If $C$ is convex then $\cup_{y\in C} B(y,r)$ is convex.

Question

I am studying for an upcoming exam on convex optimization and one of the practice exercises that I am working through is the following;

Let $C\subseteq \mathbb{R}^n$ be a convex set. Is the set

$$\mathcal{C} := \bigcup_{y \in C} B(y,r), \qquad r>0$$

also convex? If yes prove it and if not then give a counter example.

I have a hunch that this is set is convex, and I remember proving so at the start of the semester. But I have lost my working and as usual when revisiting a problem I am struggling to view it in a new instead of just trying to remember what I did.

Hints are favorable, but answers are also accepted. Cheers in advance for the help.

Thank you, that what was a typo, it should be $\mathbb{R}^n$. In $\mathbb{R}$ this is trivial since $\mathcal{C}$ is just an open interval. — Jandré Snyman, Jun 15 '19 at 10:08
You claim that $\mathcal C$ is convex, then ask if it is convex? I don't get it. — Rodrigo de Azevedo, Jun 15 '19 at 10:21

score 2 · Accepted Answer · answered Jun 15 '19 at 10:10

2

If $x_1,x_2\in\bigcup_{y\in C} B(y,r)$, then $x_i=y_i+z_i$ for $y_i\in C$, $z_i\in B(0,r)$. Now use convexity of $C$ and $B(0,r)$ to show that, for $\lambda\in[0,1]$, $$ \lambda y_2+(1-\lambda)y_1 \in C\text{ and } \lambda z_2+(1-\lambda)z_1 \in B(0,r). $$ Hence $$ \lambda x_2+(1-\lambda)x_1 = (\lambda y_2+(1-\lambda)y_1) + (\lambda z_2+(1-\lambda)z_1) \in \bigcup_{y\in C}B(y,r). $$

answered Jun 15 '19 at 10:10

user10354138

33,239

wow! That's super simple! From what I remember of I did, my solution wasn't nearly as simple. Thank you for that. – Jandré Snyman Jun 15 '19 at 10:14

If $C$ is convex then $\cup_{y\in C} B(y,r)$ is convex.

1 Answers1