Real analysis

Real analysis studies limits, continuity, differentiation, integration, and convergence of sequences and functions taking values in the real numbers. Its central question is how finite approximations determine an exact answer: what does an arbitrarily small error mean, and which operations respect that error? A derivative is a limit of quotients, while a Riemann integral is obtained from increasingly fine Riemann sums. Both need conditions that make the answer independent of the approximations chosen.

Two organizing ideas are completeness of the real numbers, which ensures that suitable approximations have a destination, and uniform convergence, which controls approximation error over an entire domain. Pointwise convergence supplies separate guarantees at separate inputs; uniform convergence supplies one guarantee for all inputs. This distinction explains why replacing an approximating function by its limit can preserve continuity or integrals under appropriate hypotheses, yet fail without them.

For historical context, Cauchy's 1821 Cours d'analyse explicitly discusses convergence of series. Its Note VII treats double series and compares summing by rows and by columns under stated convergence assumptions: controlling an infinite process was already a concrete problem. This is historical context for questions about limits, rather than an attribution of the modern definitions below. ^[ref].

Hardy's inequality bounds averages or weighted functions by their original size or derivatives. For $p>1$ and $f\ge 0$, its integral form is ${\int }_0^{\infty }(x^{-1}{\int }_0^{x}f(t)dt{)}^{p}dx\le (p/(p-1){)}^p{\int }_0^{\infty }f(x{)}^{p}dx$. Spatial weighted forms include the Hardy inequality in Euclidean space.

Let $1<p<\infty$ and let $f$ be a real or complex measurable function in Lp space $L^p((0,\infty ),dx)$, with Lebesgue measure $dx$. The Hardy operator, denoted here by $H$, takes the running averageHölder's inequality ensures ${\int }_0^x|f(t)|dt\le x^{1-1/p}\Vert f{\Vert }_p<\infty$, so every such average exists. Hardy's integral inequality statesThe real signed and complex extensions of the nonnegative statement follow from the triangle inequality $|Hf|\le H|f|$. Changing $f$ on a null set changes neither average nor Lp norm.^[ref]

Averaging over an expanding interval spreads a localized input across the whole positive half-line. Hardy's integral inequality bounds the total Lp norm of that spread. The Hardy operator is consequently a bounded linear operator on this Lp space. The factor $p/(p-1)$ bounds its operator norm; its $p$th power is the constant in the displayed integral bound. These are different constants: at $p=2$, the L2 norm bound is two and the squared L2 norm bound is four.

Take the indicator function $f=1_{(0,1)}$. Direct integration givesAlthough the function vanishes beyond one, its average remembers the unit area accumulated before one. Squaring and integrating both portions givesThus the tail accounts for half of the squared L2 norm. Averaging does not always decrease the L2 norm with this Lebesgue measure on the half-line. The same indicator function also explains the restriction $p>1$: ${\int }_1^{\infty }{x}^{-1}dx=\infty$, so its average is not in L1 space.

The indicator function example reaches a squared L2 norm ratio of two. To approach four, concentrate more strongly near zero. For $0\le a<1/2$, use the measurable functionAll integrals use Lebesgue measure $dx$. Keeping both portions of the squared L2 norm givesConsequently,Every allowed member has ratio strictly below four. For any proposed constant $C<4$, a sufficiently large $a<1/2$ has ratio greater than $C$. Combined with Hardy's integral inequality, this proves that four is the smallest possible squared L2 norm constant.^[ref]

The endpoint is excluded: at $a=1/2$, ${\int }_0^1|f_a{|}^{2}dx={\int }_0^1{x}^{-1}dx=\infty$, and the inner portion of $\Vert H{f}_a{\Vert }_2^2$ also diverges. Its tail still has finite integral four. There is no finite L2 norm ratio at this endpoint, hence no equality example obtained by substituting it. The limit four is unattained by this family. Multiplying each $f_a$ by $\sqrt{1-2a}$ makes its L2 norm one without changing the ratio.

For a nonnegative $f$, first set $g_n(x)=\min (f(x),n)1_{(1/n,n)}(x)$ and $F_n(x)={\int }_0^x{g}_n(t)dt$. These bounded functions vanish near zero and have compact support. Their averages have finite Lp norms: they vanish near zero and decay as $1/x$ beyond $n$. Integration by parts givesThe boundary term $x^{1-p}{F}_n(x{)}^p$ vanishes at zero because $F_n$ is zero there, and at infinity because $F_n$ is eventually constant and $p>1$. The inequality is Hölder's inequality with conjugate exponents $p$ and $p/(p-1)$. Divide when $I_n>0$; the zero case is immediate. Since $g_n\uparrow f$ and $H{g}_n\uparrow Hf$, the monotone convergence theorem proves the full bound without presupposing that $Hf$ has finite Lp norm. Apply it to $|f|$ for the signed or complex case.^[ref]

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.

Asymptotic analysis describes what remains dominant as an argument approaches a specified limit. It compares functions or sequences by scale: how much of a complicated expression can be replaced by a simpler one, and how large is the omitted part? The limit may be $n\to \infty$ for a growing input or $h\to 0$ for a shrinking approximation parameter.

For example, $n^2+10^{5}n$ is eventually dominated by its quadratic term, since division by $n^2$ gives $1+10^5/n\to 1$. But at $n=100$, the linear term contributes $10^7$, compared with only $10^4$ from the quadratic term. A statement about the limiting scale and a good approximation at a particular input are different claims.

Three useful comparisons keep their strengths distinct. Big O notation supplies an eventual upper bound up to a constant multiplier; big theta notation supplies both an upper bound and a lower bound up to constants; asymptotic equivalence makes the ratio tend to one. Little o notation says that a term becomes negligible relative to a specified scale. These distinctions make approximations checkable instead of merely calling one expression small.

The same language connects time complexity with Taylor remainders. In an operation-count model, one asks how a cost scales with input size. In a Taylor expansion, one asks how the approximation error scales with the increment. Neither use requires interpreting the unspecified constant as one. The asymptotic comparison of exponential functions gives another example: dividing by the largest contributing exponential reveals the terms that disappear.

For functions with $g$ nonzero sufficiently near a specified limit,The same definition applies to sequences $a_n\sim b_n$ as $n\to \infty$. Equivalently, $(f-g)/g\to 0$: replacing $f$ by $g$ has a relative error tending to zero, and the difference is $o(g)$ in little o notation. The limit and reference scale are essential parts of the statement.

For example, $n^2+n\sim n^2$ because $(n^2+n)/n^2=1+1/n\to 1$. Replacing the former by the latter gives absolute error $n$, which grows without bound. Thus asymptotic equivalence controls a ratio, rather than requiring the absolute error to vanish. This distinction is crucial when approximating a rapidly growing factorial with Stirling's formula.

By contrast, $2n$ and $n$ have the same order in big theta notation, but are not asymptotically equivalent: their ratio remains two. For positive functions, asymptotic equivalence gives both-sided constant bounds eventually, whereas constant bounds alone do not recover the leading coefficient.

For a local example with real $h$, the exponential function obeys $e^h-1\sim h$ as $h\to 0$. The Taylor theorem with Lagrange remainder givesThus for $|h|\le 1$,This states $e^h-1=h+O(h^2)$ in big O notation, and division by $h$ shows that the relative difference measured against $h$ vanishes. The diagram uses positive $h$; the bound is valid on both sides of zero. At $h=0.1$, the absolute error is approximately $0.00517092$ and the relative difference against $h$ is approximately $0.0517092$.

For positive bases $a,b$, comparison of the exponential functions $a^n$ and $b^n$ reduces to a geometric progression:If $a<b$, this ratio tends to zero, so $a^n=o(b^n)$ in little o notation. If $a=b$, the ratio is one; if $a>b$, it grows without bound.

Consequently, provided at least one coefficient is nonzero, a finite sum with distinct positive bases is asymptotically equivalent to its largest-base term whose coefficient is nonzero. For example,The relative difference against $5\cdot 3^n$ first drops below one percent at the integer $n=11$. Dividing by the proposed dominant term provides an explicit error estimate, not just a comparison of two rapidly growing numbers. If several terms have the same largest base, their coefficients must first be combined; cancellation can remove that base entirely.

Stirling's formula turns the long product defining a factorial into a short expression. For positive integers $n$, putThen $n!\sim S_n$ as $n\to \infty$. This is asymptotic equivalence: $S_n/n!\to 1$, so the relative error tends to zero. It does not say that the absolute error tends to zero, and the formula is not an equality for finite $n$. The Gamma function extends this approximation beyond integer inputs. ^[ref]

For $n=10$,The leading estimate is about $0.830\%$ low. Multiplying by the first correction factor giveswhich is about $0.003176\%$ low. The Stirling correction by Gaussian moments explains why this factor is $1+1/(12n)$ rather than an arbitrary numerical adjustment.

The usefulness is in ratios as well as in enormous numbers. Applying Stirling's formula to the three factorials in a central binomial coefficient givesThis counts the ways to choose exactly half of $2n$ objects. Equivalently, for $2n$ independent fair coin tosses, the binomial distribution assigns probability $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/4^n\sim 1/\sqrt{\pi n}$ to exactly $n$ heads. An exponentially large count becomes a simple probability scale after cancellation.

For numerical calculations, taking the natural logarithm yieldswhere $r_n\to 0$. Computing a logarithm of a large factorial avoids representing the factorial itself. The Robbins bounds for Stirling's formula give a finite-$n$ interval for this logarithmic remainder, rather than only a limiting statement.

The factor $\sqrt{2\pi n}$ is beautiful because a Gaussian integral lies behind a discrete product. The Gaussian peak behind Stirling's formula displays this connection through the Gamma integral and Laplace's method. A rigorous proof must also control the tails of the integral; the local expansion explains the geometry but is not by itself a complete proof.

James Stirling published Methodus differentialis in 1730, a treatise on summation and interpolation of infinite series. This setting connects the formula with a practical calculation problem: taking logarithms turns a long product into a sum, whose growth can be approximated. ^[ref]

For integer $N\ge 2$, write $\ln (N!)={\sum }_{k=2}^N\ln k$. Monotonicity gives ${\int }_{k-1}^k\ln xdx\le \ln k\le {\int }_k^{k+1}\ln xdx$. Summing and evaluating the endpoint integrals proves the displayed bounds. They also hold with equality when $N=1$. The lower integral starts at one and the upper at two; confusing those endpoints loses the constant in the upper bound. The bounds imply $\ln (N!)=N\ln N-N+O(\ln N)$, while Stirling's formula determines the logarithmic and constant corrections.

The positive sequence $a_n=\sqrt{n}(n/e{)}^n/n!$ has logarithmic incrementsThe rational lower bound for the natural logarithm proves $u_n>0$. For $n>1$, the Taylor series givesSince $0<(k-1)/[2k(k+1)]<1/2$, a geometric series bounds $u_n<1/[2n(n-1)]\le n^{-2}$ for $n\ge 2$. The comparison test for series therefore proves $\sum u_n<\infty$. The finite telescoping sum $\ln a_{n+1}=-1+{\sum }_{k=1}^n{u}_k$ shows that $a_n$ increases to a finite positive limit. This argument establishes the existence of the constant in Stirling's formula; its value $1/\sqrt{2\pi }$ requires an independent identification.

For every positive integer $n$, Herbert Robbins's bounds giveEquivalently, the remainder in $\log (n!)=\log S_n+r_n$ satisfies $1/(12n+1)<r_n<1/(12n)$. This both proves that $S_n$ underestimates $n!$ and squeezes $n!/S_n$ to one, establishing Stirling's formula. Robbins published the bounds in “A Remark on Stirling's Formula” in 1955. ^[ref]

For $n=10$, the relative error of $S_{10}$ lies strictly between $1-e^{-1/121}$ and $1-e^{-1/120}$, approximately $0.823\%$ and $0.830\%$. A bound specifies the accuracy of an approximation before using it in another calculation.

Repeated integration by parts gives the Gamma integralThe logarithm of its integrand is $n\log t-t$, whose unique maximum is at $t=n$. Write $t=n+\sqrt{n}s$. For fixed $s$ and large $n$, a Taylor expansion givesThe Big O notation here is for a fixed $s$, or uniformly for $s$ in a fixed bounded interval. It does not claim a uniform approximation over the entire growing integration domain. Consequently the peak, divided by its height, approaches $e^{-s^2/2}$.

The exact substitution givesReplacing the integral by the Gaussian integral ${\int }_{\mathbb{R}}{e}^{-s^2/2}ds=\sqrt{2\pi }$ produces $S_n$. Thus the estimate is peak height $n^n{e}^{-n}$ times width $\sqrt{n}$ times the area of a standard Gaussian peak. This is Laplace's method in a concrete form; controlling the integral's tails justifies the replacement. The next terms, including their products after exponentiation, produce the Stirling correction by Gaussian moments.

In Laplace method for the Gamma function, scaling $t=x(1+s/\sqrt{x})$ gives a standard Gaussian integral. The first even correction polynomial is $s^2-7s^4/12+s^6/18$. Its expectation under a unit-variance centered Gaussian is $1-7\cdot 3/12+15/18=1/12$. The amplitude expansion contributes to this coefficient as well as the phase expansion; expanding only the latter gives the wrong Stirling formula correction.

For nondecreasing real $\beta$, convergence of the displayed improper integral forces $\beta (x)/x\to 1$. For $\lambda >1$, the integral over $[X,\lambda X]$ tends to zero and is at least $(1-{\lambda }^{-1})\beta (X)/X-\log \lambda$. The integral over $[X/\lambda ,X]$ tends to zero and is at most $(\lambda -1)\beta (X)/X-\log \lambda$. These bound the upper and lower limits by $\log \lambda /(1-{\lambda }^{-1})$ and $\log \lambda /(\lambda -1)$. Letting $\lambda \downarrow 1$ squeezes both to one. The limit of the improper integral must be finite.

A function is homogeneous of degree $k$ when rescaling its arguments by a common scalar $\lambda$ rescales its value by ${\lambda }^k$, whenever the arguments and that power are defined. For integer $k$ over a vector space, one can require every nonzero scalar; a homogeneous polynomial gives an example. On a domain stable under positive real rescaling, the positive-scaling version allows real $k$. Degree-one positive homogeneity is the convention used for norms and many convex functions. The Euler theorem for homogeneous functions turns this scaling law into a differential identity for a differentiable function.

Homogeneity of degree $k$ means that multiplying all arguments by a common positive factor $a$ multiplies the output by $a^k$. A homogeneous function is the function possessing this property. This is different from degree-one positive homogeneity; investment values with constant relative risk aversion utility usually have degree $1-R$.

A function on a cone is positively homogeneous of degree one when $f(tX)=tf(X)$ for every $t>0$. If such a function is concave, thenbecause homogeneity rewrites the left-hand side as $(1+t)f((X+tY)/(1+t))$ before concavity is applied.

Let $f$ be a differentiable function on an open domain stable under positive rescaling, with $f(\lambda x)={\lambda }^{k}f(x)$ for $\lambda >0$. The chain rule differentiates this homogeneous function identity at $\lambda =1$ to giveIf $f$ is twice differentiable, differentiating once more gives ${\sum }_i{x}_i{\partial }_i{\partial }_{j}f=(k-1){\partial }_{j}f$. In degree one its Hessian matrix therefore annihilates the radial vector $x$. This is the identity used in the no-scale identity from degree-one homogeneity. For an extensive quantity, the degree-one formula also supplies the thermodynamic identity behind the Gibbs-Duhem equation.

For a twice continuously differentiable homogeneous function $h(x,y)$ of degree one, Euler theorem for homogeneous functions gives $x{h}_x+y{h}_y=h$. Differentiation first with respect to $x$ and then, separately, with respect to $y$ givesMultiplying by $x$ and $y$, respectively, shows that $x^2{h}_{xx}=y^2{h}_{yy}=-xy{h}_{xy}$. This identity can verify a second-order partial differential equation without expanding a complicated homogeneous expression twice.

Let $f$ be a twice continuously differentiable function on an open domain stable under positive scaling, with $f(\lambda x)={\lambda }^{n}f(x)$. Apply the chain rule twice to the scalar function $\lambda \mapsto f(\lambda x)$ and set $\lambda =1$. The result isIn two variables, this reads $x^2{f}_{xx}+2xy{f}_{xy}+y^2{f}_{yy}=n(n-1)f$. In matrix notation it is $x^T({\nabla }^{2}f)x=n(n-1)f$, where ${\nabla }^{2}f$ is the Hessian matrix. Equivalently, the radial differential operator $D={\sum }_i{x}_i{\partial }_i$ obeys $D^{2}f=Df+{\sum }_{i,j}{x}_i{x}_j{\partial }_i{\partial }_{j}f$; substituting $Df=nf$ explains the factor $n(n-1)$.

On an open domain invariant under positive scaling, $D{f}_x(x)=cf(x)$ implies $f(\lambda x)={\lambda }^{c}f(x)$. Indeed, $\frac{d}{d\lambda }[{\lambda }^{-c}f(\lambda x)]={\lambda }^{-c-1}[D{f}_{\lambda x}(\lambda x)-cf(\lambda x)]=0$. The converse follows by differentiating the scaling identity at $\lambda =1$. Connectedness of the whole domain is unnecessary: the argument works along each positive ray.

For a sufficiently regular homogeneous function $f$ of degree $n$ on a three-dimensional region, Euler theorem for homogeneous functions gives $x\cdot \nabla f=nf$. Hence $\nabla \cdot (fx)=(n+3)f$, and the divergence theorem gives the displayed reduction when $n\ne -3$. At degree $-3$ only the zero-flux identity follows, so division by $n+3$ is unavailable. Singular functions additionally require an excision or limiting argument; the surface contribution at a removed singularity must not be silently dropped. On the sloping side of a cone with its apex at zero, $x$ is tangent to the side and $x\cdot n=0$, so only the cap contributes to this formula.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.

A positive measurable function $\ell$ is slowly varying at infinity iffor every fixed $u>0$. It has regular variation of index zero. “Slowly” concerns relative change under multiplication of the argument; it does not mean that the function must be bounded, have a small derivative, or converge to a finite value.

Every positive constant is a slowly varying function. So are $\log t$, $1/\log t$ and $(\log t{)}^a$ for any fixed real number $a$, on $t>1$. Indeed,The first grows without bound, the second tends to zero, and the constant does neither. Another unbounded example is $\ell (t)=e^{\sqrt{\log t}}$:This function can grow much faster than $\log t$ while retaining regular variation of index zero.