# How to show convexity of a ball in metric space?

If $(X,\|\cdot\|)$ is a normed linear space, then how to show any ball $B(x,r)$ is convex?
I know that if $x,y\in A\subset V$ then $[x,y]\subset A$, where $A$ is a convex subset of vector space $V$ and $[x,y]=\{(1-t)x+ty\mid 0\leq t \leq 1\}$. Please give me some hint.

I tried the following:

Claim. $[a,b] \subset B(x,r)$
Let $a,b \in B(x,r).$ Then I get $\|x-a\|<r$ and $\|x-b\|<r$.

Is it right?

• Hint: Use the triangle inequality. – Alexander Thumm Aug 27 '12 at 15:07
• Do you mean "if $x,y\in A\subset V$, then $[x,y]\subset A$ where $A$ is a convex subset of $V$"? – Kuba Helsztyński Aug 27 '12 at 15:24
• @KubaHelsztyński yes – Siddhant Trivedi Aug 27 '12 at 15:27

Suppose $a,b \in B(x,r)$. Then $\|a-x\| < r$, and similarly for $b$. You want to show that for any $t \in [0,1]$, $ta+(1-t)b \in B(x,r)$. To do this we must show that $\|ta+(1-t)b-x\| < r$.

Using the triangle inequality we have: $$\|ta+(1-t)b-x\| = \|t(a-x)+(1-t)(b-x)\| \leq t \|a-x\|+(1-t) \|b-x\|$$ $$< t r +(1-t)r = r,$$ hence $ta+(1-t)b \in B(x,r)$.

• why u r taking b=b-x? – Siddhant Trivedi Aug 27 '12 at 15:44
• It a was a cut & paste induced typo.! Thanks for catching it. – copper.hat Aug 27 '12 at 15:55
• I want to understand please tell me. in the first step why u r taking b=b-x? – Siddhant Trivedi Aug 27 '12 at 15:59
• Can you explain where I am taking $b=b-x$ please? I had a typo. earlier, but now I don't know what you are referring to. Remember, $x=t x + (1-t)x$, and $t\geq0, 1-t \geq 0$. – copper.hat Aug 27 '12 at 16:02
• @SiddhantTrivedi $x$ was subtracted in first expression, because you want to show that distance between points $ta+(1-t)b$ and $x$ is less than $r$. That was not assuming $b=b-x$, but artifical $+tx-tx$ in order to reformulate $\|ta+(1-t)b-x\|$ in a convenient way. – Kuba Helsztyński Aug 27 '12 at 16:14

Since convexity is translation invariant, it is sufficient to show that $B(\mathbf{0},r)$ is convex.

In order to do that, take arbitary points $a,b\in B(\mathbf{0},r).$ We will show that any convex combination of $a$ and $b$, namely $ta+(1-t)b$, belongs to $B(\mathbf{0},r).$ In order to do that, fix $t\in [0,1]$ and note that $$\|ta+(1-t)b\|\stackrel{\triangle}{\le}\|ta\|+\|(1-t)b\|\stackrel{\text{homogeneity}}{=}t\|a\|+(1-t)\|b\|<tr+(1-t)r=r.$$ Thus $ta+(1-t)b\in B(\mathbf{0},r)$ for all $t\in [0,1]$.

In other words, given any points $a,b$ from the ball one has $[a,b]\subset B(\mathbf{0},r)$, which proves desired convexity.

You have listed the immediate consequences of $a,b \in B(x,r)$. Now you want to show that $[a,b] \subset B(x,r)$. You are already given in the statement of the question that an arbitrary point on the line segment $[a,b]$ can be expressed as $(1-t)a + tb; 0\leq t \leq 1$. One way to proceed is to show that $(1-t)a + tb \in B(x,r)$ for any $0\leq t \leq 1$.

You could start with a specific example for $t$, say $t=1/2$. Then write out what it would mean for $\frac{1}{2}a + \frac{1}{2}b \in B(x,r)$. See at that point if you can get the triangle inequality to work out for you given what you already know.

If you can get the proof to work out for $t=1/2$ then you should be able to write out the proof for any $t$ where $0\leq t \leq 1$. Hope this helps.