Is function $f$ convex?

Question

$$f(x) = (x_1-x_2)^2+\frac{1}{1+x_1^2+x_2^2}$$

Is $f$ convex? And why? Also, is $(1,-1)$ a local/global minimum?

Answer 1

Here is what the function looks like near $(0,0)$. It is neither convex nor concave and $1$ is a local minimum

Now to prove that we compute at $(0,0)$ after the change of variable

$$\begin{align} X=&{x_1+x_2\over\sqrt{2}}\\ Y=&{x_1-x_2\over\sqrt{2}} \end{align}$$

$$\begin{bmatrix} {\partial^2{f}\over\partial{X^2}} & {\partial^2{f}\over\partial{X}\partial{Y}}\\ {\partial^2{f}\over\partial{Y}\partial{X}} & {\partial^2{f}\over\partial{Y^2}} \end{bmatrix} =\begin{bmatrix} -1 & 0\\ 0 & 3 \end{bmatrix}$$

And we have our conclusion