A proof of a general property of norms

By the triangle inequality, we have $\| x-y+y \| \leq \| x-y \| + \| y \|$. Rearranging, we have $\| x-y \| \geq \| x \| - \| y \|$. Similarly, we have $\| y-x \| \geq \| y \| - \| x \|$. Putting these inequalities together, we have $| \| x \| - \| y \| | \leq \| x-y \|$.
  • #1
gucci1
13
0
So I have been asked to prove a result that is supposedly valid for any norm on any vector space. The statement to prove is: | ||x|| - ||y|| | <= ||x - y||

The problem is, I have no idea where to start with this proof. Maybe I'm missing some fundamental property of norms, but it seems that having to prove this for any norm on any vector space rules out any definitions that I might try to apply :-/

Sorry for not having anything to go on here, but I'm lost. Thank you all very much for any help you might be able to offer!
 
Physics news on Phys.org
  • #2
This follows from the triangle inequality. Consider the quantity $\| x-y+y \|$.
 

FAQ: A proof of a general property of norms

What is a norm in mathematics?

A norm in mathematics is a function that assigns a positive length or size to each vector in a vector space. It is a measure of the size or magnitude of a vector and is commonly used in linear algebra and functional analysis.

What is the general property of norms?

The general property of norms is that they must satisfy three conditions: positivity, homogeneity, and the triangle inequality. Positivity means that the norm of a vector must be greater than or equal to zero. Homogeneity states that multiplying a vector by a scalar will result in the norm being multiplied by the absolute value of that scalar. The triangle inequality states that the norm of a sum of two vectors cannot be greater than the sum of their individual norms.

How is the general property of norms proven?

The general property of norms is proven using mathematical proofs. These proofs involve using the axioms of vector spaces and properties of norms to show that the three conditions (positivity, homogeneity, and triangle inequality) are satisfied for any given norm. This proof can be done for any norm in any vector space.

Why is the general property of norms important?

The general property of norms is important because it allows us to use norms as a tool in various mathematical contexts. For example, norms are used in optimization problems, functional analysis, and other areas of mathematics. Understanding and proving the general property of norms helps us to better understand and utilize this concept in our mathematical work.

Are there exceptions to the general property of norms?

No, there are no exceptions to the general property of norms. This property must hold true for all norms in all vector spaces. If a function does not satisfy the three conditions, then it cannot be considered a norm. However, there are different types of norms (such as p-norms or operator norms) that may have slightly different properties, but they still must satisfy the general property of norms.

Similar threads

A Proof of the inequality of a reduced basis
Replies
7
Views
1K
I Proof that if T is Hermitian, eigenvectors form an orthonormal basis
Replies
3
Views
2K
I Shouldn't this definition of a metric include a square root?
Replies
8
Views
2K
Normed linear space vs inner product space and more
Replies
15
Views
10K
I Proof concerning the Four Fundamental Spaces
Replies
14
Views
2K
Proving 2-Norm of A: Understanding the Relationship Between u and v
Replies
6
Views
5K
I Finding the orthogonal projection of a vector without an orthogonal basis
Replies
3
Views
2K
I ##L^2## square integrable function Hilbert space
Replies
11
Views
1K
B Does the L2 norm of a vector destroy all directional info?
Replies
6
Views
2K
Solving Doubts When Showing Simple Properties of Norms
Replies
7
Views
2K

Hot Threads

Recent Insights

Back
Top