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My concern is: Why the hessian of the function $f(x,y)=\dfrac{x}{y}$ is semidefinite positive by deducing from the quadratic form method, although, its determinant is negative ??

$$\nabla^2f(x,y)=\begin{pmatrix} 0 & \dfrac{1}{y^2} \\ \dfrac{1}{y^2} & \dfrac{-2x}{y^3} \end{pmatrix}$$
$$(x,y)\nabla^2f(x,y)\begin{pmatrix} x \\ y \end{pmatrix}=0$$

My question
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1 Answers1

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The actual Hessian is the negative of what you show.

The determinant of the Hessian is negative. Therefore, the Hessian has one negative eigenvalue and one positive eigenvalue. The Hessian is indefinite. The Hessian is NOT positive semidefinite.

Therefore, $f(x,y) = \frac{x}{y}$ is neither convex nor concave.

Edit: Your mistake is using the same (x,y) in the quadratic form that you use in the calculation of the Hessian. That is not correct. The correct way is to choose an (x,y), calculate the Hessian $H$ at that (x,y), and then consider any $(x^*,y^*)$ at which to evaluate the quadratic form $(x^*;y^*)^TH(x^*;y^*)$. You will see that for some choices of $(x^*;y^*)$, the quadratic form is negative, for others it is positive, and for others, it is zero.

Mark L. Stone
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  • Yes you are right, but why the quadratic form method gives the value zero – My question Oct 19 '20 at 18:42
  • Please show exactly what you did on the "quadratic form method", and how that gives what you say is a contradctory result. – Mark L. Stone Oct 19 '20 at 18:45
  • We know that f is convex if and only if $\nabla^2f(x)$ is positive and semidefinite. which implies $x^T \nabla^2 f(x) x \geqslant 0$. If we apply this on our function we will find $x^T \nabla^2 f(x) x=0$ – My question Oct 19 '20 at 19:37
  • Your mistake is using the same (x,y) in the quadratic form that you use in the calculation of the Hessian. That is not correct. The correct way is to choose an (x,y), calculate the Hessian $H$ at that (x,y), and then consider any $(x^,y^)$ at which to evaluate the quadratic form $(x^;y^)^TH(x^;y^)$. You will see that for some choices of $(x^;y^)$, the quadratic form is negative, for others it is positive, and for others, it is zero. – Mark L. Stone Oct 19 '20 at 20:36
  • Thank you so much !! – My question Oct 19 '20 at 21:20
  • @My question You can accept the answer if you are now satisfied. – Mark L. Stone Oct 19 '20 at 23:18
  • Yes you were right and thank you – My question Oct 20 '20 at 00:24