For a nonconstant analytic function $ℝ→ℝ$, a point is a local minimum iff at that point, the order of the first nonzero derivative is even and that derivative is positive. Is there an analogous test for local minima of multivariable functions?
In a way, the answer is yes. Using decidability of the real closed field, there is an algorithm such that:
Given: An order $n$ Taylor series expansion (around a point $O$) with algebraic coefficients of a multivariable analytic function $f:ℝ^n→ℝ$.
Output: Whether $f$ has a local minimum at $O$: yes, no, inconclusive (additional derivatives needed).
However, the above answer is unilluminating, and the question is whether the test for local minima can be presented in a more intuitive and tractable manner, either in general or in special cases. A side question is whether $∀n>0 \, \liminf \limits_{x→O} \frac{f(x)-f(O)} {|x-O|^n} = 0$ (i.e. the order $n$ test is inconclusive for all $n$) implies that $O$ is a local minimum.
The best I have is the third order test.
Third order test for whether $f$ attains a local minimum at $O$:
Let $H$ be the Hessian of $f$ at $O$, and $V$ be the null space of $H$.
- nonzero $∇f(O)$ - false
- else if $H$ is positive definite - true
- else if $H$ is not positive semidefinite - false
- else if the third order expansion of $f(O+\mathbf{x})|_V$ is not just a constant - false
- else - inconclusive: a fourth order change can affect the answer.
Note that higher order derivatives can behave in subtle ways. For example, $(0,0)$ is not a local minimum of $-x^2 y + y^2$ but becomes a minimum after adding a fourth order term $x^4/4$ (or greater), hence the third order test for $-x^2 y + y^2$ is inconclusive.
