0
$\begingroup$

Proposition: each saddle point is a isolated critical point.

I can not find a counterexample to disprove it, I think it is false.

Some idea?

  • 2
    Context, please? What are your assumptions? – 2017-01-16
  • 0
    Context: real functions of several variables. I'm trying to prove that if the critical points are not isolated, the Hessian matrix is semidefinite. To do this it would return comfortable that the above proposition is true. – 2017-01-16

2 Answers 2

0

Let $f(x,y) = (x^2+y^2)(x-y)^3.$ Then $f$ has a saddle point at each point on the line $y=x.$ Proof: A simple computation shows $f$ has a critical point at each point on this line. Since $f(x,0) = x^5,$ $f$ has no local extremum at $(0,0),$ hence $(0,0)$ is a saddle point. At any $(a,a)$ with $a\ne 0,$ you can check that $f$ changes sign as you move along the circle of radius $\sqrt 2 a.$ Thus every $(a,a)$ is a saddle point for $f.$

1

For example, try $f(x,y) = x^8 (2+\sin(1/x^2)) - y^2$ for $x \ne 0$ with $f(0,y) = -y^2$.

  • 0
    Thank you! I also found that other functionality with saddle non-isolated critical points: $ f(x,y,z) = z^2(x^3-3x-y^2+z) $. At this point I am puzzled. I do not know how to prove that in the case of non-isolated points on the Hessian test fails. – 2017-01-17