Today we learned the 2nd Derivative Test of two-variable functions for testing local minimum/maximum points, and saddle points.
We have deduced the law by calculating $$D^2_{\mathbf{v}} f=D_{\mathbf{V}} \left(\nabla f \cdot \mathbf{v} \right) = \mathbf{v}^{T} H \mathbf{v},$$ where $H$ is the Hessian matrix $\left[\begin {matrix}f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{matrix}\right].$
However, we discovered that by letting $\mathbf{v}=\left<a,b\right>$ where $||\mathbf{v}||=1$, $$D_{\mathbf{v}}^2 f= f_{xx}a^2+2f_{xy}ab+f_{yy}b^2,$$ hence if $f_{xx}$ and $f_{yy}$ are both nonzero, then $D_{\mathbf{v}}^2 f$ changes its sign at most two times while $\mathbf{v}$ rotates clockwise on the unit circle. We thought that this result was absurd, so we tried to construct a counterexample of it. This is what we got: if a function $z=f(x,y)$ is twice differentiable and satisfies $f(t,0)=-At^2, f(0, t)=-Bt^2, f(t, t)=Ct^2, f(t, -t)=Dt^2$ for some $A,B,C,D>0$ and for all sufficiently small $t$, then $f_{xx}$ and $f_{yy}$ are both nonzero, but the sign of $D_{\mathbf{v}}^2 f$ changes four times as $\mathbf{v}$ rotates once.
Comparing with the previous argument, it seems like such function does not exist, but we couldn't prove it. We do not know which is true too. Please help us.