0

I'm looking for such counter-exemple. Several times I get me thinking in situations that if I was aware of such exemple I would get rid of future problems. Is there such counter-exemple?

Nomenclature:

  • $x$ is a local minimizer in direction $d$ of $f$ if $0$ is a local minimizer of $g$ such that $g(t)=f(x+td).$

1 Answers1

0

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be defined by the expression $$f (x_1,x_2) = ({x_1} − {x_2}^2 )({x_1} - \dfrac{1}{2}{x_2}^2)$$ for all pair $(x_1,x_2)\in\mathbb{R}^2$. For every fixed $d\in\mathbb{R}^2$, we have that $x = (0, 0)$ is a local minimizer for function $\psi$ defined by $\psi(\lambda) ≡ f (x + \lambda d)$, but $x$ is not a local minimizer for $f$.