an application of DCT.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 0:07




1




1




$begingroup$
The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
$endgroup$
– reuns
Dec 19 '18 at 8:12




$begingroup$
The sufficient condition is that $lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ converges uniformly for $t in U$, so that $int_b^c lim_{A,B to infty} int_{-A}^B (-2i pi x) f(x) e^{-2i pi xt}dxdt=lim_{A,B to infty} int_{-A}^B f(x) (int_b^c (-2i pi x) e^{-2i pi xt}dt)dx$ $=lim_{A,B to infty} int_{-A}^B f(x) ( e^{-2i pi cx}- e^{-2i pi bx})dx=varphi(c)-varphi(b)$ which implies that $varphi'(c) = lim_{A,B to infty} int_{-A}^B x f(x) e^{-2i pi xt}dx$ which is then continuous
$endgroup$
– reuns
Dec 19 '18 at 8:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045851%2fconditions-for-the-differentiation-of-fourier-transform-of-a-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28


















1












$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28
















1












1








1





$begingroup$

This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$






share|cite|improve this answer









$endgroup$



This is a fundamental theorem in the Lebesgue Theory of Integration :



Let ${displaystyle X}$ be an open subset of ${displaystyle mathbf {R} }$, and ${displaystyle Omega }$ be a measure space. Suppose ${displaystyle fcolon Xtimes Omega rightarrow mathbf {R} }$ satisfies the following conditions:



${displaystyle f(x,omega )}$ is a Lebesgue-integrable function of ${displaystyle omega }$ for each ${displaystyle xin X}$.
For almost all ${displaystyle omega in Omega }$, the derivative ${displaystyle f_{x}}$ exists for all ${displaystyle xin X}$.
There is an integrable function ${displaystyle theta colon Omega rightarrow mathbf {R} }$ such that ${displaystyle |f_{x}(x,omega )|leq theta (omega )}$ for all ${displaystyle xin X}$ and almost every ${displaystyle omega in Omega }$.
Then by the Dominated convergence theorem for all ${displaystyle xin X}$,
${displaystyle {frac {d}{dx}}int _{Omega }f(x,omega ),domega =int _{Omega }f_{x}(x,omega ),domega .}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 0:11









MalikMalik

1018




1018












  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28




















  • $begingroup$
    Can you please provide me reference to this result?
    $endgroup$
    – ersh
    Dec 19 '18 at 1:41






  • 1




    $begingroup$
    Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
    $endgroup$
    – Malik
    Dec 19 '18 at 2:28


















$begingroup$
Can you please provide me reference to this result?
$endgroup$
– ersh
Dec 19 '18 at 1:41




$begingroup$
Can you please provide me reference to this result?
$endgroup$
– ersh
Dec 19 '18 at 1:41




1




1




$begingroup$
Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
$endgroup$
– Malik
Dec 19 '18 at 2:28






$begingroup$
Any measure theory book like Follan : Real analysis and modern techniques, or simply visit wikipedia : en.wikipedia.org/wiki/Leibniz_integral_rule, by scrolling down you'll see the measure theoretic statement
$endgroup$
– Malik
Dec 19 '18 at 2:28




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3045851%2fconditions-for-the-differentiation-of-fourier-transform-of-a-function%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
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
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
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
Read more