L-Infinity Norm is Norm

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 $f \in \map {\mathcal L^\infty} {X, \Sigma, \mu}$ $f = 0$ $\mu$-almost everywhere.

That is:


 * $\norm f_\infty = 0$ $f \sim 0$

That is, from Equivalence Class Equivalent Statements:


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

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


 * $\norm {\eqclass f \sim}_\infty = 0$ $\norm f_\infty = 0$

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.

Property $(\text N 2)$
Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$ and $\lambda \in \R$.

Then:

So we have positive homogeneity.

Property $(\text N 3)$
Let $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.

Then:

so the triangle inequality holds.