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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.

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up vote 1 down vote accepted

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

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