# Assistance with Proving if sets are convex or not.

I'm trying to see if the following sets are convex for all $$x$$. I proved $$S_1$$, but having trouble with $$S_2$$ and $$S_3$$.

$$S_1 = \{x \in R^n : \lVert Ax \rVert \leq 1\}$$
$$S_2= \{x \in R^n : \lVert Ax \rVert \geq 1\}$$
$$S_3= \{x \in R^{2n} : \sum_{k=1}^{n} x^2_k \leq \sum_{k=n+1}^{2n} x^2_k \}$$

$$S_1$$:

$$$$\label{eq1} \begin{split} \lVert A(\alpha x + (1 - \alpha)y)\rVert & = \lVert \alpha A x + (1 - \alpha)Ay\rVert \\ & \leq \alpha\lVert A x \rVert +(1 - \alpha)\lVert Ay\rVert \\ & = \alpha(1) + (1-\alpha)(1) = 1 \end{split}$$$$ Therefore $$S_{1}$$ is convex because norms cannot be negative and therefore it must be equal to 1.

For $$S_2$$, intuitively I can see that it is concave. But how can I show this? I would prefer to see a proof rather than a counterexample.

For $$S_3$$, I don't even know how to start.

Thanks!

• It may be obvious, but which norm are we considering? A specific one or any norm? Commented Dec 15, 2019 at 16:23
• Sorry this the the L2 (euclidian norm). Commented Dec 15, 2019 at 16:25

Suppose $$S_2$$ is non-empty, and if $$x \in S_2$$, then we also have $$-x \in S_2$$, check that $$0 \notin S_2$$ and you can make a conclusion.
For $$S_3$$, let's first consider the special case for $$n=1$$, then we have $$(1,1) \in S_3$$ and $$(1,-1)\in S_3$$, check that their midpoint $$(1,0) \notin S_3$$. Use this idea to generalize to arbitrary $$n$$.