convexity of the function $f(x,y)=x^4-alpha y^2$












0












$begingroup$


To study the convexity of this function I calculate the Hessian:



$$begin{bmatrix}12x^2 & 0 \ 0 & -2alpha end{bmatrix}$$



For $alpha<0$ the matrix is semidefinite-positive so the function is convex?










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$endgroup$

















    0












    $begingroup$


    To study the convexity of this function I calculate the Hessian:



    $$begin{bmatrix}12x^2 & 0 \ 0 & -2alpha end{bmatrix}$$



    For $alpha<0$ the matrix is semidefinite-positive so the function is convex?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      To study the convexity of this function I calculate the Hessian:



      $$begin{bmatrix}12x^2 & 0 \ 0 & -2alpha end{bmatrix}$$



      For $alpha<0$ the matrix is semidefinite-positive so the function is convex?










      share|cite|improve this question









      $endgroup$




      To study the convexity of this function I calculate the Hessian:



      $$begin{bmatrix}12x^2 & 0 \ 0 & -2alpha end{bmatrix}$$



      For $alpha<0$ the matrix is semidefinite-positive so the function is convex?







      real-analysis






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 15 at 20:01









      Giulia B.Giulia B.

      505311




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          $begingroup$

          As a matter of fact, this function is strictly convex if $alpha$ is negative. Both $x^4$ and $-alpha y^2$ are strictly convex and, hence, their sum is also strictly convex.






          share|cite|improve this answer









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            $begingroup$

            As a matter of fact, this function is strictly convex if $alpha$ is negative. Both $x^4$ and $-alpha y^2$ are strictly convex and, hence, their sum is also strictly convex.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              As a matter of fact, this function is strictly convex if $alpha$ is negative. Both $x^4$ and $-alpha y^2$ are strictly convex and, hence, their sum is also strictly convex.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                As a matter of fact, this function is strictly convex if $alpha$ is negative. Both $x^4$ and $-alpha y^2$ are strictly convex and, hence, their sum is also strictly convex.






                share|cite|improve this answer









                $endgroup$



                As a matter of fact, this function is strictly convex if $alpha$ is negative. Both $x^4$ and $-alpha y^2$ are strictly convex and, hence, their sum is also strictly convex.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 15 at 20:06









                Gerhard S.Gerhard S.

                1,06529




                1,06529






























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