0
$\begingroup$

Let $X$ be a normed space over the field $\mathbb{K}$.

Use the triangle inequality to prove that if a sequence $\{x_n\}$ in $X$ converges in the norm to an element $x \in X$ then $x_n$ is a Cauchy sequence in $X$

This is the proof:

If $u$ is the limit of $\{u_n\}$ then $$||u_n -u_m|| \leq ||u_n -u|| + ||u-u_m|| \to 0$$ as $$M,n \to 0$$

Why does $||u_n -u_m|| \leq ||u_n -u|| + ||u-u_m||$ hold?

$\endgroup$
2
  • 1
    $\begingroup$ That is just the triangle inequality. $\endgroup$
    – ervx
    Commented Apr 21, 2016 at 20:44
  • 3
    $\begingroup$ $\|u_n-u_m\|=\|u_n-u+u-u_m\|=\|(u_n-u)+(u-u_m)\|\le\|u_n-u\|+\|u-u_m\|$ $\endgroup$
    – sinbadh
    Commented Apr 21, 2016 at 20:44

2 Answers 2

2
$\begingroup$

It's a common little 'trick' to add and subtract the same value from something to prove a result.

In this case - as sinbadh points it out in the comments- you have $$ u_n -u_m = u_n - u + u -u_m$$ and so $$ \| u_n -u_m \| = \| u_n - u + u -u_m \| \leq \| u_n - u\| + \|u -u_m \|$$ Where the last step is just the triangle-inequality applied to the vectors $(u_n - u)$ and $(u-u_m)$.

$\endgroup$
1
$\begingroup$

Thats easy: you add and subtract u to the original term: $$ ||u_n -u_m|| = ||u_n -u+u-u_m|| $$ Then you view $u_n-u$ and $u-u_n$ as $a,b$ and get $$ ||u_n -u_m|| = ||u_n -u+u-u_m||=||a+b||\leq||a||+||b||=||u_n -u||+||u-u_m||$$ and since both norms converge to zero we have $$ ||u_n -u_m|| \rightarrow 0$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .