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If $(X,\|\cdot\|)$ is a normed linear space, then how to show any ball $B(x,r)$ is convex? I tried the following: Claim. $[a,b] \subset B(x,r)$ Is it right? |
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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 $ta+(1-t)b \in B(x,r)$. |
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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. |
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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. |
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