I'm solving this exercise and had a question with respect to determining if the critical point is either a minimum or a maximum.
Find the local extrema of $f(x,y)=ln(1+x^2)$.
First I computed the partial derivatives:
$$\frac{\delta f}{\delta x}=\frac{2x}{1+x^2}$$
$$\frac{\delta f}{\delta y}=0$$
Followed with
$$Df(x,y)=(\frac{2x}{1+x^2},0)$$
Knowing that $Df(x_o,y_o)=(0,0)$ at $(x_o,y_o)=(0,y)$
Now to determine if this is a $max$ or $min$ of $f$, I could compute the Hessian Matrix and apply what I learnt as "Hurwitz Criteria" (calculating the determinants of the principal minors of $Hess(f)(x_o,y_o)$ to determine if the matrix is positive definite of negative definite).
As I do so, I end up with
$$Hess(f)(0,y)=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
Which I really don't know if it makes it positive or negative definite.
I know it's a minimum since $ln(1+x^2)$ is a monotone increasing function because $(1+x^2)\geqslant1, \forall x\in\mathbb{R}$ and $ln(x')\geqslant 0, \forall x'\geqslant 1$.
But does the Hessian having only $0$ entries make it positive definite? Or can I only determine it with the reasoning of $ln$ being monotone increasing?
Best Answer
Pretty sure the Hessian is $(\text{Hess}\, f){(0,y)} = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$. This is not positive-definite (but it is positive semi-definite), however, this semi-definiteness is not sufficient to deduce the type of critical point.
But your reasoning with the increasing nature of logarithm is good, because for any $(x,y) \in \Bbb{R}^2$, \begin{align} f(x,y) =\ln(1+x^2) \geq \ln(1) = f(0,y). \end{align} So, every point of the form $(0,y)$, with $y\in \Bbb{R}$ is a (global, weak) minimum point for the function $f$.