Is $f(x_1,x_2)=10 - 2(x_2 - x_1^2)^2$ a convex function on S where S = $\{ (x_1, x_2) | -22 \le x_1 \le 2, -2 \le x_2 \le 2\}$?

Question

How to prove the $f(x_1,x_2)=10 - 2(x_2 - x_1^2)^2$ a convex function on S or not, where S = $\{ (x_1, x_2) | -22 \le x_1 \le 2, -2 \le x_2 \le 2\}$?

How should I start to prove it? Thanks.

I look into the math tutorial about convexivity online, I found this:

For $\theta \in [0, 1]$, $f(\theta x + (1 - \theta)y) \le \theta f(x) + (1 - \theta)f(y)$.

AND

assume $g(x) = f(x_1, x_2)$,

$f(\theta(x_1', x_2') + (1 - \theta)(x_1, x_2))$

$=f((1 - \theta)x_1 + \theta x_1', (1 - \theta)x_2 + \theta x_2')$

$=g((1 - \theta)x_1 + \theta x_1')$

$\ge (1 - \theta)g(x) + \theta g(x')$

$=(1 - \theta)f(x_1,x_2) + \theta f(x_1', x_2')$

$=(1 - \theta)(10 - 2(x_2 - x_1^2)^2) + \theta f(x_1', x_2')$

But I don't know what's the next step and what are the values of $(x_1', x_2')$.

Answer 1

Recall that a twice differentiable function of several variables $f$ is convex in a convex set $C$ if and only if the Hessian matrix of $f$ is positive semidefinite in the interior of $C$.

In your case, the convex set $C$ is the rectangle $[-22,2]\times[-2,2]$ and the Hessian matrix is equal to $$H(x_1,x_2)=\begin{bmatrix} 8x_2-24x_1^2 & 8 x_1 \\ 8 x_1 & -4 \end{bmatrix}.$$ Notice that a symmetric matrix is positive semidefinite if and only if all eigenvalues are non-negative.

This condition does not hold in $C$. For instance, the point $(0,1)\in C$, and $H(0,1)=\begin{bmatrix} 8 & 0 \\ 0 & -4 \end{bmatrix}$ which has eigenvalues $8$ (positive) and $-4$ (negative). Hence $f$ is not convex in a neighbourhood of $(0,1)$.