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We know that if $\nabla f(x_0) = 0$ and $\nabla^2f(x_0) \succ0$ (positive definite) then $x_0$ is the local minima.

So, prove that if $\nabla f(x_0) = 0$ and $\nabla^2f(x_0) \succeq 0$ then $x_0$ couldn't be the local minimum. If possible, don't use counterexamples, but really a proof.

max_zorn
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Tiago Troccoli
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  • What does $\prec$ and $\preceq$ refer to? I'm guessing, positive definite/semi-definite respectively? – Theo Bendit Apr 08 '18 at 02:26
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    Also, I think you're going to run into trouble, since presumably$$\nabla^2 f(x_0) \succ 0 \implies \nabla^2 f(x_0) \succeq 0.$$So, if $\nabla^2 f(x_0) \succ 0$ is sufficient for $x_0$ being a local minimum, then $\nabla^2 f(x_0) \succeq 0$ will be sometimes true when $x_0$ is a local minimum. That is, a general proof that $x_0$ is never a local minimum cannot possibly exist; only counter-examples could be provided. – Theo Bendit Apr 08 '18 at 02:32
  • Yes, Theo Bendit, it's positive definite I've just edited it... – Tiago Troccoli Apr 08 '18 at 03:08
  • @TheoBendit: https://math.stackexchange.com/q/669085/52878 – Michael Grant Apr 08 '18 at 18:07
  • Tiago, your logic is wrong. Theo picked apart a bit of it. But the fact is that even if $\nabla^2 f(x_0)\not\succ 0$ and $\nabla^2 f(x_0)\succeq 0$, it is possible for $x_0$ to be a local minimum; consider, for instance, $f(x)=x^4$ and $x_0=0$. – Michael Grant Apr 08 '18 at 18:09

1 Answers1

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I'm going to give you some examples. You can decide if they answer your proof or not.

Take $f(x, y) = x^2 + y^2$. Then $\nabla f(0, 0) = (0, 0)$, and $$\nabla^2 f(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \succeq 0.$$ The fact that the matrix is positive definite means that it's also positive semi-definite. Note that $(0, 0)$ is a local (and in fact, global) minimum.

Take $g(x, y) = x^2$. Then $\nabla g(0, 0) = (0, 0)$, and $$\nabla^2 g(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ Note that $0$ is the minimal value of $g$, making $(0, 0)$ a local (and global) minimum. However, this minimum is achieved all along the line $x = 0$, e.g. at $(0, 1)$ as well.

Take $h(x, y) = x^2 + y^4$. Then $\nabla h (0, 0) = (0, 0)$, and $$\nabla^2 h(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ Again, we have a global minimum at $(x, y) = (0, 0)$, and this time it's unique. However, the Hessian was only positive semi-definite.

Finally, take $k(x, y) = x^2 - y^4$. Then $\nabla k(0, 0) = (0, 0)$, and $$\nabla^2 k(0, 0)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} \succeq 0.$$ This time, we have positive semi-definiteness in the Hessian, but no local minimum.

Theo Bendit
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