# Show scalar multiplication sequence of a Cauchy sequence is also a Cauchy sequence.

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 $$\|\lambda_n x_n - \lambda_m x_m\| < \epsilon$$

• Could you do this if $\lambda_n$ was constantly $\lambda$? – Theo Bendit Sep 11 '19 at 1:54
• @Theo Bendit: Yes. WLOG, assume $\lambda$ is not zero and take it out so we are done. – Saeed Sep 11 '19 at 1:57

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 $$(\lambda_n)$$.

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.

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:

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