I came across this exercise question from a course on optimisation. It only discussed basic aspects of convex functions.
The question asks: if the solutionn set of the following inequalities convex. The inequalities are:
$x_1^3-x_1^2+x_2^2-2x_1x_2\leq 0$
$x_1-x_2\leq 1$
$x_1\geq 1$
How do I appraoch this? There is no obvious factorisation.
Let $x$ denote the variable $x_1$, and $y$ denote the variable $x_2$. Then:
The set $x \ge 1$ is convex.
The set $x - y \le 1$ is convex.
The set $x^3 - x^2 + y^2 - 2xy \le 0$ is convex for $x \ge 1$.
To show the third of these, write $(y - x)^2 \le x^2(2 - x)$, implying that $x \le 2$ and $x - x\sqrt{2-x} \le y \le x + x\sqrt{2-x}$. Take the second derivative of the lower and upper bounds on $y$ and show that each has the correct sign.
Then, the intersection of convex sets is convex so we're done.
Yes, it is.
The last two inequalities define a triangular wedge supported at right of the vertical line passing through $x_1=1$ and above the line $x_2=x_1-1$.
Rearrange the first equation as: $(x_2-x_1)^2 \leq 2x_1^2-x_1^3=x_1^2(2-x_1)$ (*)
To have solution we need $2-x_1\geq 0$, hence $x_1\leq 2$. Thus the set is constrained to be included in the trapezoidal set between vertical lines x1=1 and x1=2, and bounded below by the line x2=x1-1.
$x_1-x_1\sqrt{(2-x_1)} \leq x2 \leq x_1 + x_1\sqrt{(2-x_1)}$ (**)
Analyze the two functions $f_{+/-}(x_1) = x_1 +/- 4x_1\sqrt{(2-x_1)}$ Note: $f_+(1)=2=f_+(2)$ , $f_-(1)=0$ and $f_-(2)=2$ Compute their first derivative, and then the second derivative. If I got it right: $f''_{+/-}(x_1) = +/- (3x_1-8)/(4(2-x_1)^3/2)$
For $1\leq x_1<2$ , $3x_1-8<0 \rightarrow f''_{+}(x_1)<0$ and $f''_-(x_1)>0$ Thus (**) defines a convex set bounded above by the concave function $f_+(x1)$ and bounded below by the convex function $f_{-}(x_1)$. These two functions have a common vertical asymptote at $x_2=2$.
