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













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      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




















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