Show if $\{x_n\}$ is a Cauchy sequence in $X$ then the sequence of norms $\{\|x_n\|\}$ is a Cauchy sequence of real numbers

Question

I am trying to prove that if $\{x_n\}$ is a Cauchy sequence in $X$ then the sequence of norms $\{\|x_n\|\}$ is a Cauchy sequence of real numbers.

This is the proof:

Since $\|u_m\| - \|u_n\| \leq \|u_m - u_n\|$ the sequence is a Cauchy sequence in $\mathbb{R}$

I am not sure what the question is trying to ask, surely the question is trivial?

I understand what a Cauchy sequence is but why does the question specify "real numbers"?

How does the proof answer the question?

Answer 1

The hypothesis that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence in $X$ means that

for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $\|x_k-x_\ell\|<\epsilon$ whenever $k,\ell\ge m$.

The desired conclusion, that $\langle\|x_n\|:n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$, means that

for each $\epsilon>0$ there is an $m\in\Bbb N$ such that $\big|\|x_k\|-\|x_\ell\|\big|<\epsilon$ whenever $k,\ell\ge m$.

As you can see, these are not the same statement: the first is a statement about a sequence of vectors in $X$, and the second is a statement about the sequence of the norms of those vectors, which is a sequence of real numbers. Thus, there is definitely something to be proved.

To prove it, suppose that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy in $X$, and let $\epsilon>0$ be arbitrary. By hypothesis there is an $m\in\Bbb N$ such that $\|x_k-x_\ell\|<\epsilon$ whenever $k,\ell\ge m$. Now use the fact, mentioned in your question, that

$$\|u\|-\|v\|\le\|u-v\|\tag{1}$$

for any $u,v\in X$: since $(1)$ holds for all $u$ and $v$, it’s equally true that

$$-(\|u\|-\|v\|)=\|v\|-\|u\|\le\|v-u\|=\|u-v\|\;.\tag{2}$$

$(1)$ and $(2)$ together imply that

$$\big|\|u\|-\|v\|\big|\le\|u-v\|\;.$$

Now apply this to any $x_k$ and $x_\ell$ with $k,\ell\ge m$:

$$\big|\|x_k\|-\|x_\ell\|\big|\le\|x_k-x_\ell\|<\epsilon\;.$$

Thus, $\big|\|x_k\|-\|x_\ell\|\big|<\epsilon$ whenever $k,\ell\ge m$, and $\langle\|x_n\|:n\in\Bbb N\rangle$ is indeed a Cauchy sequence in $\Bbb R$.

Answer 2

$$ \|u\| = \overbrace{\|(u-v)+v\| \le \|u-v\| + \|v\|}^\text{This is a basic property of norms.}; $$ therefore $$ \|u\|-\|v\| \le \|u-v\|. \tag 1 $$ But the same reasoning can show that $$ \|v\|-\|u\| \le \|v-u\|. \tag 2 $$ To say that $(1)$ and $(2)$ are both true is to say that $$ \left| \|u\| - \|v\| \right| \le \|u-v\|. $$ Consequently if you've shown that $\|u-v\| \le \varepsilon$, you can conclude that $\left| \|u\|-\|v\| \right| \le \varepsilon$.