0
$\begingroup$

Here's my working:

$\|\mathbf{u} -\mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + (- 2\, \mathbf{u}\,\bullet\mathbf{v})$

Since, by the Cauchy-Schwarz theorem, $|\mathbf{u}\,\bullet\mathbf{v}| \leq \|\mathbf{u}\| \cdot\|\mathbf{v}\|$,

$-(\mathbf{u} \cdot\mathbf{v})$ is also $\leq \|\mathbf{u}\| \cdot\|\mathbf{v}\|$ thanks to the modulus operator.

So, $\|\mathbf{u} -\mathbf{v}\|^2 \leq \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + 2\,\|\mathbf{u}\| \cdot\|\mathbf{v}\|$

From here onwards, it can be shown that indeed $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $ with a little algebra.

So, I seem to have proved that $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $, however I am wary of this. Searching in Linear Algebra books, I was not able to find this result anywhere. I found only the "reverse triangle inequality", which is not the same as this. I suspect that I have made a mistake in my proof. Is my proof correct?

$\endgroup$
  • 3
    $\begingroup$ Your proof is ok. You may have found the triangle inequality $$\lVert x+y\rVert \le \lVert x\rVert +\lVert y\rVert$$ What happens if you set $x=u$ and $y=-v$ ? $\endgroup$ – user228113 May 14 '16 at 17:54
  • $\begingroup$ So are you saying that $\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\| $ is the same as the triangle inequality? $\endgroup$ – hb20007 May 14 '16 at 17:55
  • $\begingroup$ Well, yes.${}{}$ $\endgroup$ – user228113 May 14 '16 at 17:56
  • 2
    $\begingroup$ Since $\lVert -v\rVert = \lVert v\rVert$, yes. $\endgroup$ – Clement C. May 14 '16 at 17:56
  • 1
    $\begingroup$ an interesting question is : if $\|cu\| = |c| \ \|u\|$, $\ \|u\| \ge 0$, $\ \|u\|=0 \implies u=0$ and $\|u+v\|^2 \le \|u\|^2+\|v\|^2 + 2 \|u\| \|v\|$, do we get $\|u+v\| \le \|u\| + \|v\|$ i.e. that $\|.\|$ is a norm ? en.wikipedia.org/wiki/Norm_(mathematics)#Definition $\endgroup$ – reuns May 14 '16 at 19:12
0
$\begingroup$

The proof is correct.

$\|\mathbf{u} -\mathbf{v}\|\leq \|\mathbf{u}\| + \|\mathbf{v}\|$ is the same as the triangle inequality: $\|\mathbf{x} +\mathbf{y}\|\leq \|\mathbf{x}\| + \|\mathbf{y}\|$.

Since $\|-\mathbf{v}\|= \|\mathbf{v}\|$.

Full credit goes to G. Sassatelli and Clement C. for providing the full answer in comments.

Answer derived from their comments based on SE policy here.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.