How to check convexity?












18












$begingroup$


How can I know the function $$f(x,y)=frac{y^2}{xy+1}$$ with $x>0$,$y>0$ is convex or not?






convex-analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
    $endgroup$
    – Michael Greinecker♦
    Mar 12 '12 at 20:37










  • $begingroup$
    I checked the Hessian matrix, but unfortunately it is indefinite.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 20:51








  • 7




    $begingroup$
    So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
    $endgroup$
    – Harald Hanche-Olsen
    Mar 12 '12 at 21:18










  • $begingroup$
    Thanks for your answer. I think it is a good suggestion.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 22:13
















18












$begingroup$


How can I know the function $$f(x,y)=frac{y^2}{xy+1}$$ with $x>0$,$y>0$ is convex or not?






convex-analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
    $endgroup$
    – Michael Greinecker♦
    Mar 12 '12 at 20:37










  • $begingroup$
    I checked the Hessian matrix, but unfortunately it is indefinite.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 20:51








  • 7




    $begingroup$
    So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
    $endgroup$
    – Harald Hanche-Olsen
    Mar 12 '12 at 21:18










  • $begingroup$
    Thanks for your answer. I think it is a good suggestion.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 22:13














18












18








18


11



$begingroup$


How can I know the function $$f(x,y)=frac{y^2}{xy+1}$$ with $x>0$,$y>0$ is convex or not?






convex-analysis







share|cite|improve this question











$endgroup$




How can I know the function $$f(x,y)=frac{y^2}{xy+1}$$ with $x>0$,$y>0$ is convex or not?






convex-analysis




convex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 12 '12 at 20:37







user17762

















asked Mar 12 '12 at 20:30









Xiangyu MengXiangyu Meng

2751312




2751312








  • 4




    $begingroup$
    Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
    $endgroup$
    – Michael Greinecker♦
    Mar 12 '12 at 20:37










  • $begingroup$
    I checked the Hessian matrix, but unfortunately it is indefinite.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 20:51








  • 7




    $begingroup$
    So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
    $endgroup$
    – Harald Hanche-Olsen
    Mar 12 '12 at 21:18










  • $begingroup$
    Thanks for your answer. I think it is a good suggestion.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 22:13














  • 4




    $begingroup$
    Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
    $endgroup$
    – Michael Greinecker♦
    Mar 12 '12 at 20:37










  • $begingroup$
    I checked the Hessian matrix, but unfortunately it is indefinite.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 20:51








  • 7




    $begingroup$
    So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
    $endgroup$
    – Harald Hanche-Olsen
    Mar 12 '12 at 21:18










  • $begingroup$
    Thanks for your answer. I think it is a good suggestion.
    $endgroup$
    – Xiangyu Meng
    Mar 12 '12 at 22:13








4




4




$begingroup$
Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
$endgroup$
– Michael Greinecker♦
Mar 12 '12 at 20:37




$begingroup$
Have you looked at the Hessian matrix? If it is positive semidefinite, $f$ is convex.
$endgroup$
– Michael Greinecker♦
Mar 12 '12 at 20:37












$begingroup$
I checked the Hessian matrix, but unfortunately it is indefinite.
$endgroup$
– Xiangyu Meng
Mar 12 '12 at 20:51






$begingroup$
I checked the Hessian matrix, but unfortunately it is indefinite.
$endgroup$
– Xiangyu Meng
Mar 12 '12 at 20:51






7




7




$begingroup$
So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
$endgroup$
– Harald Hanche-Olsen
Mar 12 '12 at 21:18




$begingroup$
So the two eigenvalues of the Hessian have opposite signs, meaning one eigenvalue is negative. The function will be concave in the direction of the corresponding eigenspace.
$endgroup$
– Harald Hanche-Olsen
Mar 12 '12 at 21:18












$begingroup$
Thanks for your answer. I think it is a good suggestion.
$endgroup$
– Xiangyu Meng
Mar 12 '12 at 22:13




$begingroup$
Thanks for your answer. I think it is a good suggestion.
$endgroup$
– Xiangyu Meng
Mar 12 '12 at 22:13










3 Answers
3






active

oldest

votes


















18












$begingroup$

Consider $y=x$ then we have $displaystyle g(x)=frac{x^2}{x^2+1}=1-frac 1{x^2+1}$



The second derivative of this is $g''(x)=frac{2-6x^2}{(1+x^2)^3}$
and will change sign around $x=frac 1{sqrt{3}}$ so that $g$ is convex in $(0,frac 1{sqrt{3}})$ and concave in $(frac 1{sqrt{3}},infty)$.



Your function is clearly not convex nor concave on $(mathbb{R^{+*}})^2$ but you could search more restricted sets if needed...



Here is a picture (from below) of your function (convex near $y=0$ and concave when $y$ becomes larger at least in the x=y direction, in the x=-y direction it looks convex...) :



picture






share|cite|improve this answer











$endgroup$





















    8












    $begingroup$

    The Hessian of $frac{y^2}{1+xy}$ is
    $$
    H
    =
    frac1{(1+xy)^3}begin{bmatrix}
    2y^4&-y^2(3+xy)\[12pt]
    -y^2(3+xy)&2
    end{bmatrix}
    $$
    and
    $$
    begin{bmatrix}
    u&v
    end{bmatrix}
    H
    begin{bmatrix}
    u\v
    end{bmatrix}
    =frac2{(1+xy)^3}(v-uy^2)^2-frac1{(1+xy)^2}uvy^2
    $$
    Setting $begin{bmatrix}u&vend{bmatrix}=begin{bmatrix}1&y^2end{bmatrix}$ gives
    $$
    begin{bmatrix}
    1&y^2
    end{bmatrix}
    H
    begin{bmatrix}
    1\y^2
    end{bmatrix}
    =-frac{y^4}{(1+xy)^2}
    $$
    so $frac{y^2}{1+xy}$ is not convex as long as $yne0$.






    share|cite|improve this answer









    $endgroup$





















      7












      $begingroup$

      The book "Convex Optimization" by Boyd, available free online here, describes methods to check.



      The standard definition is if f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y) for 0≤θ≤1 and the domain of x,y is also convex.



      So if you could prove that for your function, you would know it's convex.



      The Hessian being positive semi-definite, as mentioned in comments, would also show that the function is convex.



      See page 67 of the book for more.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
        $endgroup$
        – Xiangyu Meng
        Mar 12 '12 at 22:13






      • 3




        $begingroup$
        It is a very good book on the subject if you wish to go deeper than simple calculus.
        $endgroup$
        – Nick Alger
        Mar 12 '12 at 22:26










      • $begingroup$
        Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
        $endgroup$
        – Squirtle
        Jul 29 '13 at 23:47








      • 2




        $begingroup$
        @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
        $endgroup$
        – Rasputin
        Jul 17 '17 at 19:53













      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%2f119418%2fhow-to-check-convexity%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      18












      $begingroup$

      Consider $y=x$ then we have $displaystyle g(x)=frac{x^2}{x^2+1}=1-frac 1{x^2+1}$



      The second derivative of this is $g''(x)=frac{2-6x^2}{(1+x^2)^3}$
      and will change sign around $x=frac 1{sqrt{3}}$ so that $g$ is convex in $(0,frac 1{sqrt{3}})$ and concave in $(frac 1{sqrt{3}},infty)$.



      Your function is clearly not convex nor concave on $(mathbb{R^{+*}})^2$ but you could search more restricted sets if needed...



      Here is a picture (from below) of your function (convex near $y=0$ and concave when $y$ becomes larger at least in the x=y direction, in the x=-y direction it looks convex...) :



      picture






      share|cite|improve this answer











      $endgroup$


















        18












        $begingroup$

        Consider $y=x$ then we have $displaystyle g(x)=frac{x^2}{x^2+1}=1-frac 1{x^2+1}$



        The second derivative of this is $g''(x)=frac{2-6x^2}{(1+x^2)^3}$
        and will change sign around $x=frac 1{sqrt{3}}$ so that $g$ is convex in $(0,frac 1{sqrt{3}})$ and concave in $(frac 1{sqrt{3}},infty)$.



        Your function is clearly not convex nor concave on $(mathbb{R^{+*}})^2$ but you could search more restricted sets if needed...



        Here is a picture (from below) of your function (convex near $y=0$ and concave when $y$ becomes larger at least in the x=y direction, in the x=-y direction it looks convex...) :



        picture






        share|cite|improve this answer











        $endgroup$
















          18












          18








          18





          $begingroup$

          Consider $y=x$ then we have $displaystyle g(x)=frac{x^2}{x^2+1}=1-frac 1{x^2+1}$



          The second derivative of this is $g''(x)=frac{2-6x^2}{(1+x^2)^3}$
          and will change sign around $x=frac 1{sqrt{3}}$ so that $g$ is convex in $(0,frac 1{sqrt{3}})$ and concave in $(frac 1{sqrt{3}},infty)$.



          Your function is clearly not convex nor concave on $(mathbb{R^{+*}})^2$ but you could search more restricted sets if needed...



          Here is a picture (from below) of your function (convex near $y=0$ and concave when $y$ becomes larger at least in the x=y direction, in the x=-y direction it looks convex...) :



          picture






          share|cite|improve this answer











          $endgroup$



          Consider $y=x$ then we have $displaystyle g(x)=frac{x^2}{x^2+1}=1-frac 1{x^2+1}$



          The second derivative of this is $g''(x)=frac{2-6x^2}{(1+x^2)^3}$
          and will change sign around $x=frac 1{sqrt{3}}$ so that $g$ is convex in $(0,frac 1{sqrt{3}})$ and concave in $(frac 1{sqrt{3}},infty)$.



          Your function is clearly not convex nor concave on $(mathbb{R^{+*}})^2$ but you could search more restricted sets if needed...



          Here is a picture (from below) of your function (convex near $y=0$ and concave when $y$ becomes larger at least in the x=y direction, in the x=-y direction it looks convex...) :



          picture







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 13 '12 at 22:37

























          answered Mar 12 '12 at 22:11









          Raymond ManzoniRaymond Manzoni

          37.2k563117




          37.2k563117























              8












              $begingroup$

              The Hessian of $frac{y^2}{1+xy}$ is
              $$
              H
              =
              frac1{(1+xy)^3}begin{bmatrix}
              2y^4&-y^2(3+xy)\[12pt]
              -y^2(3+xy)&2
              end{bmatrix}
              $$
              and
              $$
              begin{bmatrix}
              u&v
              end{bmatrix}
              H
              begin{bmatrix}
              u\v
              end{bmatrix}
              =frac2{(1+xy)^3}(v-uy^2)^2-frac1{(1+xy)^2}uvy^2
              $$
              Setting $begin{bmatrix}u&vend{bmatrix}=begin{bmatrix}1&y^2end{bmatrix}$ gives
              $$
              begin{bmatrix}
              1&y^2
              end{bmatrix}
              H
              begin{bmatrix}
              1\y^2
              end{bmatrix}
              =-frac{y^4}{(1+xy)^2}
              $$
              so $frac{y^2}{1+xy}$ is not convex as long as $yne0$.






              share|cite|improve this answer









              $endgroup$


















                8












                $begingroup$

                The Hessian of $frac{y^2}{1+xy}$ is
                $$
                H
                =
                frac1{(1+xy)^3}begin{bmatrix}
                2y^4&-y^2(3+xy)\[12pt]
                -y^2(3+xy)&2
                end{bmatrix}
                $$
                and
                $$
                begin{bmatrix}
                u&v
                end{bmatrix}
                H
                begin{bmatrix}
                u\v
                end{bmatrix}
                =frac2{(1+xy)^3}(v-uy^2)^2-frac1{(1+xy)^2}uvy^2
                $$
                Setting $begin{bmatrix}u&vend{bmatrix}=begin{bmatrix}1&y^2end{bmatrix}$ gives
                $$
                begin{bmatrix}
                1&y^2
                end{bmatrix}
                H
                begin{bmatrix}
                1\y^2
                end{bmatrix}
                =-frac{y^4}{(1+xy)^2}
                $$
                so $frac{y^2}{1+xy}$ is not convex as long as $yne0$.






                share|cite|improve this answer









                $endgroup$
















                  8












                  8








                  8





                  $begingroup$

                  The Hessian of $frac{y^2}{1+xy}$ is
                  $$
                  H
                  =
                  frac1{(1+xy)^3}begin{bmatrix}
                  2y^4&-y^2(3+xy)\[12pt]
                  -y^2(3+xy)&2
                  end{bmatrix}
                  $$
                  and
                  $$
                  begin{bmatrix}
                  u&v
                  end{bmatrix}
                  H
                  begin{bmatrix}
                  u\v
                  end{bmatrix}
                  =frac2{(1+xy)^3}(v-uy^2)^2-frac1{(1+xy)^2}uvy^2
                  $$
                  Setting $begin{bmatrix}u&vend{bmatrix}=begin{bmatrix}1&y^2end{bmatrix}$ gives
                  $$
                  begin{bmatrix}
                  1&y^2
                  end{bmatrix}
                  H
                  begin{bmatrix}
                  1\y^2
                  end{bmatrix}
                  =-frac{y^4}{(1+xy)^2}
                  $$
                  so $frac{y^2}{1+xy}$ is not convex as long as $yne0$.






                  share|cite|improve this answer









                  $endgroup$



                  The Hessian of $frac{y^2}{1+xy}$ is
                  $$
                  H
                  =
                  frac1{(1+xy)^3}begin{bmatrix}
                  2y^4&-y^2(3+xy)\[12pt]
                  -y^2(3+xy)&2
                  end{bmatrix}
                  $$
                  and
                  $$
                  begin{bmatrix}
                  u&v
                  end{bmatrix}
                  H
                  begin{bmatrix}
                  u\v
                  end{bmatrix}
                  =frac2{(1+xy)^3}(v-uy^2)^2-frac1{(1+xy)^2}uvy^2
                  $$
                  Setting $begin{bmatrix}u&vend{bmatrix}=begin{bmatrix}1&y^2end{bmatrix}$ gives
                  $$
                  begin{bmatrix}
                  1&y^2
                  end{bmatrix}
                  H
                  begin{bmatrix}
                  1\y^2
                  end{bmatrix}
                  =-frac{y^4}{(1+xy)^2}
                  $$
                  so $frac{y^2}{1+xy}$ is not convex as long as $yne0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 5 '16 at 8:24









                  robjohn♦robjohn

                  268k27308634




                  268k27308634























                      7












                      $begingroup$

                      The book "Convex Optimization" by Boyd, available free online here, describes methods to check.



                      The standard definition is if f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y) for 0≤θ≤1 and the domain of x,y is also convex.



                      So if you could prove that for your function, you would know it's convex.



                      The Hessian being positive semi-definite, as mentioned in comments, would also show that the function is convex.



                      See page 67 of the book for more.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                        $endgroup$
                        – Xiangyu Meng
                        Mar 12 '12 at 22:13






                      • 3




                        $begingroup$
                        It is a very good book on the subject if you wish to go deeper than simple calculus.
                        $endgroup$
                        – Nick Alger
                        Mar 12 '12 at 22:26










                      • $begingroup$
                        Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                        $endgroup$
                        – Squirtle
                        Jul 29 '13 at 23:47








                      • 2




                        $begingroup$
                        @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                        $endgroup$
                        – Rasputin
                        Jul 17 '17 at 19:53


















                      7












                      $begingroup$

                      The book "Convex Optimization" by Boyd, available free online here, describes methods to check.



                      The standard definition is if f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y) for 0≤θ≤1 and the domain of x,y is also convex.



                      So if you could prove that for your function, you would know it's convex.



                      The Hessian being positive semi-definite, as mentioned in comments, would also show that the function is convex.



                      See page 67 of the book for more.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                        $endgroup$
                        – Xiangyu Meng
                        Mar 12 '12 at 22:13






                      • 3




                        $begingroup$
                        It is a very good book on the subject if you wish to go deeper than simple calculus.
                        $endgroup$
                        – Nick Alger
                        Mar 12 '12 at 22:26










                      • $begingroup$
                        Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                        $endgroup$
                        – Squirtle
                        Jul 29 '13 at 23:47








                      • 2




                        $begingroup$
                        @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                        $endgroup$
                        – Rasputin
                        Jul 17 '17 at 19:53
















                      7












                      7








                      7





                      $begingroup$

                      The book "Convex Optimization" by Boyd, available free online here, describes methods to check.



                      The standard definition is if f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y) for 0≤θ≤1 and the domain of x,y is also convex.



                      So if you could prove that for your function, you would know it's convex.



                      The Hessian being positive semi-definite, as mentioned in comments, would also show that the function is convex.



                      See page 67 of the book for more.






                      share|cite|improve this answer









                      $endgroup$



                      The book "Convex Optimization" by Boyd, available free online here, describes methods to check.



                      The standard definition is if f(θx + (1 − θ)y) ≤ θf(x) + (1 − θ)f(y) for 0≤θ≤1 and the domain of x,y is also convex.



                      So if you could prove that for your function, you would know it's convex.



                      The Hessian being positive semi-definite, as mentioned in comments, would also show that the function is convex.



                      See page 67 of the book for more.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Mar 12 '12 at 21:24









                      Jan GorznyJan Gorzny

                      792914




                      792914












                      • $begingroup$
                        Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                        $endgroup$
                        – Xiangyu Meng
                        Mar 12 '12 at 22:13






                      • 3




                        $begingroup$
                        It is a very good book on the subject if you wish to go deeper than simple calculus.
                        $endgroup$
                        – Nick Alger
                        Mar 12 '12 at 22:26










                      • $begingroup$
                        Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                        $endgroup$
                        – Squirtle
                        Jul 29 '13 at 23:47








                      • 2




                        $begingroup$
                        @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                        $endgroup$
                        – Rasputin
                        Jul 17 '17 at 19:53




















                      • $begingroup$
                        Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                        $endgroup$
                        – Xiangyu Meng
                        Mar 12 '12 at 22:13






                      • 3




                        $begingroup$
                        It is a very good book on the subject if you wish to go deeper than simple calculus.
                        $endgroup$
                        – Nick Alger
                        Mar 12 '12 at 22:26










                      • $begingroup$
                        Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                        $endgroup$
                        – Squirtle
                        Jul 29 '13 at 23:47








                      • 2




                        $begingroup$
                        @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                        $endgroup$
                        – Rasputin
                        Jul 17 '17 at 19:53


















                      $begingroup$
                      Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                      $endgroup$
                      – Xiangyu Meng
                      Mar 12 '12 at 22:13




                      $begingroup$
                      Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach.
                      $endgroup$
                      – Xiangyu Meng
                      Mar 12 '12 at 22:13




                      3




                      3




                      $begingroup$
                      It is a very good book on the subject if you wish to go deeper than simple calculus.
                      $endgroup$
                      – Nick Alger
                      Mar 12 '12 at 22:26




                      $begingroup$
                      It is a very good book on the subject if you wish to go deeper than simple calculus.
                      $endgroup$
                      – Nick Alger
                      Mar 12 '12 at 22:26












                      $begingroup$
                      Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                      $endgroup$
                      – Squirtle
                      Jul 29 '13 at 23:47






                      $begingroup$
                      Using the standard definition is almost always completely useless. The only time it is useful is if you have a function which is not continuous in its second derivative (or it doesn't exist) then you can rule out it is convex if you can numerically find a counter-example simply by randomly evaluating SEVERAL points. But this isn't generally practical or fun, nor can it ever prove a function IS convex, only that it isn't if you happen to find a counter-example.
                      $endgroup$
                      – Squirtle
                      Jul 29 '13 at 23:47






                      2




                      2




                      $begingroup$
                      @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                      $endgroup$
                      – Rasputin
                      Jul 17 '17 at 19:53






                      $begingroup$
                      @Squirtle, it can absolutely prove a function is convex: if you show a function satisfies that condition, it is convex. As a basic example, let f(x)=5x and you will see that that condition is always an equality. Therefore f(x)=5x is convex everywhere.
                      $endgroup$
                      – Rasputin
                      Jul 17 '17 at 19:53




















                      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%2f119418%2fhow-to-check-convexity%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

                      Mario Kart Wii

                      Image
                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                      Read more

                      What does “Dominus providebit” mean?

                      Image
                      5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                      Read more

                      Antonio Litta Visconti Arese

                      Image
                      Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                      Read more