# Extrema of $f:\mathbb{R}^n\rightarrow \mathbb{R}$

My question is about finding the extrema of a multidimensional function, $f:\mathbb{R}^n\rightarrow \mathbb{R}$. From lecture I know that

$H_f(x_0) < 0$ implies a isolated maximum

$H_f(x_0) > 0$ implies a isolated minimum

$H_f(x_0)$ indefinite implies a saddle point

Where $H_f(x_0)$ is the Hesse-Matrix at point $x_0$. So how can I identify non isolated maxima? What about the cases where $H_f$ is positive (or negative) semidefinite? What if $H_f = 0$?

I know of a specialized test for the $\mathbb{R^2}$, which answers the above questions, but I wonder about the $\mathbb{R^n}$. What rules can be applied there?

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What does it mean for a matrix to be larger then a number? –  Thomas Rot Jun 12 '11 at 12:43
Probably only ad-hoc reasoning. If your Hessian is semidefinite, it means in some directions the directional derivative is zero. Find eigenvalues for those directions, then study your function in those directions as functions of one variable. –  GEdgar Jun 12 '11 at 12:44
matrix > 0 means: matrix is positive definite. And < 0 means it is negative definite. –  GEdgar Jun 12 '11 at 12:45
If $H_f=0$ that criterion doesn't make a statement. You have to study the function in that region "by hand". –  Listing Jun 12 '11 at 12:49
@ftiaronsem: there is a criterion involving higher derivatives to handle some cases wherw $H_f$ is zero. It can be found in oldish calculus books (I can only think of Rey Pastor's Análisis Matemático right now, which is in Spanish...) There is no 100% effective criterion to decide which only involves derivatives (of any order) at a point: one can construct examples of functions all of whose derivatives vanish at a point and which have there a maximum, or a minimum, or whatever you like. –  Mariano Suárez-Alvarez Jun 16 '11 at 13:47

## 1 Answer

Assume that $f$ is of class $C^2$ in a ball $B_r(x_0)$, and 1) $\nabla f(x_0) = 0$, 2) $H_f(x) \geq 0$ for every $x\in B_r(x_0)$. Then $x_0$ is a local minimum. Indeed the second assumption implies that $f$ is convex in $B_r(x_0)$, so that condition 1) is sufficient in order to have a minimum point at $x_0$. (A similar statement holds true if the Hessian matrix is negative semidefinite in a neighborhood of $x_0$.)

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$H_f(x)>0$ for every $x$ near $x_0$ shows local minimum. But $\ge 0$ does not. Not even in 1 dimension. –  GEdgar Jun 16 '11 at 16:30
I could hardly find such a counter-example... –  Rigel Jun 16 '11 at 17:52
You are right, a whole neighborhood will do. It is just possibly not a strict local minimum. If a function is convex, then the tangent plane at a point is below (or on) the graph. –  GEdgar Jun 16 '11 at 19:09