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Let $(x_n)$ be a Cauchy sequence in a normed vector space $V$ and let $(\lambda_n)$ be a convergent sequence in $\mathbb{R}$. Show sequence $(\lambda_n x_n)$ is also Cauchy sequence.

My try: Since $(x_n)$ is a Cauchy sequence, for any given $\epsilon$, there is $N \in \mathbb{N}$ such that for all $n,m \geq N$ we have $$ \|x_n - x_m\| < \epsilon $$ We need to show $$ \|\lambda_n x_n - \lambda_m x_m\| < \epsilon $$

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  • $\begingroup$ Could you do this if $\lambda_n$ was constantly $\lambda$? $\endgroup$ – Theo Bendit Sep 11 '19 at 1:54
  • $\begingroup$ @Theo Bendit: Yes. WLOG, assume $\lambda$ is not zero and take it out so we are done. $\endgroup$ – Saeed Sep 11 '19 at 1:57
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Use the usual trick. Consider \begin{align*} \|\lambda_n x_n - \lambda_m x_m\| &= \|\lambda_n x_n - \lambda_n x_m + \lambda_n x_m - \lambda_m x_m\| \\ &\le |\lambda_n| \|x_n - x_m\| + |\lambda_n - \lambda_m|\|x_m\|. \end{align*} Now, recall that Cauchy sequences are bounded. There must exist $A, B$ such that $\|x_n\| \le A$ and $|\lambda_n| \le B$ for all $n$. Thus, $$\|\lambda_n x_n - \lambda_m x_m\| \le A\|x_n - x_m\| + B|\lambda_n - \lambda_m|.$$ Both terms in this sum can be made less than $\varepsilon / 2$ easily using the Cauchiness of the two sequences $(x_n)$ and $(\lambda_n)$.

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Consider the following identity: $$x_ny_n - x_my_m = (x_n-x_m)(y_n - y_m) + x_m(y_n - y_m) + y_m(x_n - x_m) $$ Everything should follow from this pretty quickly.

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Let $(V,\|.\|)$ be normed vector space, and $(x_n)_{n \in \mathbb N} \in V^\mathbb N$ be a cauchy sequence. Moreover, let $(\lambda_n)_{n \in \mathbb N} \in \mathbb R^\mathbb N$ be a convergent sequence with limit $\lambda \in \mathbb R$.

First claim: Both cauchy and convergent sequences are bounded.

For convergent, we have that for some $\epsilon > 0$ there exists $N(\epsilon)$ such that $|\lambda_n - \lambda | < \epsilon$ for $n > N(\epsilon)$.

Clearly then for every $n \in \mathbb N$ we have $|\lambda_n| \le \max\{|\lambda_1|,...,|\lambda_n|, |\lambda| + \epsilon \}$.

For cauchy one, if it wasn't bounded, then we can choose a subsequence $(n_k)_{ k \in \mathbb N}$ such that $|x_{n_k}| > k$ for every $k \in \mathbb N$. Now taking any $\epsilon > 0$, we can't find such $N(\epsilon)$, such that $|x_{n_k} - x_{n_r}| < \epsilon$, for $k,r \ge N(\epsilon)$ because when we take any $N(\epsilon)$, then there exist such $r \in \mathbb N$ so that $|x_{n_r}| > |x_{n_{N(\epsilon)}} | + \epsilon + 1$.

Now as we have this, we can proceed as follows:

Take any $\epsilon > 0$, and let $B$ be a bound for cauchy sequence. We want to show, there exists $N(\epsilon)$, such that for $n,m > N(\epsilon)$ we have $\| \lambda_n x_n - \lambda_m x_m \| < \epsilon$.

Firsly take such $M_1(\epsilon)$, so that for $n > M_1(\epsilon)$ we have $\lambda_n = \lambda + r_n$, where $|r_n| < \frac{\epsilon}{3B}$.

Secondly, take such $M_2(\epsilon)$ so that $\| x_n - x_m \| < \frac{\epsilon}{3\lambda}$ for $n,m > M_2(\epsilon)$

Take $N(\epsilon) = \max\{M_1(\epsilon),M_2(\epsilon)\}$

Then we have:

$ \|x_n\lambda_n - x_m\lambda_m\| = \| x_n(\lambda + r_n) - x_m(\lambda + r_m) \| \le \lambda \|x_n - x_m\| + \|x_nr_n\|+ \|r_mx_m\| \le \lambda \frac{\epsilon}{3\lambda} + B\frac{\epsilon}{3B} + B \frac{\epsilon}{3B} =\epsilon$

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