= L-infinity norm
{c}
{title2=$\lVert f\rVert_\infty=\operatorname{ess\,sup}_\mu|f|$}
= Essential supremum norm
{synonym}
On a <measure space> of positive total <measure>, the <L-infinity norm> of an <essentially bounded> <measurable function> is $\lVert f\rVert_\infty=\operatorname{ess\,sup}_\mu|f|$. The <absolute value> matters: a constant <function> $f=-3$ has <L-infinity norm> three, even though its <essential supremum> is $-3$. The <norm> belongs to the <equivalence class> in <Lp space>; different representatives equal <almost everywhere> give the same answer. <Measure Theory by John K. Hunter/Lp spaces (Chapter 7 of Measure Theory)>[Hunter, Measure Theory, Definition 7.4: the L-infinity norm]{ref}.
For the <function> $g$ in <essential infimum and essential supremum>, \b[$\lVert g\rVert_\infty=1$ under <Lebesgue measure>, whereas its <supremum norm> is four]. Replacing the exceptional values by the ordinary values $x$ leaves every finite <Lp norm> unchanged as well: $\lVert g\rVert_p=(\int_0^1x^p\,dx)^{1/p}=(p+1)^{-1/p}$. This tends to one as $p\to\infty$. That agreement explains why the endpoint with <exponent> $p=\infty$ uses an essential bound rather than a pointwise bound.
A finite <probability space> makes the distinction between zero and small <measure> explicit. Let $f(a)=1$, $f(b)=3$, and $f(c)=10$. Assign weights $\mu(a)=1/2-\varepsilon$, $\mu(b)=1/2$, $\mu(c)=\varepsilon$, where $0\leq\varepsilon<1/2$. For $1\leq p<\infty$,
$$
\lVert f\rVert_p=\left((1/2-\varepsilon)+\tfrac12 3^p+\varepsilon 10^p\right)^{1/p}.
$$
When $\varepsilon=0$, the <singleton> $\{c\}$ is a <null set> and $\lVert f\rVert_\infty=3$. For every $\varepsilon>0$, however small, $\lVert f\rVert_\infty=10$. For example, $\varepsilon=0.001$ gives $\lVert f\rVert_1=2.009$ and $\lVert f\rVert_2=\sqrt{5.099}$, yet the large-$p$ <Lp norms> approach ten. The ordinary pointwise <supremum norm> is ten in both cases.
\Image[/lp-norm-null-versus-rare-value.png]
{title=A zero-weight value disappears from every <Lp norm>, while a rare positive-weight value determines the limiting <L-infinity norm>}
{description=The original three-point example has values $1,3,10$. The curves use weights $(0.5,0.5,0)$ and $(0.499,0.5,0.001)$, respectively. Dashed levels show the limiting norms three and ten; the plotted finite <exponents> do not themselves reach those limits.}
Why can a rare value dominate? For a bounded <function> on a <probability space>, an <almost everywhere> bound $|f|\leq M$ gives $\lVert f\rVert_p\leq M$. If $0<t<\lVert f\rVert_\infty$, the <set> $A_t=\{|f|>t\}$ has positive <probability>, so $\lVert f\rVert_p\geq t\,\mu(A_t)^{1/p}$. As $p\to\infty$, the factor $\mu(A_t)^{1/p}$ tends to one. These two bounds squeeze the <limit> to $\lVert f\rVert_\infty$. The <probability measure> assumption keeps total <measure> finite; this explanation does not assert the same finite-$p$ integrability on arbitrary infinite-measure spaces.
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