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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.

Community
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user162343
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    Can't you just prove that the property of convexity is invariant under scaling and translation? – oxeimon Nov 08 '15 at 18:41
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    It should be sufficient to consider balls centered at origin. – AJY Nov 08 '15 at 18:41
  • 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? – user162343 Nov 08 '15 at 18:43
  • I have found the following. It might help: http://math.stackexchange.com/questions/137738/proving-that-closed-and-open-balls-are-convex – shabo Nov 08 '15 at 18:50
  • what have you found ? – user162343 Nov 08 '15 at 18:51
  • This follows from the convexity of the norm defining the ball (here the $\ell_2$-norm). – dohmatob Nov 09 '15 at 03:02

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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 $x_0$.

Hagen von Eitzen
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