analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
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For $p \in \mathbb{R}$, $p \geq 1$, the $p$-norm is a norm on suitable real vector spaces given by the $p$th root of the sum (or integral) of the $p$th-powers of the absolute values of the vector components. With due care the definition makes sense for non-finite dimensional vector spaces such as sequence spaces and Lebesgue spaces, making them into normed vector spaces, hence metric spaces.
For $p = 1$ the $p$-norm is the Taxicab norm or Manhattan norm.
For $p = 2$ the $p$-norm is the standard Euclidean norm, defining Euclidean spaces and Hilbert spaces of square integrable functions.
For $p = \infty$ the $p$-norm (found by taking the limit $p \to \infty$) is the supremum (or essential supremum in the continuous case) of the absolute values of the components of vectors, then called the supremum norm.
For $0 \leq p \lt 1$ one may still make sense of the formulas that define $p$-norms for $p \geq 1$ (see at Generalizations below), but the resulting concepts are no longer genuine norms.
The concept of $p$-norm makes sense in increasing generality,
For $n \in \mathbb{N}$, $p \in \mathbb{R}$, $p \gt 0$, the $p$-norm $\Vert - \Vert_p$ is the norm on the real finite dimensional vector space $\mathbb{R}^n$ given by the $p$th root of the sum of the $p$-powers of the absolute value of the components of a given vector $\vec x = (x)_{i = 1}^n \in \mathbb{R}^n$:
Equipping it with this norm makes $\mathbb{R}^n$ a normed vector space.
For $p = 2$ this is the Euclidean norm, the standard norm that defines Euclidean space.
For $p = \infty$ one takes the supremum over the absolute values of the components
The graphics on the right (grabbed from Wikipedia) shows unit circles in $\mathbb{R}^2$ with respect to various p-norms.
For $p \in \mathbb{R}$, write $\ell^p$ for the vector space of those sequences $(x_i)_{i \in \mathbb{N}}$ in $\mathbb{R}$ for which the series
(the sum of the $p$th powers of the absolute value of the components of the sequence) converges.
For $p \geq 1$ the the function
defines a norm on this real vector space. This normed vector space is complete, hence a Banach space. This is called the sequence space.
For $p = \infty$ one takes $\ell^\infty$ to be the space of bounded sequences and
to be the supremum over the absolute values of the components of the sequence. This is also called the supremum norm.
More generally, for $(X,\mu)$ a measure space, write $L^p(X)$ for the vector space of equivalence classes of those measurable functions $f \colon X \to \mathbb{R}$, for which the integral
exists, and where two such functions are regarded as equivalent, $f_1 \sim f_2$, if
On this space the function
defines a norm. The triangle inequality holds due to Minkowski's inequality. The normed vector space $(L^p(X), {\Vert- \Vert_p})$ is also called a Lebesgue space.
For $0 \leq p \lt 1$, the above definitions for $\Vert {-}\Vert_p$ still make sense in themselves, but the result is no longer a norm, as Minkowski's inequality (the triangle inequality for $p$-norms) fails.
A variant definition for $0 \lt p \leq 1$ (which agrees with the usual definition for $p = 1$, preserving continuity in $p$) leaves out the $p$th root; then the result satisfies the triangle inequality (and indeed is a metric) but fails to be a norm because it is not positive-homogeneous of degree $1$ (but of degree $p$ instead). Such a thing is called an F-norm.
For $p = 0$, we might try to take the limit as $p \searrow 0$. For the unmodified $p$-norm (with the root), this is infinite if there is more than one nonzero entry and is the absolute value of the one nonzero entry if there is only one (or 0 if there is none); for the modified $p$-norm (without the root), it is the (possibly infinite) number of nonzero entries. In either case, however the triangle inequality fails. Therefore, there is a further modified $0$-norm, given by
for $l^0$, and this is an $F$-norm. (But I don't know what is the justification for thinking of this as a $p$-norm for $p = 0$.)
Wikipedia, Lp space – The p-norm in finite dimensions
German Wikipedia, p-Norm
Last revised on December 10, 2022 at 11:06:42. See the history of this page for a list of all contributions to it.