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UPDATE: Let $x=(x_{n})$ and $y=(y_{n}) \in A$ with $A:=\{x=(x_{n})\in \ell^{2}| \phantom{x} \|x\|\leq 1\}$. Prove that $d:A\times A \rightarrow \mathbb{R}_{+}$ defined by $$d(x,y)=\sum_{n=1}^{\infty}(1/3)^{n}|x_{n}-y_{n}|$$ is bounded.

Let $x,y\in A$ we get

\begin{align*} \sum_{n=1}^{\infty}(1/3)^{n}|x_{n}-y_{n}| &\leq \sum_{n=1}^{\infty}(1/3)^{n}(|x_{n}|+|y_{n}|) \phantom{x} (1) \\ &= \sum_{n=1}^{\infty}(1/3)^{n}|x_{n}|+\sum_{n=1}^{\infty}(1/3)^{n}|y_{n}| \phantom{x} (2)\\ &\leq \sqrt{\sum_{n=1}^{\infty}(1/3)^{n}} \sqrt{\sum_{n=1}^{\infty}|x_{n}|^{2}} + \sqrt{\sum_{n=1}^{\infty}(1/3)^{n}} \sqrt{\sum_{n=1}^{\infty}|y_{n}|^{2}} \phantom{x} (3) \\ &= \left(\frac{1}{1-1/3}\right)^{1/2} \|x_{n}\|_{2}+\left(\frac{1}{1-1/3}\right)^{1/2} \|y_{n}\|_{2} \phantom{x} (4)\\ &= \sqrt{3/2} \|x_{n}\|_{2}+\sqrt{3/2} \|y_{n}\|_{2}.\phantom{x} (5) \end{align*}

Thus, $d(x,y)$ is bounded.

What is used in these steps: (1) Triangle inequality ? (2) What is used here? (3) Cauchy schwarz? (4) and (5) geometric series and standard calculation.

Lech121
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1 Answers1

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(1) Triangle inequality: $|a-b|=|a+(-b)|\leqslant |a|+|-b|=|a|+|b|$.

(2) Linearity of the summation: the limit of a sum is the sum of the limits.

(3) Yes, it's Cauchy-Schwarz inequality.

(4) and (5) Yes.

Davide Giraudo
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  • Thank you very much @Davide. I was just about to update my own answer. I still have one question though about the mapping being well defined. I would really appreciate it if you could give me some feedback on that. – Lech121 Sep 19 '13 at 15:16