Infimum and supremum ($\inf$, $\sup$)

For a subset $A$ of a partially ordered set, its infimum is its greatest lower bound and its supremum is its least upper bound. An upper bound only has to lie above every member; the supremum is the smallest such bound. A lower bound lies below every member; the infimum is the largest such bound. These definitions identify the exact edges of a set even when neither edge belongs to it.

For example, $(0,1)$ and $[0,1]$ both have infimum zero and supremum one. Only $[0,1]$ has a minimum and a maximum: those must be actual members of the set. The set $C=\{1/n:n\in {\mathbb{N}}_{>0}\}$ has infimum zero without a minimum, but its supremum one is also its maximum. Such examples distinguish a bound from an attained extremal value. ^[ref].

For a nonempty subset of the real numbers, completeness of the real numbers guarantees a real supremum when it is bounded above, and a real infimum when it is bounded below. The bound and the ambient partially ordered set both matter: $\{q\in \mathbb{Q}:q\ge 0,q^2<2\}$ has supremum $\sqrt{2}$ in the real numbers, but no supremum in the rational numbers. The detailed explanation belongs to completeness of the real numbers.

Why are unattained bounds useful? They specify a precise limiting target. In mathematical optimization, $f(x)=x$ on $(0,1)$ has infimum zero but no global minimizer. The inputs $x_n=1/n$ for $n\ge 2$ give a minimizing sequence: their objective values approach zero without ever reaching it. Bounds therefore describe what approximation can achieve even when no feasible input is best. The approximation property of an infimum and approximation property of a supremum express this idea quantitatively.

For a nonempty bounded array of real numbers, taking the supremum of each row and then the supremum of those values gives the supremum of all entries. Every row supremum is bounded by the global supremum; conversely, every entry is bounded by the iterated supremum. The same argument with lower bounds proves ${\inf }_m{\inf }_n{x}_{m,n}={\inf }_{m,n}{x}_{m,n}$. Mixed infima and suprema satisfy the weak minimax inequality rather than an equality in general.

For a nonempty bounded array of real numbers, each column infimum is at most every row supremum, because ${\inf }_k{x}_{k,n}\le x_{m,n}\le {\sup }_j{x}_{m,j}$ for all $m,n$. Taking the supremum over columns and then the infimum over rows proves the displayed inequality. Equality need not hold: for $x_{m,n}=1_{\{m=n\}}$ on positive integer indices, every column infimum is zero and every row supremum is one. This elementary bound is the order-theoretic starting point for stronger minimax theorems with additional hypotheses.

The essential infimum and essential supremum describe the lower and upper bounds of a measurable function after disregarding null sets for a specified measure. A pointwise infimum or supremum sees every input; an essential bound sees whether values below or above a proposed threshold occur on a set of positive measure. Thus "essential" means insensitive to changes on a null set, rather than insensitive to unusual values. This distinction makes essential bounds compatible with Lebesgue integrals and the equivalence classes used in Lp spaces. ^[ref].

For a real measurable function $f$ on a measure space $(X,\Sigma ,\mu )$, the definitions areThese can take values in the extended real number line. On a space of positive total measure, finite essential bounds obey ${\mathrm{essinf}}_{\mu }f\le f\le {\mathrm{esssup}}_{\mu }f$ almost everywhere. If $s={\mathrm{esssup}}_{\mu }f$ is finite, every threshold $b<s$ has $\mu (\{f>b\})>0$; the corresponding lower-edge statement is $\mu (\{f<a\})>0$ for every $a>{\mathrm{essinf}}_{\mu }f$. These are the measure-based counterparts of the approximation property of a supremum and approximation property of an infimum.

Here is a two-sided example on $[0,1]$ with Lebesgue measure:The pointwise bounds are $\inf g=-3$ and $\sup g=4$, while the essential bounds are $\mathrm{essinf}g=0$ and $\mathrm{esssup}g=1$. Both exceptional singletons are null sets. But for $0<a<1$, the inputs where $g<a$ still have measure $a$, and those where $g>a$ have measure $1-a$: deleting or adding two individual inputs cannot change those lengths. Hence neither essential bound can be moved inward.

The choice of measure is part of the answer. The same $g$ has essential bounds $-3$ and $4$ under counting measure, because every nonempty set has positive measure. Under the Dirac measure at $1/4$, both essential bounds equal $-3$. A null set for one measure need not be a null set for another.

Nor must the exceptional inputs be isolated. The Dirichlet function, equal to one at rational numbers and zero at irrational numbers, has pointwise supremum one on $[0,1]$, but essential supremum zero for Lebesgue measure: the rational numbers form a countable set that is a null set, even though they are dense. Conversely, a value ten on an interval of length $0.001$ affects the essential supremum just as decisively as it would on half the domain. Small positive measure is not zero measure.

The essential infimum of a real measurable function is ${\mathrm{essinf}}_{\mu }f=-{\mathrm{esssup}}_{\mu }(-f)$. It is the largest lower bound that holds almost everywhere, with values allowed in the extended real number line. Unlike a pointwise infimum, it is unchanged by altering a representative on a null set.

For the function $g$ in essential infimum and essential supremum, the value $-3$ at one input changes the pointwise infimum, but the essential infimum remains zero under Lebesgue measure. If instead a function is $-3$ throughout an interval of any positive length and zero elsewhere, its essential infimum is $-3$. The distinction concerns the measure of the low-value set, rather than the magnitude of the low value.

For a nonempty set $A\subseteq \mathbb{R}$, a real number $s$ is its supremum precisely when it is an upper bound and no smaller real number is an upper bound. It is unique: two least upper bounds must each be no greater than the other.

For $A=(0,1)$, every number at least one is an upper bound, and none smaller than one bounds the whole set. Thus $\sup A=1$, although $1\notin A$. For $A=[0,1]$, that same supremum is also the maximum. The approximation property of a supremum supplies witnesses arbitrarily close to an unattained upper edge.

Let $A,B$ be nonempty sets of real numbers bounded above. Their Minkowski sum is $A+B=\{a+b:a\in A,b\in B\}$, allowing every pair from the Cartesian product $A\times B$. ThenThe sum of the two suprema is an upper bound for every permitted sum. For the reverse bound, use the approximation property of a supremum to choose $a>\sup A-\epsilon /2$ and $b>\sup B-\epsilon /2$. Their sum is within $\epsilon$ of the claimed value. This reasoning does not require either supremum to be a maximum.

Separate choices matter. On $X=[0,1]$, choose the functions $f(x)=x$ and $g(x)=1-x$. Each has supremum one, but their sum at a shared input is always one:The maximum of $f$ occurs at $x=1$, while that of $g$ occurs at $x=0$. A common input cannot realize both choices. With separate inputs, choosing $x=1$ and $y=0$ gives $f(x)+g(y)=2$. The allowed pairs of output values are the entire unit square for separate inputs, but only the line segment $a+b=1$ for the shared input.

For any two bounded functions on the same nonempty domain of a function, the useful general estimate isThe inequality can be strict, as the example shows. Equality does not require attained maxima, but it does require that the two large values can be approached at compatible inputs. For example, $f(x)=g(x)=x$ on $(0,1)$ gives equality with value two, although neither function reaches its supremum. This distinction is useful when an objective function combines terms sharing a decision variable: adding the best possible termwise scores gives an upper bound, not necessarily a feasible combined score.

Suppose $f,g:X\to \mathbb{R}$ are bounded functions on the same nonempty set, and the supremum norm of their difference is at most $\delta$. Then their optimal values differ by at most the same amount:Indeed, $f(x)\le g(x)+\delta \le {\sup }_{X}g+\delta$ for every $x$, so ${\sup }_{X}f\le {\sup }_{X}g+\delta$; exchange $f$ and $g$ for the other direction. Adding a constant shifts a supremum by that constant, a basic compatibility between upper bounds and addition.^[ref]

For a chosen example on $[-1,1]$, take $f(x)=-x^2$ and $g(x)=-x^2+0.2x$. Their supremum error is $0.2$, since $|g(x)-f(x)|=0.2|x|$. Completing the square givesThus $f$ has maximum zero at $x=0$, while $g$ has maximum $0.01$ at $x=0.1$. The change in optimal value, $0.01$, is smaller than the certified bound $0.2$. For the vertical translation $h(x)=f(x)+0.2$, the bound is exact: its supremum error and optimal-value change are both $0.2$.

A stable optimal value does not guarantee a stable maximizing input. On $[-1,1]$, the two linear functions $f_{\epsilon }(x)=\epsilon x$ and $g_{\epsilon }(x)=-\epsilon x$ satisfyFor every positive $\epsilon$, their unique maximizing inputs remain two units apart, at $1$ and $-1$. Both functions approach the flat zero function, which has many maximizing inputs. Consequently an accurate objective function value is not by itself a certificate that the optimizing decision is accurate.

The estimate also gives a convenient consequence of uniform convergence. If $f_n$ converges uniformly to a bounded function $f$ on a nonempty $X$, then ${\sup }_X{f}_n\to {\sup }_{X}f$ for the bounded tail of the sequence. Uniform control handles the changing input that defines the best value. Pointwise convergence alone supplies no such control; the existing examples under uniform convergence illustrate that distinction.

Let $A\subseteq \mathbb{R}$ be nonempty and bounded above. An upper bound $s$ equals $\sup A$ exactly when, for every $\epsilon >0$, some $a\in A$ satisfies $s-\epsilon <a\le s$. If no such member existed, $s-\epsilon$ would be a smaller upper bound. Conversely, the condition defeats every proposed upper bound $u<s$ by taking $\epsilon =s-u$. ^[ref].

For $(0,1)$ and $\epsilon =0.2$, $a=0.9$ is one witness: $0.8<0.9<1$. Choosing $a_n\in A$ with $s-1/n<a_n\le s$ produces a sequence converging to $s$ by the squeeze theorem. The choices need not form a monotone sequence.

If a nonempty real set $A$ has finite supremum $s$ and $s\notin A$, then every interval $(s-\epsilon ,s)$ contains infinitely many elements of $A$. Otherwise its finitely many elements have a largest value $m<s$; all remaining elements are at most $s-\epsilon$, making $\max (m,s-\epsilon )<s$ an upper bound. The interval cannot be empty either, by the defining approximation property of a supremum. Thus failure to attain a finite supremum forces accumulation from below.

For a nonempty set $A\subseteq \mathbb{R}$, a real number $\ell$ is its infimum precisely when it is a lower bound and every other lower bound is at most $\ell$. It is the greatest lower bound, not necessarily the minimum.

For $C=\{1/n:n\in {\mathbb{N}}_{>0}\}$, zero is a lower bound. Every positive candidate $\epsilon$ fails: the Archimedean property gives an integer $n>1/\epsilon$, hence $1/n<\epsilon$. Thus $\inf C=0$, although zero is never a member of $C$. This is the lower-edge counterpart of approximating an unattained supremum.

Let $A\subseteq \mathbb{R}$ be nonempty and bounded below. A lower bound $\ell$ equals $\inf A$ exactly when every $\epsilon >0$ admits $a\in A$ with $\ell \le a<\ell +\epsilon$. Otherwise $\ell +\epsilon$ would be a larger lower bound. Taking tolerances $1/n$ gives a sequence of members converging to $\ell$. For a real-valued objective with finite infimum, this supplies a minimizing sequence without assuming that a global minimizer exists.