Check if set is convex

Question

Check if set is convex: $$V = \{x \in \mathbb{R^n} : \lVert x\rVert = r \}$$

Obviously this set is non convex, but if there is a method to show that numerically instead of by logical thinking. I mean, in my class we prove the set is convex by: $$V = \{x \in \mathbb{R^n} : \lVert x\rVert \le r \}$$
$\lVert \theta x + (1 - \theta)y \rVert = \lVert \theta x \rVert + \lVert (1 - \theta)y \rVert = \theta \lVert x \rVert + (1 - \theta) \lVert y \rVert \le \theta r + (1 - \theta)r = r$

which implies that V is convex. But why it does not hold for the first set? $\lVert \theta x + (1 - \theta)y \rVert = \lVert \theta x \rVert + \lVert (1 - \theta)y \rVert = \theta \lVert x \rVert + (1 - \theta) \lVert y \rVert = \theta r + (1 - \theta)r = r$

For me this looks valid, but for sure it should not. I know that if I find a subset that does not satisfy this inequality then it will be a proof that this set is non convex. But besides having intuiton that given set might non be convex a then looking for some subset is there a way to check numerically without guessing that it might not be convex?

Answer 1

You can easily see that points $[r,0,\dots,0]$ and $[-r,0,\dots,0]$ are in the set, but its average, which is $[0,0,\dots,0]$, is definitely not in the set. And that's it.

Answer 2

Because this statement: $$\lVert \theta x + (1 - \theta)y \rVert = \lVert \theta x \rVert + \lVert (1 - \theta)y \rVert = \theta \lVert x \rVert + (1 - \theta) \lVert y \rVert \le \theta r + (1 - \theta)r = r$$

is false. In particular, it is not true in general that $\lVert \theta x + (1 - \theta)y \rVert = \lVert \theta x \rVert + \lVert (1 - \theta)y \rVert$. Instead, all we can say is

$$\lVert \theta x + (1 - \theta)y \rVert \le \lVert \theta x \rVert + \lVert (1 - \theta)y \rVert$$

This is enough to prove the convexity of your second set, but not your first set.