L-Infinity Norm is Norm

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ vector space on $\struct {X, \Sigma, \mu}$.

Let $\norm \cdot_\infty$ be the $L^\infty$ norm.


Then $\norm \cdot_\infty$ is a norm on $\map {L^\infty} {X, \Sigma, \mu}$.


Proof

Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space.

Let $\sim$ be the $\mu$-almost everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.

Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$.

Then, we have by the definition of the $L^\infty$ norm we have:

$\norm {\eqclass f \sim}_\infty = \norm f_\infty$

From P-Seminorm is Seminorm, (in the case $p = \infty$) we have:

$\norm f_\infty \ge 0$

so:

$\norm {\eqclass f \sim}_\infty \ge 0$

So $\norm \cdot_\infty$ is a map from $\map {L^\infty} {X, \Sigma, \mu}$ to the non-negative real numbers.

It remains to verify the norm axioms.

Property $(\text N 1)$

From P-Seminorm of Function Zero iff A.E. Zero, we have:

$\norm f_\infty = 0$ for $f \in \map {\mathcal L^\infty} {X, \Sigma, \mu}$ if and only if $f = 0$ $\mu$-almost everywhere.

That is:

$\norm f_\infty = 0$ if and only if $f \sim 0$

That is, from Equivalence Class Equivalent Statements:

$\norm f_\infty = 0$ if and only if $\eqclass f \sim = \eqclass 0 \sim = 0_{\map {L^p} {X, \Sigma, \mu} }$.

Then, for $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$, we have:

$\norm {\eqclass f \sim}_\infty = 0$ if and only if $\norm f_\infty = 0$

from the definition of the $L^\infty$ norm.

This is in turn equivalent to $\eqclass f \sim = 0_{\map {L^\infty} {X, \Sigma, \mu} }$

So we have positive definiteness.

$\Box$


Property $(\text N 2)$

Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$ and $\lambda \in \R$.

Then:

\(\ds \size \lambda \norm {\eqclass f \sim}_\infty\) \(=\) \(\ds \size \lambda \norm f_\infty\) Definition of $L^\infty$ Norm
\(\ds \) \(=\) \(\ds \norm {\lambda f}_\infty\) P-Seminorm is Seminorm
\(\ds \) \(=\) \(\ds \norm {\eqclass {\lambda f} \sim}_\infty\) Definition of $L^\infty$ Norm
\(\ds \) \(=\) \(\ds \norm {\lambda \cdot \eqclass f \sim}_\infty\) Definition of Pointwise Scalar Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality

So we have positive homogeneity.

$\Box$


Property $(\text N 3)$

Let $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.

Then:

\(\ds \norm {\eqclass f \sim + \eqclass g \sim}_\infty\) \(=\) \(\ds \norm {\eqclass {f + g} \sim}_\infty\) Definition of Pointwise Addition on Space of Real-Valued Measurable Functions Identified by A.E. Equality
\(\ds \) \(=\) \(\ds \norm {f + g}_\infty\) Definition of $L^\infty$ Norm
\(\ds \) \(\le\) \(\ds \norm f_\infty + \norm g_\infty\) Minkowski's Inequality on Lebesgue Space
\(\ds \) \(=\) \(\ds \norm {\eqclass f \sim}_\infty + \norm {\eqclass g \sim}_\infty\) Definition of $L^\infty$ norm

so the triangle inequality holds.

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=L-Infinity_Norm_is_Norm&oldid=579551"