Inmathematics, the concepts ofessential infimum andessential supremum are related to the notions ofinfimum and supremum, but adapted tomeasure theory andfunctional analysis, where one often deals with statements that are not valid forall elements in aset, but ratheralmost everywhere, that is, except on aset of measure zero.

While the exact definition is not immediately straightforward, intuitively the essential supremum of a function is the smallest value that is greater than or equal to the function values everywhere while ignoring what the function does at a set of points of measure zero. For example, if one takes the function${\displaystyle f(x)}$ that is equal to zero everywhere except at${\displaystyle x=0}$ where${\displaystyle f(0)=1,}$ then the supremum of the function equals one. However, its essential supremum is zero since (under theLebesgue measure) one can ignore what the function does at the single point where${\displaystyle f}$ is peculiar. The essential infimum is defined in a similar way.

As is often the case in measure-theoretic questions, the definition of essential supremum and infimum does not start by asking what a function${\displaystyle f}$ does at points${\displaystyle x}$ (that is, theimage of${\displaystyle f}$), but rather by asking for the set of points${\displaystyle x}$ where${\displaystyle f}$ equals a specific value${\displaystyle y}$ (that is, thepreimage of${\displaystyle y}$ under${\displaystyle f}$).

Let${\displaystyle f:X\to \mathbb{R}}$ be areal valuedfunction defined on a set${\displaystyle X.}$ Thesupremum of a function${\displaystyle f}$ is characterized by the following property:${\displaystyle f(x)\le supf\le \mathrm{\infty }}$ forall${\displaystyle x\in X}$ and if for some${\displaystyle a\in \mathbb{R}\cup \{+\mathrm{\infty }\}}$ we have${\displaystyle f(x)\le a}$ forall${\displaystyle x\in X}$ then${\displaystyle supf\le a.}$ More concretely, a real number${\displaystyle a}$ is called anupper bound for${\displaystyle f}$ if${\displaystyle f(x)\le a}$ for all${\displaystyle x\in X;}$ that is, if the set$${\displaystyle f^{-1}(a,\mathrm{\infty })=\{x\in X:f(x)>a\}}$$isempty. Let$${\displaystyle U_f=\{a\in \mathbb{R}:f^{-1}(a,\mathrm{\infty })=\varnothing \}}$$be the set of upper bounds of${\displaystyle f}$ and define theinfimum of the empty set by${\displaystyle inf\varnothing =+\mathrm{\infty }.}$ Then the supremum of${\displaystyle f}$ is$${\displaystyle supf=inf{U}_f.}$$By definition, if the set of upper bounds${\displaystyle U_f}$ is empty, we have${\displaystyle supf=+\mathrm{\infty }}$.

Now assume in addition that${\displaystyle (X,\mathrm{\Sigma },\mu )}$ is ameasure space and, for simplicity, assume that the function${\displaystyle f}$ ismeasurable. Similar to the supremum, the essential supremum of a function is characterised by the following property:${\displaystyle f(x)\le \mathrm{ess}supf\le \mathrm{\infty }}$ for${\displaystyle \mu }$-almost all${\displaystyle x\in X}$ and if for some${\displaystyle a\in \mathbb{R}\cup \{+\mathrm{\infty }\}}$ we have${\displaystyle f(x)\le a}$ for${\displaystyle \mu }$-almost all${\displaystyle x\in X}$ then${\displaystyle \mathrm{ess}supf\le a.}$ More concretely, a number${\displaystyle a}$ is called anessential upper bound of${\displaystyle f}$ if the measurable set${\displaystyle f^{-1}(a,\mathrm{\infty })}$ is a set of${\displaystyle \mu }$-measure zero,^[a] That is, if${\displaystyle f(x)\le a}$ for${\displaystyle \mu }$-almost all${\displaystyle x}$ in${\displaystyle X.}$ Let$${\displaystyle U_f^{\mathrm{ess}}=\{a\in \mathbb{R}:\mu (f^{-1}(a,\mathrm{\infty }))=0\}}$$be the set of essential upper bounds. Then theessential supremum is defined similarly as$${\displaystyle \mathrm{ess}supf=inf{U}_f^{\mathrm{e}\mathrm{s}\mathrm{s}}}$$if${\displaystyle U_f^{\mathrm{ess}}\ne \varnothing ,}$ and${\displaystyle \mathrm{ess}supf=+\mathrm{\infty }}$ otherwise.

Exactly in the same way one defines theessential infimum as the supremum of theessential lower bounds, that is,if the set of essential lower bounds is nonempty, and as${\displaystyle -\mathrm{\infty }}$ otherwise; again there is an alternative expression as${\displaystyle \mathrm{ess}inff=sup\{a\in \mathbb{R}:f(x)\ge a\text{ for almost all }x\in X\}}$ (with this being${\displaystyle -\mathrm{\infty }}$ if the set is empty).

On the real line consider theLebesgue measure and its corresponding𝜎-algebra${\displaystyle \mathrm{\Sigma }.}$ Define a function${\displaystyle f}$ by the formula

The supremum of this function (largest value) is 5, and the infimum (smallest value) is −4. However, the function takes these values only on the sets${\displaystyle \{1\}}$ and${\displaystyle \{-1\},}$ respectively, which are of measure zero. Everywhere else, the function takes the value 2. Thus, the essential supremum and the essential infimum of this function are both 2.

As another example, consider the functionwhere${\displaystyle \mathbb{Q}}$ denotes therational numbers. This function is unbounded both from above and from below, so its supremum and infimum are${\displaystyle \mathrm{\infty }}$ and${\displaystyle -\mathrm{\infty },}$ respectively. However, from the point of view of the Lebesgue measure, the set of rational numbers is of measure zero; thus, what really matters is what happens in the complement of this set, where the function is given as${\displaystyle \arctan x.}$ It follows that the essential supremum is${\displaystyle \pi /2}$ while the essential infimum is${\displaystyle -\pi /2.}$

On the other hand, consider the function${\displaystyle f(x)=x^3}$ defined for all real${\displaystyle x.}$ Its essential supremum is${\displaystyle +\mathrm{\infty },}$ and its essential infimum is${\displaystyle -\mathrm{\infty }.}$

Lastly, consider the functionThen for any${\displaystyle a\in \mathbb{R},}$${\displaystyle \mu (\{x\in \mathbb{R}:1/x>a\})\ge \frac{1}{|a|}}$ and so${\displaystyle U_f^{\mathrm{ess}}=\varnothing }$ and${\displaystyle \mathrm{ess}supf=+\mathrm{\infty }.}$

If${\displaystyle \mu (X)>0}$ then$${\displaystyle inff\le \mathrm{ess}inff\le \mathrm{ess}supf\le supf.}$$ and otherwise, if${\displaystyle X}$ has measure zero then^[1]$${\displaystyle +\mathrm{\infty }=\mathrm{ess}inff\ge \mathrm{ess}supf=-\mathrm{\infty }.}$$

If${\displaystyle f}$ and${\displaystyle g}$ are measurable, then

$${\displaystyle \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}(f+g)\le \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}f+\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}g}$$

and

$${\displaystyle \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{f}(f+g)\ge \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{f}f+\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{f}g.}$$

If${\displaystyle f}$ and${\displaystyle g}$ are measurable and if${\displaystyle f\le g}$ almost everywhere, then

$${\displaystyle \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}f\le \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{s}\mathrm{u}\mathrm{p}g}$$

and

$${\displaystyle \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{f}f\le \mathrm{e}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{f}g.}$$

If the essential supremums of two functions${\displaystyle f}$ and${\displaystyle g}$ are both nonnegative, then${\displaystyle \mathrm{ess}sup(fg)\le (\mathrm{ess}supf)(\mathrm{ess}supg).}$

The essential supremum of a function is not just the infimum of the essential lower bounds, but also their minimum. A similar result holds for the essential infimum.

Given ameasure space${\displaystyle (S,\mathrm{\Sigma },\mu ),}$ thespace${\displaystyle {\mathcal{L}}^{\mathrm{\infty }}(S,\mu )}$ consisting of all of measurable functions that are bounded almost everywhere is aseminormed space whoseseminormis the essential supremum of a function's absolute value when${\displaystyle \mu (S)\ne 0.}$^[b]

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