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Let $X$ be a normed nonempty space and $x \in X$. We define the set:$$B(x,r)=\{y \in X:\|y-x\|<r\}$$; I have to show that:$$B(x+x',r+r')=B(x,r)+B(x',r')$$. I proved an inclusion like that: let $a \in B(x,r)+B(x',r')$ ; so $a=a_{1}+a_{2}$, with $a_{1}\in B(x,r)$ and $a_{2}$ in the other ball; we have that $$\|x-a_1\|<r, \|x'-a_2\|<r'$$; suming up, we obtain: $\|x+x'-a_1-a_2\|\le \|x-a_1\|+\|x-a_2\|<r+r'$; so $\|x+x'-a\|<r+r'$ and we obtain that $a \in B(x+x',r+r')$; for the reverse inclusion, I dont see a solution at all; let $a \in B(x+x',r+r'$; so we have that $\|x+x'-a\|<r+r'$; I have to find $a_1,a_2 $, such that $a=a_1+a_2$ and $a_1\in B(x,r)$ and $a_2$ in the other ball. Is a useful idea to use that these balls are convex?

Michael Hardy
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user308560
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1 Answers1

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For any $a \in B(x+x', r + r')$, let $a = x+x' + y$, where $\|y\| \leq r + r'$, then $a$ can also be written as $$a = x + x' + \dfrac{r}{r+r'}y + \dfrac{r'}{r+r'}y$$ you can easily check $x + \dfrac{r}{r+r'}y \in B(x, r)$ and $x' + \dfrac{r'}{r+r'}y \in B(x',r')$

Petite Etincelle
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