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Let $\{x_n\}$ a Cauchy sequence in a normed vector space $X$. Is $$y_n = \frac{x_n}{\|x_n\|}$$ another Cauchy sequence in $D = \{x\in X : \|x\| = 1\}$?

Remark: The idea is prove that if $D$ is complete, then $X$ is also complete. Thanks so much.

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    $\begingroup$ notice that $|\!|x_n|\!|$ is cauchy in $\mathbb{R}$ so it converges. if $\lim _{n\rightarrow \infty}|\!|x_n|\!| \not \rightarrow 0$ then you have $y_n$ is cauchy, otherwise as pointed by Hourieh it is uncertain. $\endgroup$ – clark Mar 27 '13 at 22:07
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    $\begingroup$ But otherwise $x_n\to 0\in X$, so that's ok. Can you spell out the proof that $y_n$ is Cauchy if $\lim\|x_n\|\ne 0$? $\endgroup$ – Berci Mar 27 '13 at 22:21
  • $\begingroup$ If I understand right. When $\|x_n\|\rightarrow a\neq 0$, then exists a subsequence $x_{n_k} \neq 0$, and I can consider $$y_k = \frac{x_{n_k}}{\|x_{n_k}\|}, \mbox{ good define!}$$ Later, (taking $y_n = \frac{x_n}{\|x_n\|}$ for brevity) $$\|y_n-y_m\| = \left\|\frac{x_n}{\|x_n\|} - \frac{x_m}{\|x_m\|}\right\| = \frac{1}{\|x_m\|\|x_m\|}\left\|x_n\|x_m\| - x_m\|x_n\|\right\| \leq \frac{2}{\|x_m\|}\|x_n-x_m\|$$ where I used the triangular inequality and the fact of $\left|\|x_n\| \frac{\phantom{x}}{ } \|x_m\|\right| \leq \|x_n-x_m\|$. But later, I don't know :S $\endgroup$ – FASCH Mar 27 '13 at 22:34
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Consider: $$ x_n = \frac{(-1)^n}{n} $$

Then: $$ y_n = (-1)^n $$

$x_n$ is Cauchy in $\Bbb R$ as it converges to $0$. However, $y_n$ keeps alternating between $1$ and $-1$ and it's not Cauchy in $\{-1, 1\}$.

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    $\begingroup$ I see, do you have any idea to prove that X is complete if D is complete? Thanks so much. $\endgroup$ – FASCH Mar 27 '13 at 22:09
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    $\begingroup$ @FASCH as you see in my comment you have two possibilities the norm converges to zero then $x_n \rightarrow 0$ and otherwise it is cauchy in $D$ which is complete so it also converges. $\endgroup$ – clark Mar 27 '13 at 22:14
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Let $f(x)=\frac{x}{\|x\|}$. Let $r >0$, then $f$ is uniformly continuous on $B(0,r)^c$. To see this, suppose $\|x-y\| < \delta$. Then $|\|x\|-\|y\|| \leq \|x-y\| < \delta$ as well, and we have: \begin{eqnarray} \|f(x)-f(y)\|&=& \left\Vert \frac{x}{\|x\|} - \frac{y}{\|x\|} +\frac{y}{\|x\|} - \frac{y}{\|y\|} \right\Vert \\ &\le& \frac{1}{\|x\|}\|x-y \| + \left| \frac{1}{\|x\|} - \frac{1}{\|y\|} \right| \|y \| \\ &=& \frac{1}{\|x\|}\|x-y \| + \frac{1}{\|x\|} |\|x\|-\|y\| | \\ &\le& 2 \frac{\delta}{r} \end{eqnarray}

Note that uniformly continuous functions map Cauchy sequences into Cauchy sequences. Also, if $x_n$ is Cauchy, then $\lim_n \|x_n\|$ converges, as $\mathbb{R}$ is complete. (Also, note that $\|x_n\|$ is bounded.)

Now suppose $x_n$ is Cauchy. If $l=\liminf_n \|x_n\| >0$, then for $n$ sufficiently large, $x_n \in B(0,\frac{l}{2} )^c$, and then $f(x_n)$ is Cauchy, hence $f(x_n) \to \hat{f}$ for some $\hat{f} \in D$, and $n = \lim_n \|x_n\|$ exists. Then we have \begin{eqnarray} \|x_n - n \hat{f}\| &=& \left\Vert \|x_n\| f(x_n) - n \hat{f} \right\Vert \\ &=& \left\Vert \|x_n\| f(x_n) -n f(x_n) + n f(x_n)- n \hat{f} \right\Vert \\ &\le& \left| \|x_n\|-n \right| + n \|f(x_n)-\hat{f}\| \end{eqnarray} and it follows that $\lim_n x_n = n \hat{f}$.

If $l=0$, pick some element $d \in D$, let $B$ be an upper bound for $\|x_n\|$ and let $x_n'=x_n+(B+1)d$. Then $\|x_n'\| \ge 1$, and $x_n'$ is also Cauchy. By the above result, $x_n' \to \hat{y}$ for some $\hat{y}$. Hence $\lim_n x_n = \hat{y}-(B+1)d$.

It follows that $X$ is complete.

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  • $\begingroup$ @AymanHourieh: Thanks! My Latex is slowly evolving... $\endgroup$ – copper.hat Mar 27 '13 at 23:08
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I can not see clear the solution at all. That's why I am trying to use the fact (hint of my professor) of if a Cauchy sequence have a convergent subsequence, then the complete Cauchy sequence converges.

Thus, in the case of $\lim\limits_{n\rightarrow +\infty}\|x_n\| \neq 0$, I considered a subsequence $\{x_{n_k}\}$ that hold $x_{n_k}\neq 0, \forall k\in\mathbb{N}$ and then the new sequence $$y_k = \frac{x_{n_k}}{\|x_{n_k}\|}$$ is good defined.

Later, $\{x_{n_k}\}$ is also a Cauchy sequence and $y_k \neq 0$ for all $k\in\mathbb{N}$. So, the next step, is prove that $y_k$ is Cauchy in $D$ and for the fact, I need to prove that exists of a $M > 0$ that $M \leq x_{n_k}$ for all $k\in\mathbb{N}$. But I don't know how I can justify the existence of $M$. Please help me. Thanks

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