---
tag: 4KX81Q85DT
kind: definition
name: Seminorm
msc: [46A03]
defines: FP5GVCN8GV
uses: [05EGHEQZAW, 00S7C68D51]
sources:
  - url: https://en.wikipedia.org/wiki/Seminorm
---

## Definition

Let $\mathbb{K}$ be $\R$ or $\C$, and let $V$ be a [[vector space|05EGHEQZAW]] over $\mathbb{K}$.
A *seminorm* on $V$ is a map $p \colon V \to \R$ such that:

1. $p(x + y) \le p(x) + p(y)$ for all $x, y \in V$.
2. $p(\lambda x) = \abs{\lambda} \, p(x)$ for all $\lambda \in \mathbb{K}$ and $x \in V$, where $\abs{\lambda}$ is the [[absolute value|00S7C68D51]].

## Remarks

The two conditions imply $p(0) = 0$ and $p(x) \ge 0$ for all $x$.
Indeed $p(0) = p(0 \cdot 0) = 0 \cdot p(0) = 0$, and then $0 = p(x - x) \le p(x) + p(-x) = 2 p(x)$.
A seminorm is a [[norm|08NYZ2FS6B]] exactly when $p(x) = 0$ implies $x = 0$.
Example: on the space $C([0, 1])$ of continuous real functions, $p(f) = \sup_{0 \le t \le 1/2} \abs{f(t)}$ is a seminorm; both conditions follow from the same inequalities for the absolute value at each point.
It is not a norm, since it vanishes on a nonzero function supported in $(1/2, 1]$.
Non-example: on $\R$, $p(x) = x^2$ is not a seminorm, since $p(2 \cdot 1) = 4 \ne 2 = \abs{2} \, p(1)$.
The kernel $\set{x : p(x) = 0}$ is a linear subspace, and $p$ induces a norm on the quotient of $V$ by it.