3
$\begingroup$

In a normed space and given three points $x,y,z $ the norm satisfies the triangle inequality

$$\|x-y \| \le \|x-z \| + \|z-y \|$$

But is also the intuitively obvious fact from $\mathbb R^2$ true that if $z$ is a point on the line $x+t(x-y)$, $0 < t < 1 $, between $x $ and $y $ then the distances betewwen $x $ and $z $ , and $z $ and $y $ add up to the distance between $x $ and $y $

$$\|x-y\|=\|x-z\|+\|z-y\|$$

? And how would we prove it?

Most grateful for any help!

$\endgroup$
3
$\begingroup$

Since $z = x + t(x-y)$ then the expression

$$ \| x - y \| = \| x - z \| + \| z - y \| $$ becomes $$ \| x - y \| = \| x - x - t(x- y) \| + \| x + t(x-y) - y \| = |t| \| x - y \| + |1 - t| \| x - y \| \equiv \| x - y \|, $$ where the last equality follows from the fact that $t \in (0, 1)$ and hence $|t| = t$; $|1 - t| = 1 - t$.

QED.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Yes this holds, here's the proof :

Let $t \in [0;1]$, $z:= x + t (y-x)$.
One has $z - x = t(y-x)$, and also $y - z = (1-t)(y-x)$. Hence, $$\lVert x - z\rVert + \lVert z - y\rVert = \lVert z - x\rVert + \lVert y - z\rVert =\\ |t|\lVert y - x \rVert + |1-t| \lVert y - x \rVert = \lVert y - x \rVert$$

| cite | improve this answer | |
$\endgroup$

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.