I want to conclude that $B_r(x_0)$ is convex from the fact that $B_1(0)$ is convex. So I was trying to use the answer provided here Proving that closed (and open) balls are convex

But the thing is that I don't know how to conclude, I think that they may be need that $B_1(0)$ is convex to get the result. Can someone help me to get this result please?

Thanks a lot in advance.

  • 2
    $\begingroup$ Can't you just prove that the property of convexity is invariant under scaling and translation? $\endgroup$
    – oxeimon
    Nov 8, 2015 at 18:41
  • 1
    $\begingroup$ It should be sufficient to consider balls centered at origin. $\endgroup$
    – AJY
    Nov 8, 2015 at 18:41
  • $\begingroup$ Right, but how do you prove that convexity is invariant? I thought is was easier using those inequalities given in that post :), but can you prove the invariance? And why we just need to consider the balls at the origin? $\endgroup$
    – user162343
    Nov 8, 2015 at 18:43
  • $\begingroup$ I have found the following. It might help: math.stackexchange.com/questions/137738/… $\endgroup$
    – shabo
    Nov 8, 2015 at 18:50
  • $\begingroup$ what have you found ? $\endgroup$
    – user162343
    Nov 8, 2015 at 18:51

1 Answer 1


Let $C\subseteq \Bbb R^n$ be convex, let $f\colon \Bbb R^n\to\Bbb R^n$ be an affine linear map (i.e., $f(x)=Ax+b$ with an $n\times n$ matrix $A$ and $b\in\Bbb R^n$). Then $f(C)$ is convex. Indeed, if $x,y\in f(C)$, say $x=f(u), y=f(v)$, then any convex combination $tx+(1-t)y$, $0\le t\le 1$, is also in $f(C)$, namely $$\begin{align}f(tu+(1-t)v)&=A(tu+(1-t)v)+b\\&=tAu+(1-t)Av+tb+(1-t)b\\&=t(Au+b)+(1-t)(Av+b)\\&=tx+(1-t)y.\end{align}$$

Now note that $B_r(x_0)$ is obtained from $B_1(0)$ by (linearly) scaling with $r$ and translating by $x_0$.

  • $\begingroup$ Right, but what if we are not in R? $\endgroup$
    – user162343
    Nov 8, 2015 at 18:53

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