For $C$ open, bounded, and convex, is it true that $x+r C\subset x + 3rC$, $x\in \mathbb R^d, r>0$?

Question

I have a question regarding transformations of convex sets. Given an open, bounded and convex set $C$, is it true that $$ x+r C\subset x + 3rC \qquad x\in \mathbb R^d, \ r>0 \quad? $$

The reason I am asking is that the set $(x - a, x + a)$ is contained in $(x - 3a, x+ 3a)$. Is this situation true more generally for convex sets $C$ in $\mathbb R^d $?

Further, if the claim in the title is true, is the conditions of openness and boundedness of $C$ be necessary?

Thanks in advance!

score 2 · Accepted Answer · answered Jun 15 '20 at 09:00

2

No. Take $d=1, C=(1,2), x=0$ and any $r>0$.

answered Jun 15 '20 at 09:00

Kavi Rama Murthy

311,013

For $C$ open, bounded, and convex, is it true that $x+r C\subset x + 3rC$, $x\in \mathbb R^d, r>0$?

1 Answers1