I read on my textbook that when finding the extremum of a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$, under the constraints $F^j(x)=0,\; j=1,…,m$ (which define a surface $S$). We define the
Lagrangian
$$L(x;\lambda)=f(x)-\sum\lambda_j F^j(x).$$
Then a necessary condition for $x_0$ being a local extremum is that there exists $\lambda\in \mathbb{R}$ for which $grad\; L(x_0,\lambda)=0$. Fix this $\lambda$ and we can see $L$ as a function of $x$. Then a sufficient condition for $x_0$ being a local extremum is that the restriction to the tangent of $S$ at $x_0$ of the Hessian of $L(x)$ is positive definite or negative definite. However, we cannot obtain a sufficient condition with the definiteness of the restriction to the tangent of $S$ at $x_0$ of the Hessian of $f(x)$ itself.
My question is, is the positive (resp. negative) definiteness of the Hessian of $f(x)$ (and not its restriction to the tangent space) a sufficient condition for local minimum (resp. maximum)? Does the positive or negative definiteness of the
Hessian of $f(x)$ imply that of the same nature for the restriction of the Hessian of $L(x)$ to the tangent space?
Thanks in advance!
Best Answer
The answer is negative.
For example, let the $F(x,y)=x^2+y^2-2$ and $f(x,y)=x^2+y^2+(x-1)^3$. At the point $(1,1)$ the conditions are satisfied with $\lambda=1$. The Hessian of $f$ is positive definite, but $L(x,y,1)=(x-1)^3+2$ does not have any local extremum.