Let $f:\mathbb{R^2}\to \mathbb{R}$ be defined by
$$f(x,y)= \alpha x^2-y$$
I am trying to find the values of $\alpha $ such that $f(x,y)$ is convex over $$X=\{(x,y)\in\mathbb{R^2}:y\geq x^2-1\, , y\leq 2\}$$
I found it's Hessian matrix
$$H(x,y)=\left[\begin{array}{cc}2\alpha &0\\0 & 0\end{array}\right]$$
So $f$ is convex if $\alpha \geq 0$
Is this correct? What is the rol of $X$ in the convexity study of $f$.
Now I have to find the values of $\alpha$ such that $(0,-1)$ is a KKT.
Let
$$L(x,y,u_1,u_2)=\alpha x^2-y+u_1(x^2-y-1)+u_2(y-2)$$
One condition to be a KKT point is
$dL/dx=2\alpha x+2 u_1x=0$
So for $(0,-1)$ this is always true, then $\alpha \in \mathbb{R} ?$
Thank you.
Yes, the function $f$ is convex if $\alpha \geq 0$. To answer your first question: A set $X = \{ x \in \mathbb{R}^n \mid g(x) \leq 0 \}$ is convex, if $g : \mathbb{R}^n \to \mathbb{R}$ is a convex function. A function $f : X \subseteq \mathbb{R}^{n} \to \mathbb{R}$ is called convex, if $f$ is a convex function on a convex set $X$. So you need to check that $g_1(x, y) = x^2-y - 1$ and $g_2(x, y) = y - 2$ are convex functions. Since $g_2$ is an affin-linear function, it is convex. For $g_1$ you can check again the Hessian matrix $\nabla^2 g_1(x)$.