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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.

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 range of a real or complex measurable function $m$ consists of the values $z$ whose every neighborhood receives inputs of positive measure:It depends on the measure as well as the function. This definition concerns values arbitrarily near $z$, not necessarily inputs at which $m=z$ exactly. It therefore retains meaningful limiting values while ignoring values occurring only on null sets. ^[ref].

For $m(x)=x$ on $(0,1)$ with Lebesgue measure, the ordinary image of a function is $(0,1)$, but the essential range is $[0,1]$. Neither endpoint is attained, yet inputs with $|m(x)|<\epsilon$ or $|m(x)-1|<\epsilon$ have positive measure for every $\epsilon >0$. Every interior value also belongs to the essential range, although its exact fiber is a singleton of zero measure. Positive measure of a fiber is sufficient, but is not necessary, for a value to belong to the essential range.

The two-sided example $g$ from essential infimum and essential supremum illustrates what disappears. It follows $g(x)=x$ on $[0,1]$, except $g(1/4)=-3$ and $g(3/4)=4$. Its ordinary image of a function is $([0,1]\setminus \{1/4,3/4\})\cup \{-3,4\}$, whereas its essential range remains $[0,1]$. The missing ordinary values $1/4$ and $3/4$ still have arbitrarily close neighbors with positive measure; the exceptional values $-3$ and $4$ can each be isolated by a sufficiently small neighborhood whose preimage is a null set.

For the three-point example in L-infinity norm, the values are $1,3,10$. With weights $(0.5,0.5,0)$ the essential range is $\{1,3\}$; with weights $(0.499,0.5,0.001)$ it is $\{1,3,10\}$. A small but positive probability retains ten, so an essential range does not discard a value merely because it is rare. For a real function on a space of positive total measure, the essential infimum and essential supremum are the lower and upper edges of its essential range, allowing infinite bounds. The L-infinity norm instead takes the upper edge of the essential range of $|m|$.

A useful operator connection is the spectrum of a real multiplication operator: on a nonzero L2 space over a sigma-finite measure, multiplication by a real measurable function $m$ has spectrum equal to its essential range, with the natural domain when $m$ is unbounded. Outside that range, $|m-z|$ has a positive lower bound almost everywhere, so multiplication by $1/(m-z)$ gives an inverse function that is a bounded linear operator. Near an essential value, this lower bound fails. Thus discarding null sets is precisely what makes the range match an operator acting on almost everywhere equivalence classes. ^[ref].

The essential supremum is the upper bound for a measurable function after disregarding null sets for a specified measure $\mu$. Its definition iswith values allowed in the extended real number line. A supremum controls every value; an essential supremum controls values almost everywhere. It therefore depends on the chosen measure, as well as on the function. ^[ref].

For a worked example, use Lebesgue measure on $[0,1]$ and putIts supremum is ten, attained at $1/2$. But its essential supremum is one: $f\le 1$ outside the singleton $\{1/2\}$, which is a null set. Every candidate $M<1$ fails, because the interval $(\max (0,M),1)$ has positive Lebesgue measure and includes values larger than $M$; changing one point cannot remove that interval. In particular, $\mu (\{f>0.8\})=0.2$, while $\mu (\{f>1\})=0$.

Why is this useful? Lebesgue integrals ignore changes on null sets, and Lp spaces identify functions that agree almost everywhere. The L-infinity norm uses $\mathrm{esssup}|f|$ so it respects that identification. For this nonnegative example the supremum norm is ten and the L-infinity norm is one; replacing $f(1/2)$ by $1/2$ changes the former while leaving the latter unchanged. ^[ref].

The measure qualification is essential. For $\mu ={\delta }_{1/2}$, the Dirac measure concentrated at the exceptional input, that singleton has measure one and this function has essential supremum ten. A set ignored by one measure need not be ignored by another. The lower counterpart is the essential infimum, with $\mathrm{essinf}f=-\mathrm{esssup}(-f)$.

The exponential integral of a fixed measurable function is controlled, on its logarithmic scale, by the function's essential infimum. Suppose $0<{\int }_A{e}^{-\varphi }dx<\infty$ and $\alpha ={\mathrm{essinf}}_A\varphi$ is finite. For $\theta \ge 1$, the integral is at most $e^{-(\theta -1)\alpha }{\int }_A{e}^{-\varphi }dx$. For every $\epsilon >0$, the set $A_{\epsilon }=\{x\in A:\varphi (x)<\alpha +\epsilon \}$ has positive finite Lebesgue measure, and the integral is at least $|A_{\epsilon }|e^{-\theta (\alpha +\epsilon )}$. Taking logarithms, dividing by $\theta$, and then letting $\epsilon \downarrow 0$ proves the displayed limit. The argument needs no isolated or nondegenerate minimizer. The Laplace principle for probability measures treats families whose distributions also change with the scaling parameter.

Let $M={\mathrm{esssup}}_{\Omega }u$. The standard local boundedness of weak elliptic subsolutions gives a finite upper bound on every relatively compact ball. In particular, the assumed equality of essential suprema makes $M$ finite. Equivalently, when using the maximum principle for an upper-bounded weak subsolution, this preliminary fact is already part of the hypothesis.

Set $v=M-u$. Because the divergence-form elliptic operator here annihilates constants, $v\ge 0$ and $Lv=-Lu\le 0$. The hypothesis says ${\mathrm{essinf}}_{B}v=0$. Cover the compact closure of $B$ by finitely many balls $D_j$ whose doubled balls lie in $\Omega$. At least one $D_j$ has essential infimum zero: otherwise the minimum of their finitely many positive lower bounds would be a positive lower bound on $B$. The Weak Harnack inequality with zero forcing now givesConsequently $v=0$ almost everywhere on $D_j$. If another admissible ball overlaps $D_j$ in an open set, its essential infimum of $v$ is also zero, and the same argument makes $v$ vanish there. Since a domain is connected, a chain of overlapping interior balls reaches every point of $\Omega$. This propagation of zeros by the weak Harnack inequality proves $u=M$ almost everywhere on $\Omega$, which is the claimed strong maximum principle. Notice why the absence of the $b,d$ terms matters: it is precisely what makes $L(M-u)=-Lu$.

Assume $u\in W^{1,2}(B_2(0))$ is nonnegative almost everywhere, and that the real coefficient matrix $A=(a^{ij})$ is measurable and satisfies, for almost every $x$ and every $\xi \in {\mathbb{R}}^n$,The operator norm bound controls all coefficients, including any antisymmetric part. For a symmetric matrix, these assumptions are exactly the usual lower and upper quadratic-form bounds. The weak solution condition isThen the Harnack inequality for uniformly elliptic divergence-form equations isA nonnegative weak solution has its supremum controlled by its infimum on the smaller ball. The De Giorgi-Nash-Moser theorem supplies a Hölder continuous function representative, for which ordinary supremum and infimum can replace the essential supremum and essential infimum. Nonnegativity and the larger ball are essential hypotheses; there is no differentiability requirement on the coefficients.

Adopt the sign convention that a weak supersolution for $Lu=D_i(a^{ij}{D}_{j}u)$ satisfies $Lu\le 0$, or $\int a^{ij}{D}_{j}u{D}_i\phi \ge 0$ for every nonnegative compactly supported test function. For a nonnegative supersolution, the Weak Harnack inequality gives some $q>0$ and $C_H\ge 1$, depending only on $n,\lambda ,\Lambda$, such thatHere $|E{|}^{-1}{\int }_E$ denotes the average over $E$; equivalently the right side gives a lower bound for the essential infimum. One may take a sufficiently small positive exponent, so no assumption $q\ge 1$ is needed. The inequality is unchanged under translations and dilations of the Euclidean balls.

Since a Sobolev space element is defined up to changes on a null set, the infimum is initially essential. After choosing a continuous representative it is the ordinary infimum. Without choosing such a representative, an arbitrary pointwise infimum cannot be bounded by an integral.

For a nonnegative weak supersolution with zero forcing, the Weak Harnack inequality makes zero essential infimum on an interior ball imply vanishing almost everywhere on that ball. An overlapping ball then has zero essential infimum too. Chains of overlapping interior balls propagate the vanishing through a connected domain.