From: wcw@math.psu.edu (William C Waterhouse)
Newsgroups: sci.math
Subject: Re: minima of functions of several variables
Date: 27 Dec 1996 20:14:34 GMT
In article 851314991@gold.missouri.edu, stephen@gold.missouri.edu
(Stephen Montgomery-Smith) writes:
> Suppose that f:R^n -> R is a differentiable function such that
>
> 1) f has a local minimum at the origin
> 2) grad f is not zero except at the origin.
>
> Does it follow that f has a global minimum at the origin?
>
> Where would I read to find out about such things?
The answer to the first question is no. (An example follows.)
The answer to the second question, unfortunately, is
"in the hard exercises or passing comments in a good textbook."
Positive results natually tend to get much more emphasis
than the counterexamples to other ideas that turn out to be false.
Once you begin to imagine possible shapes for the surface
z = f(x,y), you can begin to get an intuitive idea of how a
counterexample might look. Explicit ones are always fairly
messy, I think. Here`s one from _Vector Calculus_ by
J. E. Marsden and A. J. Tromba, Section 4.2, Exercise 39:
f = -y^4 - e^{-x^2} + (2y^2)(e^x + e^{-x^2})^(1/2).
[Sketch of proof:
1) The condition that the y-partial be 0
gives either y = 0 or y^2 = (e^x + e^{-x^2})^)(1/2).
The x-partial reduces to e^x when we substitute that value
for y^2, so that never gives a critical point. The x-partial
reduces to 2xe^{-x^2} when we set y = 0, and thus x = y = 0
is the only critical point.
2) The mixed second partial has a factor of y and thus vanishes
at (0,0). We have f(0,y) = -y^4 -1 +2y^2 (2)^(1/2), so the second
y-partial at (0,0) is just 4 (2)^(1/2), a positive number. Similarly,
f(x,0) = -e^{-x^2}, so the second x-partial at (0,0) is just 2.
Thus the origin is a local minimum.
3) It is easy to see that f(0,y) takes on arbitrarily large
negative values. For instance, f(0, 100) is much less than
-1 = f(0,0).]
William C. Waterhouse
Penn State