Proving that closed (and open) balls are convex

Question

Let $X$ be a normed linear space, $x\in X$ and $r>0$. Define the open and closed ball centered at $x$ as $$ B(x, r) = \{y \in X : \Vert x − y\Vert < r\} $$ $$ \overline{B}(x, r) = \{y \in X : \Vert x − y\Vert \leq r\}. $$ Then $B(x, r)$ and $\overline{B}(x, r)$ are convex.

I tried to prove this, but either my calculation is incorrect, or I am on the wrong path:

I aim to show for the closed ball $\overline{B}(x,r)$ (for open ball I assume the proof is similar). Suppose $y,z \in \overline{B}(x, r)$. Then $\Vert x − y\Vert \leq r$ and $\Vert x − z\Vert \leq r$. We must show that $t \in [0,1]$ implies $ty + (1-t)z \in \overline{B}(x,r)$. But $t \in [0,1]$ implies $$ \Vert ty + (1-t)z - x\Vert = \Vert t(y-z) + z - x\Vert \leq |t| \Vert y-z\Vert + \Vert z-x\Vert \leq |t|(\Vert y-x\Vert + \Vert x-z\Vert) + \Vert z-x\Vert < |t|(2r) + r = r(2|t| + 1), $$ which is not necessarily $\leq r$. We probably wanted to end up with $< |t|r$ or $\leq |t|r$ as our final inequality.

Thanks in advance.

score 31 · Accepted Answer · answered Apr 27 '12 at 17:55

31

Hint: $$ ty+(1-t)z-x=t(y-x)+(1-t)(z-x) $$

answered Apr 27 '12 at 17:55

Norbert

56,803

1

That trick is a bit too sneaky for my liking, haha. Thanks – Adam Rubinson Apr 27 '12 at 18:10
8

In mathmatics there are much more neat tricks. This one is standard. – Norbert Apr 27 '12 at 18:39
I'm sure I've seen this trick before. Just didn't recognize it in this situation... – Adam Rubinson Apr 27 '12 at 18:45
Sorry, but how do you conclude? Thanks a lot :) – user162343 Nov 08 '15 at 18:19
Right, but have you used that $B_1(0)$ is convex? – user162343 Nov 08 '15 at 19:14
I proved this $\phantom{}$ – Norbert Nov 08 '15 at 19:16
by the way is there some source to find some neat and standart tricks? – Micheal Brain Hurts May 18 '20 at 18:48
@Anil - Not sure. I think in general you learn each individual "trick" when you come across them. It's a bit like the saying, "I'll cross that bridge when I get there". But there may be some sort of list of these types of algebra tricks, I'm not sure. Maybe ask that as a new question. – Adam Rubinson Jul 24 '20 at 21:30

score 8 · Answer 2 · edited May 03 '20 at 22:29

8

Suppose $x, y$ are in a unit closed ball. Consider $$||z||=||(1-a)x+ay||\leq (1-a)||x||+a||y||\leq 1-a+a=1.$$ So $||z||\leq 1$, so $z$ is in unit close ball.

edited May 03 '20 at 22:29

Peter_Pan

1,836

answered Dec 05 '16 at 15:56

MS KHAN

81

Please use MathJax/LaTeX markup to improve readability! Thanks! – The Count Dec 05 '16 at 16:02

Tunde Ishola · Answer 3 · 2018-02-16T15:25:05.163

2

Let $ B(a,r)=\{y \in X: \|y-a\| < r \}$ be an open ball in a Banach space $X$. Let $x,y \in B(x,r)$ then $\| x-a\| < r $ and $\| y-a\| < r $. $$\|x\|-\|a\| <\| x-a\| < r ~~ and~~ \|y\|-\|a\| <\| y-a\| < r$$ which implies \begin{equation} \|x\| < r + \|a\|,~ \|y\| < r + \|a\|~~~~~~ (1) \end{equation} for any $t \in [0,1]$; $x,y \in X$ using $(1)$ we have $$ \|tx+(1-t)y\| \leq t\|x\|+(1-t)\|y < t(r+\|a\|)+(1-t)(r+\|a\|)=r+\|a\|\\ $$ Since $a$ is the centre (Origin) of the ball $\|a\|=0$. Hence, the result $\square$

edited Feb 16 '18 at 15:25

answered Feb 16 '18 at 15:19

Tunde Ishola

31

1

If $a$ is centre of the ball, then isn't $|a-a|=0$ instead? – Alexy Vincenzo Jan 13 '20 at 04:53

Proving that closed (and open) balls are convex

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