Convexity of $ \{x_1,x_2 \in \mathbb{R} : (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 \le 1 \}$ set

Question

For which conditions of $\alpha_1 $ and $\alpha_2 $ the following set is convex.

$ \{x_1,x_2 \in \mathbb{R} : (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 \le 1 \}$

I have started from finding min/max/saddle point.

By defining the equation:

$f(x_1,x_2) = (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 $

and deriving by $x_1,x_2$

$f'_{x_1} = 2(\alpha_1 +2)x_1 +2 \alpha_2 x_2 $

$f'_{x_2} = \alpha_1 x_2 +2 \alpha_2 x_1 $

Now I compeer it to $0$

It's easy to see that $ x_1=x_2=0$ minimum point

Additionally if $ x_1=0$ and $ \alpha_1 = \alpha_2 = 0 $. $x_2$ can get any value,

and if $ x_2=0$ and $ \alpha_1 = -2, \alpha_2 = 0 $. $x_1$ can get any value.

Otherwise $ \alpha_1^2+2 \alpha_1 = \alpha_2^2 $ satisfies the equation.

But from here I'm now sure what I should do, any help?

*Maybe all what I did is wrong

Answer 1

I assume you mean the set $$S=\{\color{red}(x_1,x_2\color{red}) \in \mathbb{R}^2 : (\alpha_1 +2)x_1^2 +\alpha_1 x_2^2 +2 \alpha_2 x_1 x_2 \le 1 \}.$$

Assume first $\alpha_1\ne 0$. Then

$$S=\left\{(x_1,x_2) \in \mathbb{R}^2 : \alpha_1\left(x_2+\frac{\alpha_2}{\alpha_1}x_1\right)^2+ \left(\alpha_1-\frac{\alpha_2^2}{\alpha_1}+2\right)x_1^2\le 1\right\}.$$

The following cases are possible.

1) $\alpha_1<0$, $\alpha_1-\frac{\alpha_2^2}{\alpha_1}+2\le 0$. Then $S$ is the plane $\Bbb R^2$, so $S$ is convex.

2) $\alpha_1\cdot \left(\alpha_1-\frac{\alpha_2^2}{\alpha_1}+2\right)<0$. Then $S$ is a domain bounded by a hyperbola, so $S$ is non-convex.

3) $\alpha_1>0$, $\alpha_1-\frac{\alpha_2^2}{\alpha_1}+2>0$. Then $S$ is a domain bounded by an ellipse, so $S$ is convex.

4) $\alpha_1>0$, $\alpha_1-\frac{\alpha_2^2}{\alpha_1}+2=0$. Then $S$ is a strip, so $S$ is convex.

If $\alpha_1=0$ then

$$S=\left\{ (x_1,x_2) \in \mathbb{R}^2 : 2\left(x_1 +\frac{\alpha_2}2 x_2\right)^2-\frac{\alpha_2^2}2 x_2^2 \le 1\right \}.$$ Thus the remaining cases are.

5) $\alpha_1=0$, $\alpha_2\ne 0$. Then $S$ is a domain bounded by a hyperbola, so $S$ is non-convex.

6) $\alpha_1=0$, $\alpha_2=0$. Then $S$ is a strip, so $S$ is convex.

Answer 2

The set is written as $\{x\in \mathbb{R}^2: \ x^T A x \le 1\}$ where $$A = \left( \begin{array}{cc} \alpha_1 + 2 & \alpha_2 \\ \alpha_2 & \alpha_1 \\ \end{array} \right). $$ The set is convex if and only if $A$ is positive semidefinite, or $A$ is negative semidefinite. Thus, the set is convex if and only if \begin{align} \left\{\begin{array}{r} \alpha_1 + 2 \ge 0 \\ \alpha_1 \ge 0 \\ (\alpha_1 + 2)\alpha_1 - \alpha_2^2 \ge 0 \end{array} \right. \quad \mathrm{or}\quad \left\{\begin{array}{r} \alpha_1 + 2 \le 0 \\ \alpha_1 \le 0 \\ (\alpha_1 + 2)\alpha_1 - \alpha_2^2 \ge 0 \end{array} \right. \end{align} which is equivalent to $(\alpha_1 + 2)\alpha_1 - \alpha_2^2 \ge 0$.

In general, we have the following results.

Fact 1: Let $\lambda_1, \lambda_2, \cdots, \lambda_n$ be real numbers. Let $S_1 = \{x \in \mathbb{R}^n: \ \lambda_1 x_1^2 + \lambda_2 x_2^2 + \cdots + \lambda_n x_n^2 \le 1\}$. Then $S_1$ is convex if and only if $\lambda_i \le 0, i=1, 2, \cdots, n$, or $\lambda_i \ge 0, i= 1, 2, \cdots, n$.

Fact 2: Let $A$ be a $n\times n$ (real) symmetric matrix. Let $S_2 = \{x\in \mathbb{R}^n: \ x^T A x \le 1\}$. Then $S_2$ is convex if and only if $A \succeq 0$ or $A \preceq 0$.