Proving That Every Norm Defines a Metric

A proof of a claim I made in my previous entry.

Written by Giu · 07 Feb 2012

At the end of my previous entry I wrote the following sentence (emphasis mine):

And exactly this is the reason why a norm function ||·|| of a normed vector space (V, ||·||) always induces a metric[...]

This is quite a bold statement to make, and as it is with such a mathematical statement, you need to mathematically proof it.

And so, as an excercise I set myself to work and wrote a proof of the statement on paper (see bottom of this entry), and typed it out afterwards because of possible readability issues.

The statement has been already proven, mind you, and it is not one of the more complicated proofs, but still, it was a nice exercise.

Proof: Every Norm Defines a Metric

Let (V, ||·||) be a normed vector space, with ||·||: V x V → R being the norm and R being the set of real numbers.

By definition of a norm, ||·|| satisfies the following properties:

1. ||a · v|| = |a| · ||v||, a ∈ R, v ∈ V
2. ||u + v|| ≤ ||u|| + ||v|| (triangle inequality)
3. ||v|| ≥ 0 and ||v|| = 0 ⇔ v is the zero vector

We define D_V: V x V → R as D_V(x,y) = ||x - y||.

D_V is a metric if and only if it satisfies the following properties:

1. D_V(x,y) ≥ 0 and D_V(x,y) = 0 ⇔ x = y (positive definite)
2. D_V(x,y) = D_V(y,x) (symmetry)
3. D_V(x,z) ≤ D_V(x,y) + D_V(y,z) (triangle inequality)

In the following steps we will proof that D_V satisfies the properties of a metric.

Proof of 1.): Positive Definite Form

||v|| ≥ 0 ⇒ (Let v = x - y) ||x - y|| ≥ 0

D_V(x,y) = 0 ⇔ (Def. of D_V) ||x-y|| = 0 ⇔ (Def. of norm, 3.)) x-y is the zero vector ⇔ x - y = 0 ⇔ x = y

(Note: While going through my notes again I realized that I can simplify the proof for this step; the proof on paper is a little bit longer)

It follows that D_V is positive definite.

Proof of 2.): Symmetry

D_V(x,y) = ||x-y|| = ||(-1) · (y - x)|| = (Def. of norm, 1.)) |(-1)| · ||y - x|| = ||y - x|| = D_V(y,x)

It follows that D_V is symmetric.

Proof of 3.): Triangle Inequality

D_V(x,z) = ||x - z|| = ||x - y + y - z|| ⇒ (Def. of norm, triangle inequality) ||x - y + y - z|| ≤ ||x - y|| + ||y - z|| ⇒ ||x - z|| ≤ ||x - y|| + ||y - z|| ⇒ D_V(x,y) ≤ D_V(x,y) + D_V(y,z)

It follows that D_V satisfies the triangle inequality.

By having proofed all the three properties of a metric for D_V, it follows that D_V is a metric, and more importantly, that every norm ||·|| defines a metric D_V.

Q.E.D.

Addendum: Proof On Paper