How to proof that these two statements are equivalent?

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

 

should be: <=>

 

You dropped this :crown:

 

You dropped this :crown: ??? what do you mean by that

 

Means thanks for the help king, thats why a crown

 

then should be queen

 

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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