Determining if a set is convex, and if not, finding a convex relaxation

Question

This is for an advanced course in Chemical Engineering, and I do not have much previous experience with matrix mathematics so this question has me stumped.

The way the material was presented, it seems like we should use the Hessian to determine if the set is convex (basically, if the Hessian is greater than zero for all variables?). These are the first two of five problems:

1. $\{(x,y)|x^2 + y^2 \le 1\}$,

2. $\{(x,y)|x^2 + y^2 = 1\}$.

I think I'm doing the Hessian correctly $$\begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}$$

But what throws me off is the difference between the two: how do I incorporate the inequality versus the equality that sets the two apart when trying to determine if each set is convex?

Answer 1

The theorem you are supposed to use is:

A set $X$ is convex iff for all continuous functions $f$ from $X$ to $\{0,1\}$ (with the discrete topology), $f$ is a constant function.

But I am not quite sure how to use this approach. Instead consider using the definition of a convex set.

A set $X$ is convex iff $\forall a,b\in X: tb+(1-t)a\in X, \text{where } t\in[0,1]$.

With this definition and noting that the set $\{(x,y):x^2+y^2\le1\}=\{\vec{x}\in\Bbb R^2:||\vec{x}||\le1\}=:B_1\left(0\right)$, consider $a,b\in B_1\left(0\right)$. Then: $$||tb+(1-t)a||\le||tb||+||(1-t)a||=t||b||+(1-t)||a||\le t(1)+(1-t)(1)=1$$ Now it follows that $tb+(1-t)a\in B_1\left(0\right)$ and since $a,b$ were arbitrary, the set is convex.