If $f$ is convex function and $f$ is continuous on $(a,b)$.When is $f$ continuous at $a$ and $b$? [closed]












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Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?






real-analysis continuity convex-analysis







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closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500

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    -1












    $begingroup$


    Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?






    real-analysis continuity convex-analysis







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1


      1



      $begingroup$


      Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?






      real-analysis continuity convex-analysis







      share|cite|improve this question











      $endgroup$




      Let $f:[a,b]to mathbb{R}$. We know that if $f$ is convex function then $f$ is continuous in $(a,b)$. Ιn which cases is $f$ continuous at $a$ and $b$?






      real-analysis continuity convex-analysis




      real-analysis continuity convex-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 22:41









      Will Fisher

      4,0381032




      4,0381032










      asked Dec 18 '18 at 22:23









      Chris RafaelChris Rafael

      106




      106




      closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Did, Saad, KReiser, Lord_Farin, user91500 Dec 27 '18 at 9:09


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Saad, KReiser, Lord_Farin, user91500

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:19










          • $begingroup$
            Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:30










          • $begingroup$
            I up voted your answer because of that last statement you made in the comment!
            $endgroup$
            – RRL
            Dec 18 '18 at 23:31






          • 1




            $begingroup$
            @ChrisRafael right, my bad, it should be $[0,1]$
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:34










          • $begingroup$
            Also, thanks! @RRL
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:35



















          2












          $begingroup$

          First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.



          Consider the functions $f:x mapsto x^2$ on $[0,1]$ and



          $$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$



          Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.



          What does this tell you?



          The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me one exaple with only f ?
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:27










          • $begingroup$
            What do you mean by "example with only $f$"?
            $endgroup$
            – RRL
            Dec 18 '18 at 23:32










          • $begingroup$
            An example with only one function
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:33










          • $begingroup$
            Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
            $endgroup$
            – RRL
            Dec 18 '18 at 23:41




















          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:19










          • $begingroup$
            Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:30










          • $begingroup$
            I up voted your answer because of that last statement you made in the comment!
            $endgroup$
            – RRL
            Dec 18 '18 at 23:31






          • 1




            $begingroup$
            @ChrisRafael right, my bad, it should be $[0,1]$
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:34










          • $begingroup$
            Also, thanks! @RRL
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:35
















          2












          $begingroup$

          I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:19










          • $begingroup$
            Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:30










          • $begingroup$
            I up voted your answer because of that last statement you made in the comment!
            $endgroup$
            – RRL
            Dec 18 '18 at 23:31






          • 1




            $begingroup$
            @ChrisRafael right, my bad, it should be $[0,1]$
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:34










          • $begingroup$
            Also, thanks! @RRL
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:35














          2












          2








          2





          $begingroup$

          I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.






          share|cite|improve this answer











          $endgroup$



          I think the best you can do, without knowing more about $f$, is say that $f:[a,b]rightarrowmathbb{R}$ is continuous at $a$ if $f(a)=lim_{xto a^+}f(x)$ and likewise for $f(b)$. For example, consider $f:[0,1]rightarrowmathbb{R}$ s.t. $f(0)=1$, $f(x)=-sqrt{x}$ for $x>0$. Then f is convex on $[0,1]$ but $f$ is not continuous at its left endpoint.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 23:34

























          answered Dec 18 '18 at 22:58









          R.JacksonR.Jackson

          1688




          1688












          • $begingroup$
            You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:19










          • $begingroup$
            Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:30










          • $begingroup$
            I up voted your answer because of that last statement you made in the comment!
            $endgroup$
            – RRL
            Dec 18 '18 at 23:31






          • 1




            $begingroup$
            @ChrisRafael right, my bad, it should be $[0,1]$
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:34










          • $begingroup$
            Also, thanks! @RRL
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:35


















          • $begingroup$
            You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:19










          • $begingroup$
            Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:30










          • $begingroup$
            I up voted your answer because of that last statement you made in the comment!
            $endgroup$
            – RRL
            Dec 18 '18 at 23:31






          • 1




            $begingroup$
            @ChrisRafael right, my bad, it should be $[0,1]$
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:34










          • $begingroup$
            Also, thanks! @RRL
            $endgroup$
            – R.Jackson
            Dec 18 '18 at 23:35
















          $begingroup$
          You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:19




          $begingroup$
          You can have convex functions that either are or aren’t continuous at their endpoints as both answers have shown so knowing only that a function is convex, the best you can do is say it’s continuous at its endpoints when it’s continuous at its endpoints.
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:19












          $begingroup$
          Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:30




          $begingroup$
          Yes , i undestrand what you mean.Anyway in your example you take f:[0,1)→R but we must show for f:[0,1]→R to take this as example
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:30












          $begingroup$
          I up voted your answer because of that last statement you made in the comment!
          $endgroup$
          – RRL
          Dec 18 '18 at 23:31




          $begingroup$
          I up voted your answer because of that last statement you made in the comment!
          $endgroup$
          – RRL
          Dec 18 '18 at 23:31




          1




          1




          $begingroup$
          @ChrisRafael right, my bad, it should be $[0,1]$
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:34




          $begingroup$
          @ChrisRafael right, my bad, it should be $[0,1]$
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:34












          $begingroup$
          Also, thanks! @RRL
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:35




          $begingroup$
          Also, thanks! @RRL
          $endgroup$
          – R.Jackson
          Dec 18 '18 at 23:35











          2












          $begingroup$

          First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.



          Consider the functions $f:x mapsto x^2$ on $[0,1]$ and



          $$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$



          Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.



          What does this tell you?



          The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me one exaple with only f ?
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:27










          • $begingroup$
            What do you mean by "example with only $f$"?
            $endgroup$
            – RRL
            Dec 18 '18 at 23:32










          • $begingroup$
            An example with only one function
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:33










          • $begingroup$
            Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
            $endgroup$
            – RRL
            Dec 18 '18 at 23:41


















          2












          $begingroup$

          First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.



          Consider the functions $f:x mapsto x^2$ on $[0,1]$ and



          $$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$



          Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.



          What does this tell you?



          The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can you tell me one exaple with only f ?
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:27










          • $begingroup$
            What do you mean by "example with only $f$"?
            $endgroup$
            – RRL
            Dec 18 '18 at 23:32










          • $begingroup$
            An example with only one function
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:33










          • $begingroup$
            Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
            $endgroup$
            – RRL
            Dec 18 '18 at 23:41
















          2












          2








          2





          $begingroup$

          First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.



          Consider the functions $f:x mapsto x^2$ on $[0,1]$ and



          $$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$



          Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.



          What does this tell you?



          The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.






          share|cite|improve this answer











          $endgroup$



          First, if $f$ is convex on $(a,b)$ then finite or infinite limits $lim_{x to a+} f(x)$ and $lim_{x to b-} f(x)$ always exist.



          Consider the functions $f:x mapsto x^2$ on $[0,1]$ and



          $$g(x) = begin{cases}f(x), & 0 leqslant x < 1,\ 2, & x= 1 end{cases}$$



          Both are convex, but $f$ is continuous at $x=1$ and $g$ is not.



          What does this tell you?



          The one-sided limit at an endpoint is always a finite number or $pm infty$. The limit cannot fail to exist (as with an oscillating function). As long as limits are not infinite, the function may or may not be continuous at the endpoints, but it can be made continuous by changing values to match the one-sided limits.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 23:16

























          answered Dec 18 '18 at 22:55









          RRLRRL

          51.3k42573




          51.3k42573












          • $begingroup$
            Can you tell me one exaple with only f ?
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:27










          • $begingroup$
            What do you mean by "example with only $f$"?
            $endgroup$
            – RRL
            Dec 18 '18 at 23:32










          • $begingroup$
            An example with only one function
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:33










          • $begingroup$
            Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
            $endgroup$
            – RRL
            Dec 18 '18 at 23:41




















          • $begingroup$
            Can you tell me one exaple with only f ?
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:27










          • $begingroup$
            What do you mean by "example with only $f$"?
            $endgroup$
            – RRL
            Dec 18 '18 at 23:32










          • $begingroup$
            An example with only one function
            $endgroup$
            – Chris Rafael
            Dec 18 '18 at 23:33










          • $begingroup$
            Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
            $endgroup$
            – RRL
            Dec 18 '18 at 23:41


















          $begingroup$
          Can you tell me one exaple with only f ?
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:27




          $begingroup$
          Can you tell me one exaple with only f ?
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:27












          $begingroup$
          What do you mean by "example with only $f$"?
          $endgroup$
          – RRL
          Dec 18 '18 at 23:32




          $begingroup$
          What do you mean by "example with only $f$"?
          $endgroup$
          – RRL
          Dec 18 '18 at 23:32












          $begingroup$
          An example with only one function
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:33




          $begingroup$
          An example with only one function
          $endgroup$
          – Chris Rafael
          Dec 18 '18 at 23:33












          $begingroup$
          Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
          $endgroup$
          – RRL
          Dec 18 '18 at 23:41






          $begingroup$
          Your question was in which cases is $f$ continuous at $a$ and $b$. Between the two posts here you have the answer to that. If $f(x) to pm infty$ as $x to b-$ then it is not only discontinuous at $x = b$, but the discontinuity can never be removed. If $f(x) to c neq pm infty$, then $f$ may or may not be continuous at $b$. It will be continuous if $f(b) = c$.
          $endgroup$
          – RRL
          Dec 18 '18 at 23:41





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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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