# Thread: ||x-y||+||y-z||=||x-z|| iff y= ax + (1-a)z

1. ## ||x-y||+||y-z||=||x-z|| iff y= ax + (1-a)z

Dear mathematician,

I need to show that:

||x-y||+||y-z||=||x-z||

if and only if

y= ax + (1-a)z with 0<=a<=1

In other words, if x,y and z are on the same line, that relation between the norms must be satisfied and the other way around.

I didn't come really far. Writing out the inproducts didn't give me any insight.

Can someone help me please? Would be great!

Cheers, Eva

2. Originally Posted by Eva BSc
Dear mathematician,

I need to show that:

||x-y||+||y-z||=||x-z||

if and only if

y= ax + (1-a)z with 0<=a<=1

In other words, if x,y and z are on the same line, that relation between the norms must be satisfied and the other way around.

I didn't come really far. Writing out the inproducts didn't give me any insight.

Can someone help me please? Would be great!

Cheers, Eva
You have
$\parallel x-y\parallel+\parallel y-z\parallel = \parallel x-z\parallel$
Now square both sides:
$(\parallel x-y\parallel+\parallel y-z\parallel)^2 = \parallel x-z\parallel^2$
Then bring x-y and y-z into play on the right side, like this
$\parallel x-y\parallel^2+2\parallel x-y\parallel \parallel y-z\parallel +\parallel y-z\parallel^2=\parallel (x-y)+(y-z)\parallel^2$
Then multiply out (using the inner product on the right)
$\parallel x-y\parallel^2+2\parallel x-y\parallel \parallel y-z\parallel = \parallel x-y\parallel^2+2+\parallel y-z\parallel^2$
Thus it follows:
$\parallel x-y\parallel \parallel y-z\parallel =
$

(I am assuming a vector space over $\mathbb{R}$ here.) Now consider this last equation: it should remind you of the Cauchy-Schwarz inequality. So you are asked to prove under what condition equality holds in the Cauchy-Schwarz inequality. Quite possibly, you can simply refer to your lecture notes on the Cauchy-Schwarz inequality.

3. Hi Failure,

Thanks, that really helped me!

From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

So (x-y)= b(y-z) with a real number b.

Now why does this imply that

y= ax + (1-a)z with 0<=a<=1 ???

In R2 it is obvious, but how to show this for Rn ??

Cheers, Eva

4. Originally Posted by Eva BSc
Hi Failure,

Thanks, that really helped me!

From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

So (x-y)= b(y-z) with a real number b.

Now why does this imply that

y= ax + (1-a)z with 0<=a<=1 ???

In R2 it is obvious, but how to show this for Rn ??

Cheers, Eva
Linear dependence of x-y and y-z is not equivalent to y= ax + (1-a)z with 0<=a<=1. But you know more than that x-y and y-z are linearly dependent: in addition, you now that $ \geq 0$, because the right side of my last equation was not the absolute value of that inner product but the value of the inner product itself.

5. Originally Posted by Eva BSc
Hi Failure,

Thanks, that really helped me!

From Cauchy-Schwartz it follows that (x-y) and (y-z) are lineary dependent.

So (x-y)= b(y-z) with a real number b.
Actually, that is not quite true. Linear dependence of x-y and y-z means that there exists a b such that x-y=b(y-z) or that b(x-y)=y-z.

Now why does this imply that

y= ax + (1-a)z with 0<=a<=1 ???

In R2 it is obvious, but how to show this for Rn ??

Cheers, Eva
Let me add to my first reply to this question: from $\geq 0$, together with the two linear dependence cases, you can further infer that $b\geq 0$.
Then you solve the two cases, x-y=b(y-z) or b(x-y)=y-z, for y, which gives $y=\tfrac{1}{b+1}x+\tfrac{b}{b+1}z$ or $y=\tfrac{b}{b+1}x+\tfrac{1}{b+1}z$. Finally you set $a := \tfrac{1}{b+1}$ and $a := \tfrac{b}{b+1}$, respectively.

And these two cases are indeed equivalent to there existing an a, 0 <= a <= 1, such that y=ax+(1-a)z.