Non-unique Hahn Banach extension for a functional defined on a closed subspace of $l^1$












3












$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43


















3












$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43
















3












3








3





$begingroup$


I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems







share|cite|improve this question









$endgroup$




I'm trying to solve the following exercise, but I'm stuck on the possible conclusion.




Let $Y={ {x_k}_k in l^1: x_{2015} = 0 } subset l^1, quad T in Y'$. Characterize all the linear and bounded extensions $Phi in (l^1)'$ which preserves the dual norm.




I've tried so far to use the duality and write $Phi: l^1 rightarrow mathbb{R}$ as $Phi(x)= sum_{k=1}^{infty} x_k c_k$ where ${c_k } in l^{infty}$, but I don't know how to get equality of dual norms.






sequences-and-series functional-analysis duality-theorems




sequences-and-series functional-analysis duality-theorems






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 16:27









VoBVoB

734413




734413












  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43




















  • $begingroup$
    Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 17:15












  • $begingroup$
    I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
    $endgroup$
    – VoB
    Dec 19 '18 at 17:43










  • $begingroup$
    So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
    $endgroup$
    – VoB
    Dec 19 '18 at 17:45












  • $begingroup$
    You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
    $endgroup$
    – SmileyCraft
    Dec 19 '18 at 18:22










  • $begingroup$
    Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
    $endgroup$
    – VoB
    Dec 19 '18 at 18:43


















$begingroup$
Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:15






$begingroup$
Maybe first try a finite dimensional case: $Y={(x,0):xinmathbb{R}}subsetmathbb{R}^2$ where the norm on $mathbb{R}^2$ and $Y$ are $|(x,y)|_1=|x|+|y|$. Try to solve this first.
$endgroup$
– SmileyCraft
Dec 19 '18 at 17:15














$begingroup$
I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
$endgroup$
– VoB
Dec 19 '18 at 17:43




$begingroup$
I found that $Phi(e_2) < T(e_1)$, where $e_i, quad i=1,2$ are the -th element of the canonical base
$endgroup$
– VoB
Dec 19 '18 at 17:43












$begingroup$
So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
$endgroup$
– VoB
Dec 19 '18 at 17:45






$begingroup$
So, applying the same argument to the above exercise I would get that $Phi(e_{2015}) leq sum_{k ne 2015} T(e_k)$. Could it be right? Is this the right characterization
$endgroup$
– VoB
Dec 19 '18 at 17:45














$begingroup$
You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
$endgroup$
– SmileyCraft
Dec 19 '18 at 18:22




$begingroup$
You're getting somewhere. Maybe also try $mathbb{R}^3$ and you will see the correct pattern.
$endgroup$
– SmileyCraft
Dec 19 '18 at 18:22












$begingroup$
Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
$endgroup$
– VoB
Dec 19 '18 at 18:43






$begingroup$
Well, it seems I have to put $Phi(e_{2015}) leq sum_{k ne 2015} |T(e_k)| $
$endgroup$
– VoB
Dec 19 '18 at 18:43












1 Answer
1






active

oldest

votes


















1












$begingroup$

Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
$$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
$$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



Now we have
$$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



so $|Phi e_{2015}| le |T|$.



We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






share|cite|improve this answer









$endgroup$













    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%2f3046581%2fnon-unique-hahn-banach-extension-for-a-functional-defined-on-a-closed-subspace-o%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












    $begingroup$

    Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
    $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
    so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
    $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
    so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



    Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



    Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



    Now we have
    $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



    so $|Phi e_{2015}| le |T|$.



    We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
      $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
      so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
      $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
      so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



      Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



      Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



      Now we have
      $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



      so $|Phi e_{2015}| le |T|$.



      We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
        $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
        so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
        $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
        so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



        Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



        Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



        Now we have
        $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



        so $|Phi e_{2015}| le |T|$.



        We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.






        share|cite|improve this answer









        $endgroup$



        Notice that $|T| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Indeed, if $T$ is bounded then in particular $(Te_n)_n$ is a bounded sequence in the field and for $x = sum_{substack{n=1\ nne 2015}}^infty x_ne_n in Y$ we have
        $$|Tx| = left|sum_{substack{n=1\ nne 2015}}^infty x_n Te_nright| le sum_{substack{n=1\ nne 2015}}^infty |x_n||Te_n| le left(sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|right) |x|_1$$
        so $|T| le sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n|$. Conversely, for all $n in mathbb{N}, n ne 2015$ we have
        $$|T| ge frac{|Te_n|}{|e_n|_1} = |Te_n|$$
        so $|T| ge sup_{ninmathbb{N}, nne 2015} |Te_n|$.



        Similarly (or using the $(ell^1)' leftrightarrow ell^infty$ duality) we get that $|Phi| = sup_{ninmathbb{N}} |Phi e_n|$.



        Now, since $Phi$ extends $T$, for every $n in mathbb{N}, n ne 2015$ it follows that $Te_n = Phi e_n$.



        Now we have
        $$sup_{substack{ninmathbb{N}\ nne 2015}} |Phi e_n| = sup_{substack{ninmathbb{N}\ nne 2015}} |Te_n| = |T| = |Phi| = sup_{ninmathbb{N}} |Phi e_n|$$



        so $|Phi e_{2015}| le |T|$.



        We also see that $Phi e_{n} = Te_n, forall n ne 2015$ and $|Phi e_{2015}| le |T|$ is a sufficient condition that $Phi in (ell^1)'$ is a Hahn-Banach extension of $T$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 20:53









        mechanodroidmechanodroid

        27.6k62447




        27.6k62447






























            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%2f3046581%2fnon-unique-hahn-banach-extension-for-a-functional-defined-on-a-closed-subspace-o%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

            Ellipse (mathématiques)

            Image
            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
            Read more

            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

            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