How to proof that these two statements are equivalent?

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Hey there, i'm having a lot of trouble with this task right here. Let W be a real vector space and d a metric on it. Proof that the two statements are equivalent: (1) The metric d is both homogeneous and translation invariant and (2) There is exactly one norm ∥ · ∥ on W such that the metric induced by this norm with d agrees, i.e. d(x, y) := ∥x−y∥ for all x, y ∈ W.

What i found out already:

The metric d is homogeneous if d(λx,λy) = |λ|d(x, y) for all x, y ∈ W and all λ ∈ R

Also, we call the metric d translation-invariant, if for all x, y,z ∈ W we have d(x + z, y + z) = d(x, y)

I already spoke to many of my classmates about this but noone seemed to have a clue on how to proof this. I'd be very thankfull for support on this task :)

Best regards
 
d(x, y) = ||x−y|| for all x, y ∈ W.

Proof:
Given a norm, the function d(x, y) = ||x − y|| is translation in-variant in the sense that d(x + a, y + a) = ||x + a − y − a|| = d(x, y), scalar homogeneous in the sense that d(λx, λy) = |λ|d(x, y).

Conversely if d(x, y) is translation invariant, then d(x, y) = d(x − y, y − y) = d(x − y, 0) and we
can define ||x|| = d(x, 0) so that d(x, y) = ||x − y||.

We will now show that the axioms for the norm correspond precisely to the axioms for a distance.

d satisfies the triangle inequality <=> the norm ||.|| does

d(x, y) ≤ d(x, z) + d(z, y)

corresponds to ||x − y|| ≤ ||x − z|| +||z − y||.

Similarly d(x, y) = d(y, x) corresponds to ||x − y|| = ||y − x||;
d(x, y) ≥ 0 is ||x − y|| ≥ 0,

while d(x, y) =0 <=> x = y becomes ||x − y|| = 0 <=> x − y = 0.
The scale-homogeneity of the metric supplies the final axiom for the norm. This invariance under translations and scaling has the following easy con-sequences.
 
I think i get the idea of the proof, but whats with the image in row 8 and 13? Its just a cross. Is it important?
Otherwise, thank you very much for helping me out :D
 
To prove the equivalence of the two statements, we need to show that:

  1. If the metric d is both homogeneous and translation invariant, then there is exactly one norm ∥⋅∥∥⋅∥ on W such that the metric induced by this norm with d agrees.
  2. If there is exactly one norm ∥⋅∥∥⋅∥ on W such that the metric induced by this norm with d agrees, then the metric d is both homogeneous and translation invariant.
Let's prove each direction:

Proof of (1) implies (2):
Assume d is both homogeneous and translation invariant.

Firstly, let's define the norm ∥⋅∥∥⋅∥ induced by the metric d as follows: ∥x∥=d(x,0)

We'll prove that this ∥⋅∥∥⋅∥ satisfies the properties of a norm:

  1. Non-negativity: ∥x∥=d(x,0)≥0 for all x∈W, with equality if and only if x=0.
  2. Homogeneity: ∥λx∥=d(λx,0)=∣λ∣d(x,0)=∣λ∣∥x∥ for all x∈W and λ∈R.
  3. Triangle Inequality: ∥x+y∥=d(x+y,0)≤d(x,0)+d(y,0)=∥x∥+∥y∥ for all x,y∈W.
Now, let's show uniqueness. Suppose there is another norm ∥~∥~ such that d(x,y)=∥ x−y∥ for all x,y∈W. Then for any x∈W, d(x,0)=∥x−0∥=∥~(x−0)=∥~(x)

Thus, the norm induced by d is unique.

Proof of (2) implies (1):
Assume there is exactly one norm ∥⋅∥∥⋅∥ on W such that the metric induced by this norm with d agrees.

For homogeneity, we already know that the norm ∥⋅∥∥⋅∥ induced by d satisfies homogeneity.

For translation invariance, let's use the definition of the norm induced by ∥x−y∥=d(x,y)

Let's consider z=y−x. Then x+z=y, and by translation invariance of d, we have: d(x+z,y+z)=d(y,x)=∥y−x∥=∥z∥

Thus, d(x+z,y+z)=∥z∥, which implies translation invariance.

Hence, both properties hold.

Therefore, we have shown both directions, and the equivalence between the two statements is proven.

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To prove the equivalence between the two statements, let's tackle it step by step.

(1) The metric d is both homogeneous and translation invariant.

(2) There is exactly one norm ∥⋅∥ on W such that the metric induced by this norm with d agrees, i.e., d(x,y):=∥x−y∥ for all x,y∈W.

Let's start by proving that statement (1) implies statement (2).

Assume that d is both homogeneous and translation invariant. We aim to show that there exists exactly one norm∥⋅∥ on W such that d(x,y)=∥x−y∥ for all x,y∈W.

Define the norm ∥x∥:=d(x,0) for all x∈W, where 0 is the zero vector in W.

First, we need to show that this norm is well-defined, meaning it satisfies the properties of a norm:

∥x∥≥0 for all x∈W: This follows from the non-negativity of the metric d.

∥x∥=0 if and only if x=0: This follows from the properties of the metric d.

∥λx∥=∣λ∣∥x∥ for all x∈W and λ∈R: This follows from the homogeneity property of the metric d.

∥x+y∥≤∥x∥+∥y∥ for all x,y∈W: This follows from the triangle inequality property of the metric d.

Now, we need to show that any norm induced by d satisfies the condition

d(x,y)=∥x−y∥ for all x,y∈W.

Let ∥⋅∥′ be any norm on W such that d(x,y)=∥x−y∥′ for all x,y∈W.

Consider ∥x∥′=∥x−0∥′=d(x,0)=∥x∥ for all x∈W,

where we have used the translation invariance property of d. Hence, ∥⋅∥′=∥⋅∥, and there is exactly one norm ∥⋅∥ on W such that the metric induced by this norm with d agrees.

Now, let's prove that statement (2) implies statement (1).

Assume that there exists exactly one norm ∥⋅∥ on W such that the metric induced by this norm with d agrees. We aim to show that d is both homogeneous and translation invariant.

Let ∥⋅∥ be the unique norm satisfying d(x,y)=∥x−y∥ for all x,y∈W.

Homogeneity: For any λ∈R and x,y∈W, we have:

d(λx,λy)=∥λx−λy∥=∣λ∣∥x−y∥=∣λ∣d(x,y)

Translation invariance: For any x,y,z∈W, we have:

d(x+z,y+z)=∥(x+z)−(y+z)∥=∥x−y∥=d(x,y)

Therefore, we have shown that statement (2) implies statement (1).

Conversely, statement (1) implies statement (2). Hence, the two statements are equivalent, and the proof is complete.


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