# Quasinorm

In linear algebra, functional analysis and related areas of mathematics, a quasinorm is similar to a norm in that it satisfies the norm axioms, except that the triangle inequality is replaced by $\displaystyle{ \|x + y\| \leq K(\|x\| + \|y\|) }$ for some $\displaystyle{ K \gt 1. }$

## Definition

A quasi-seminorm[1] on a vector space $\displaystyle{ X }$ is a real-valued map $\displaystyle{ p }$ on $\displaystyle{ X }$ that satisfies the following conditions:

1. Non-negativity: $\displaystyle{ p \geq 0; }$
2. Absolute homogeneity: $\displaystyle{ p(s x) = |s| p(x) }$ for all $\displaystyle{ x \in X }$ and all scalars $\displaystyle{ s; }$
3. there exists a real $\displaystyle{ k \geq 1 }$ such that $\displaystyle{ p(x + y) \leq k [p(x) + p(y)] }$ for all $\displaystyle{ x, y \in X. }$
• If $\displaystyle{ k = 1 }$ then this inequality reduces to the triangle inequality. It is in this sense that this condition generalizes the usual triangle inequality.

A quasinorm[1] is a quasi-seminorm that also satisfies:

1. Positive definite/Point-separating: if $\displaystyle{ x \in X }$ satisfies $\displaystyle{ p(x) = 0, }$ then $\displaystyle{ x = 0. }$

A pair $\displaystyle{ (X, p) }$ consisting of a vector space $\displaystyle{ X }$ and an associated quasi-seminorm $\displaystyle{ p }$ is called a quasi-seminormed vector space. If the quasi-seminorm is a quasinorm then it is also called a quasinormed vector space.

Multiplier

The infimum of all values of $\displaystyle{ k }$ that satisfy condition (3) is called the multiplier of $\displaystyle{ p. }$ The multiplier itself will also satisfy condition (3) and so it is the unique smallest real number that satisfies this condition. The term $\displaystyle{ k }$-quasi-seminorm is sometimes used to describe a quasi-seminorm whose multiplier is equal to $\displaystyle{ k. }$

A norm (respectively, a seminorm) is just a quasinorm (respectively, a quasi-seminorm) whose multiplier is $\displaystyle{ 1. }$ Thus every seminorm is a quasi-seminorm and every norm is a quasinorm (and a quasi-seminorm).

### Topology

If $\displaystyle{ p }$ is a quasinorm on $\displaystyle{ X }$ then $\displaystyle{ p }$ induces a vector topology on $\displaystyle{ X }$ whose neighborhood basis at the origin is given by the sets:[2] $\displaystyle{ \{x \in X : p(x) \lt 1/n\} }$ as $\displaystyle{ n }$ ranges over the positive integers. A topological vector space with such a topology is called a quasinormed topological vector space or just a quasinormed space.

Every quasinormed topological vector space is pseudometrizable.

A complete quasinormed space is called a quasi-Banach space. Every Banach space is a quasi-Banach space, although not conversely.

### Related definitions

A quasinormed space $\displaystyle{ (A, \| \,\cdot\, \|) }$ is called a quasinormed algebra if the vector space $\displaystyle{ A }$ is an algebra and there is a constant $\displaystyle{ K \gt 0 }$ such that $\displaystyle{ \|x y\| \leq K \|x\| \cdot \|y\| }$ for all $\displaystyle{ x, y \in A. }$

A complete quasinormed algebra is called a quasi-Banach algebra.

## Characterizations

A topological vector space (TVS) is a quasinormed space if and only if it has a bounded neighborhood of the origin.[2]

## Examples

Since every norm is a quasinorm, every normed space is also a quasinormed space.

$\displaystyle{ L^p }$ spaces with $\displaystyle{ 0 \lt p \lt 1 }$

The $\displaystyle{ L^p }$ spaces for $\displaystyle{ 0 \lt p \lt 1 }$ are quasinormed spaces (indeed, they are even F-spaces) but they are not, in general, normable (meaning that there might not exist any norm that defines their topology). For $\displaystyle{ 0 \lt p \lt 1, }$ the Lebesgue space $\displaystyle{ L^p([0, 1]) }$ is a complete metrizable TVS (an F-space) that is not locally convex (in fact, its only convex open subsets are itself $\displaystyle{ L^p([0, 1]) }$ and the empty set) and the only continuous linear functional on $\displaystyle{ L^p([0, 1]) }$ is the constant $\displaystyle{ 0 }$ function (Rudin 1991). In particular, the Hahn-Banach theorem does not hold for $\displaystyle{ L^p([0, 1]) }$ when $\displaystyle{ 0 \lt p \lt 1. }$