Minimum where Hessian is positive semidefinite without continuous second derivatives?

Question

I knew that, if function $f:A\to\mathbb{R}$ of class $C^2(A)$ in an open set $A\subset \mathbb{R}^n$ has a maximum, or respectively the minimum, in $x_0\in A$, then the Hessian matrix is positive semidefinite or respectively negative semidefinite. Analogously if the Hessian matrix of $f\in C^2(A)$ is definite positive, or respectively negative, then $f$ has a minimum, or respectively a maximum, in $x_0\in A$.

I read in Kolmogorov-Fomin's (p. 504 here) that if function $f(x_1,...,x_n)$ has a minimum in point $(x_1^0,...,x_n^0)$ then in this point $d^2f\ge 0$. (Analogously, if in point $(x_1^0,...,x_n^0)$ we have a maximum, then $df^2\le 0$) and that if in point $(x_1^0,...,x_n^0)$ we have that $df=0$ and $d^2f$ is definite positive then $f(x)$ has a minimum in this point (analogously a maximum if $d^2f<0$).

I think that the continuity of all second order derivatives is to be implicitly assumed in an open set containing $(x_1^0,...,x_n^0)$: is that so, or can we relax this assumption? I thank you very much for any answer confirming my thought or proving the contrary!

Answer 1

The problem with partial derivatives is that when they are not continuous, they tell us little about what the function does in directions other than coordinate axes. I quote two counterexamples from this article by Allan A. Struthers: $$f(x,y) = \frac{(x-y)^4}{x^2+y^2} +(x+y)^4$$ has strict local minimum at the origin, but the test says "saddle point". Also, $$f(x,y) = \frac{(x^2-y^2)^2}{x^2+y^2} +x^3y^3 $$ has saddle point at the origin, but the test says "local minimum".

For the purpose of 2nd derivative test we need the function to be twice differentiable at $a$, not just to have second-order partials there. (Incidentally, the Wikipedia page Second partial derivative test misses this point).

• First-order differentiability means that as $x\to a $, $$f(x)=f(a)+\nabla f(a)\cdot (x-a)+o(|x-a|) \tag{1}$$ where $\nabla f$ is the vector of first order derivatives.

• Second-order differentiability means that $\nabla f$ is differentiable: $$\nabla f(x)=\nabla f(a)+D^2 f(a) (x-a)+o(|x-a|) \tag{2}$$ where $D^2$ is the matrix of second order derivatives.

The continuity of second-order partials implies (2), just as the continuity of first-order partials implies (1).

If you have (2), the second derivative test works: $\nabla f=0$ and $D^2f$ being positive definite imply $f$ has local minimum, etc. One way to see it is to obtain the following consequence of (2): $$ f(x)=f(a)+\nabla f(a)\cdot (x-a) + \frac12 (x-a)^t D^2 f(a) (x-a)+o(|x-a|^2) \tag{3}$$ The property (3) directly leads to the 2nd derivative test: when the linear term vanishes, the sign of the quadratic term determines the sign of $f(x)-f(a)$.

To get (3) from (2), use the mean value theorem: $$ f(x)-f(a) = \nabla f(\xi)\cdot (\xi-a) \tag{4} $$ for some $\xi$ between $a$ and $x$. Then plug (2) into (4).