# Proof (metric spaces)

• September 3rd 2010, 07:03 PM
raphw
Proof (metric spaces)
Hello everybody,
I am stuck with a proof of which I think that I have an idea of how to solve it.

I have given two vectors $x,y \in R^k$ with $k >= 3$. I know that $|x-y| = d > 0$ and $r > 0$.

Now I am supposed to show a couple of things and what I am left to do is to prove that there is only one $z \in R^k$ s.t. $2r = d$ and $|z-x| = |z-y| = r$.

OK, now here is what I have done. I know from the Schwarz inequality that $|x-z + z - y| = |x-z| + |z-y|$ is possible in general if I chose the right z. I get this same equality if I set the given definition of d into my other constraint. Also should I be able to solve for x since I have one unknown z and three requirements for z. However, whenever I solve for z I get some nonsence. I am not exactly sure about what I am doing wrong or how else I can show that there is only one z for that all these requirements hold. Can one of you help me. Even a hint would be more than enough. Thanks a ton! Cheers.
• September 3rd 2010, 09:28 PM
Failure
Quote:

Originally Posted by raphw
Hello everybody,
I am stuck with a proof of which I think that I have an idea of how to solve it.

I have given two vectors $x,y \in R^k$ with $k >= 3$. I know that $|x-y| = d > 0$ and $r > 0$.

Now I am supposed to show a couple of things and what I am left to do is to prove that there is only one $z \in R^k$ s.t. $2r = d$ and $|z-x| = |z-y| = r$.

OK, now here is what I have done. I know from the Schwarz inequality that $|x-z + z - y| = |x-z| + |z-y|$ is possible in general if I chose the right z. I get this same equality if I set the given definition of d into my other constraint. Also should I be able to solve for x since I have one unknown z and three requirements for z. However, whenever I solve for z I get some nonsence. I am not exactly sure about what I am doing wrong or how else I can show that there is only one z for that all these requirements hold. Can one of you help me. Even a hint would be more than enough. Thanks a ton! Cheers.

What condition needs to be satisfied for equality to hold in the Cauchy-Schwarz inequality? Use that additional information about x-z and z-y to determine z.
• September 4th 2010, 12:54 PM
raphw
I am not sure if I got your hint. However, here is what I have done.
First, for convenience, I defined: $x' = x - y$ and $y' = y - y = 0$ what should, in my opinion, only "shift" the problem but not change it.

I know that: $|z| = |z - x'| = d/2$ and therefore ${x'}^2 = 2zx'$. Since $d^2 = |x'|^2 = 2zx'$, I know that $z = d^2/2 * x'^{-1}$ and can therefore not be any other value.

Q.E.D.? I'm still confused.
• September 4th 2010, 04:19 PM
Iondor
Quote:

Originally Posted by raphw
I am not sure if I got your hint. However, here is what I have done.
First, for convenience, I defined: $x' = x - y$ and $y' = y - y = 0$ what should, in my opinion, only "shift" the problem but not change it.

I know that: $|z| = |z - x'| = d/2$ and therefore ${x'}^2 = 2zx'$. Since $d^2 = |x'|^2 = 2zx'$, I know that $z = d^2/2 * x'^{-1}$ and can therefore not be any other value.

Q.E.D.? I'm still confused.

No, that's no correct, because the inverse of x' $x'^{-1}$ is not generally defined. x' is a vector. Vectors don't usually have multiplicative inverses.
You have to use the Cauchy-Schwarz inequality here to conclude that x' and z must be collinear:
$d^2 = |x'|^2 = 2zx'$ => $d^2/2 = zx'$
C-S inequality says that $zx' \leq |z||x'|$ with equality iff z and x' are collinear. In this case we have equality . So there is a $\lambda \in \mathbb{R}$, such that $x' \lambda = z$.
From this it follows that $|x'| |\lambda| =|z| => d |\lambda| = d/2 => \lambda = +-\frac{1}{2}$. So $z= \frac{1}{2} x'$ is the only possible value for z.
qed