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An antiderivative recovers a function from its local rate of change. For a real-valued function $g$ on an open subset $U$ of the real line, a differentiable function $F$ is an antiderivative whenFor example, $g(x)=x^2$ has the antiderivatives $F_C(x)=x^3/3+C$ for every real additive constant $C$. Their derivatives agree, while their values can differ. On an interval, the constant of integration describes the entire freedom in the answer. ^[ref]

The indefinite-integral notation $\int g(x)dx=F(x)+C$ records this family on an interval. For the illustrated family, the extra condition $F(1)=2$ selects $C=5/3$. Thus reversing differentiation requires both a rate and a reference value if a unique function is wanted.

The fundamental theorem of calculus supplies a construction when $g$ is continuous on an interval $I$. Choose $x_0\in I$ and set $F(x)={\int }_{x_0}^{x}g(t)dt$; then $F^{\prime }=g$. Integrability alone does not guarantee a classical antiderivative. The Heaviside step function is Riemann integrable on bounded intervals, but cannot be a derivative throughout an interval containing its jump: Darboux theorem for derivatives requires every intermediate value. ^[ref]

For four real-valued functions on a finite Cartesian product, let $\Lambda (f_{00},f_{01},f_{10},f_{11})={\mathbb{E}}_{x_0,x_1,y_0,y_1}{\prod }_{i,j=0}^1{f}_{ij}(x_i,y_j)$. Repeated Cauchy-Schwarz inequalities giveFirst separate the two $y$ averages and apply Cauchy-Schwarz inequality in $(x_0,x_1)$. Each resulting squared factor is ${\mathbb{E}}_{y,y^{\prime }}({\mathbb{E}}_{x}f(x,y)f(x,y^{\prime }))({\mathbb{E}}_{x}g(x,y)g(x,y^{\prime }))$, bounded by $\Vert f{\Vert }_{\square }^2\Vert g{\Vert }_{\square }^2$ by another Cauchy-Schwarz inequality. Expanding the four factors of $f+g$ and applying this inequality to each of the sixteen terms gives the triangle inequality for the box norm.

For nonempty finite sets $X,Y$ and a function $f:X\times Y\to \mathbb{C}$ on their Cartesian product, using uniform expectations, defineThis equals ${\mathbb{E}}_{y,y^{\prime }}|{\mathbb{E}}_{x}f(x,y)\overline{f(x,y^{\prime })}{|}^2$. Hence it is nonnegative, and vanishing forces $f=0$ by taking $y=y^{\prime }$. Absolute homogeneity and the box Cauchy-Schwarz inequality establish that it is a norm. For a real-valued function the complex conjugates can be omitted. The bilinear correlation bound for the box norm also holds for complex functions.

For a real-valued function $H$ on a nonempty finite Cartesian product, let $T_{H}v(x)={\mathbb{E}}_{y}H(x,y)v(y)$ with uniform inner products. The box norm obeys $\Vert H{\Vert }_{\square }^4=\mathrm{Tr}((T_H{T}_H^{\ast }{)}^2)={\sum }_j{\sigma }_j^4$, where ${\sigma }_j$ are its singular values. For a biregular graph this separates the constant component ${\gamma }^4$ from the other fourth powers.

Continuous real-valued functions on a space with a countable dense subset are determined by their restrictions to that subset. The equality set of two such functions is closed and contains the dense subset. Hence there are at most $|\mathbb{R}{|}^{{\aleph }_0}=2^{{\aleph }_0}$ such functions. If the domain is nonempty, distinct constant functions give the matching lower bound.

Catastrophic cancellation occurs when subtraction of nearly equal approximations removes their matching leading digits and exposes earlier roundoff error. The subtraction can itself be exact on the stored operands: the information may already have been lost while obtaining them. Goldberg distinguishes this from cancellation of accurately known quantities, which can be harmless.^[ref]

For $x>0$, consider the single real-valued functionThe rationalization of a square-root difference proves equality in exact arithmetic. The two evaluations use the same input $x$, but the direct version first forms a value near one and then subtracts one. The rearranged version keeps the small input in its numerator and divides by a denominator near two.

For binary64 x = 1e-16, the rounded sum 1.0 + x is one, so sqrt(1.0 + x) - 1.0 is zero. The rationalized evaluation returns approximately $5\times 10^{-17}$. A 100-digit Decimal calculation using the exact stored input gives relative errors of one and about $2.50\times 10^{-17}$ respectively. At x = 1e-12, the direct result is about $5.0004445029\times 10^{-13}$, with relative error $8.89\times 10^{-5}$; the rearranged result is about $4.99999999999875\times 10^{-13}$, with relative error $1.21\times 10^{-17}$.

The plot uses Decimal.from_float(x) to preserve the binary input exactly, computes $x/(\sqrt{1+x}+1)$ at 100 decimal digits, and verifies the reference at 160 digits. Exact conversion of a stored binary value is documented by Python.^[ref] Each error is $|\hat{f}-f|/|f|$, with the computed result converted exactly to Decimal before comparison. Hence the curves measure evaluation error for one common input, rather than a mixture of input conversion error and evaluation error. The plotted interval $10^{-16}\le x\le 10^{-1}$ is far above the subnormal number range.

The original problem is well conditioned with respect to relative changes in $x$. Using the derivative $f^{\prime }(x)=1/(2\sqrt{1+x})$ and rationalization of a square-root difference, its relative condition number isIt tends to one as $x$ tends to zero. The direct evaluation instead introduces the intermediate difference $s-1$ with $s=\sqrt{1+x}$; an error of order unit roundoff in $s$ is large relative to $f(x)\approx x/2$. This is a failure of the chosen algorithm's accuracy despite a small condition number for the same real-valued function. The rationalized evaluation avoids that intermediate catastrophic cancellation, retaining relative accuracy near the unit roundoff scale throughout the plotted interval. Recognizing this structure can improve a calculation far more than merely printing extra digits.

For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

A subset is Cb-embedded if every bounded continuous real-valued function on it extends to a bounded continuous real-valued function on the ambient space. A bounded extension can be clipped to preserve the original bound. This is weaker than C-embeddedness: the natural numbers in their Stone-Čech compactification have the bounded extension property, but their unbounded identity function cannot extend to a compact space.

A subset $Y\subseteq X$ is C-embedded if every continuous real-valued function on its subspace topology extends continuously to $X$. Closed subsets of normal spaces have this property by the Tietze extension theorem. The condition also makes sense for nonclosed subsets.

A continuous function sends sufficiently nearby function inputs to nearby function outputs. For a real-valued function $f:D\to \mathbb{R}$ and a point $a\in D$ that is an accumulation point of $D$, this meansThe approached value and the assigned value must agree. Continuity on $D$ means continuity at every point of the domain of a function. At an endpoint of an interval, only approaches inside that domain of a function count. The epsilon-delta definition of continuity below also applies at an isolated point, where continuity holds automatically.^[ref]

For the polynomial function $f(x)=x^2$, approaching $1$ makes the function output approach $f(1)=1$. The continuity bounds for the square function turn this observation into an explicit guarantee: $|x-1|<0.2$ implies $|x^2-1|<0.6$. This is useful whenever a small error in an function input must produce a controlled error in an function output. The required input tolerance can depend on the point and on the chosen output tolerance.

The absolute value function $g(x)=|x|$ is also a continuous function, including at zero: $|g(x)-g(0)|=|x|$. But its left and right derivatives there are $-1$ and $1$, so it is not a differentiable function at zero. Continuity controls values without requiring a tangent line. Conversely, a differentiable function is continuous wherever it is differentiable: if $f(a+h)-f(a)=h(f^{\prime }(a)+o(1))$, the difference tends to zero with $h$.

Several useful consequences reuse this one local idea. Addition, multiplication and function composition preserve continuity; a quotient of functions is continuous where its denominator is nonzero. On an interval, the intermediate value theorem turns continuity into a guarantee that no intermediate function output is skipped. On a compact set, the extreme value theorem guarantees attained maximum and minimum values. Those extra hypotheses matter: continuity alone on an unbounded domain of a function does not provide a global bound or an attained global maximum.

A rational function such as $x\mapsto 1/x$ is continuous on its domain of a function $\mathbb{R}\setminus \{0\}$. Zero is excluded from that domain of a function; the missing point prevents a continuous extension there, but it does not make the existing map discontinuous at one of its actual inputs. Always state the domain of a function before judging continuity.

For a real-valued function $H$ on a nonempty finite Cartesian product $X\times Y$, its normalized cut norm is the maximum of $|{\mathbb{E}}_{x,y}H(x,y)1_A(x)1_B(y)|$ over $A\subseteq X,B\subseteq Y$. It measures all rectangular subset discrepancies. The convention using arbitrary test functions bounded by one is equivalent within a factor of four, by decomposing their positive and negative parts into indicator functions.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

A real-valued function that is continuous on a nonempty compact space attains both a global minimum and a global maximum. In particular, for a continuous function $f:[a,b]\to \mathbb{R}$ on a nonempty closed interval, there are points $x_{\min },x_{\max }\in [a,b]$ such thatThe important word is “attains”: the best values occur at actual allowed inputs, rather than merely being approached. The interval version is ^[ref].

For a worked example, take $f(x)=(x-1{)}^2$ on $[-1,2]$. Its global minimum is $f(1)=0$, and its global maximum is $f(-1)=4$. Nonnegativity gives the first answer; $|x-1|\le 2$ on the specified closed interval gives the second. The endpoint $x=-1$ is essential: looking only for a zero derivative would find the minimum and miss the maximum.

For comparison, the continuous function $g(x)=x$ on the open interval $(0,1)$ has infimum $0$ and supremum $1$, but neither is a value of $g$. Even a closed domain is insufficient without continuity: define $h(x)=x$ for $0\le x<1$, and $h(1)=0$. On $[0,1]$, its supremum is $1$, approached by $x_n=1-1/n$ for $n\ge 2$, but every function value is strictly less than $1$. This endpoint change makes $h$ discontinuous, preventing attainment. On the unbounded real line, the continuous function $x\mapsto x$ has neither a global minimum nor a global maximum. These examples show what can fail when a hypothesis is dropped; functions outside the hypotheses may still have extrema.

The extreme value theorem is an existence tool for optimization: first establish that a continuous objective function on a nonempty compact feasible set has a best value, then investigate where it occurs. It neither locates the optimizing point nor promises uniqueness or differentiability. For example, the absolute value function $|x|$ on $[-1,2]$ has its global minimum at zero although no derivative exists there; a constant function has every input as a minimizer. This role in optimization is explained in University of Toronto MAT237 notes, section 1.4: compactness and the extreme value theorem.

A short proof outline identifies the mechanism. The continuous image of a compact space is compact; a nonempty compact subset of the real line contains its infimum and supremum. For a closed interval, one can instead choose inputs whose outputs approach the supremum, extract a convergent subsequence by the Bolzano-Weierstrass theorem, and use continuity to make its limiting output equal that supremum. The closed interval retains the limiting input. Repeating the argument for the infimum supplies the minimum.

The fundamental period is the least positive period of a function, when one exists. For $f(x)=\sin (2\pi x)$ it is one; for $f(x)={\sin }^2(2\pi x)$ it is $1/2$. A periodic function need not have a fundamental period: a constant function has every positive period, and periodic function without a least positive period is a nonconstant exception. Every nonconstant real-valued function that is continuous and periodic on the real line does have a least positive period.

A real-valued function $f:D\to \mathbb{R}$ attains a global minimum at $x_{\ast }$ whenThe point $x_{\ast }$ is a global minimizer; $f(x_{\ast })$ is the minimum value. A global maximum reverses the inequality throughout the same domain of a function. Unlike a local minimum or local maximum, a global minimum or global maximum compares every permitted input, however distant. The domain of a function is part of the claim.

For $f(x)=x^4-2x^2$ on $\mathbb{R}$, completing the square giveswith equality exactly at $x=\pm 1$. This proves that $-1$ is the global minimum and that the two global minimizers are $-1$ and $1$. Neither is a strict global minimizer, because the other attains the same value; both are strict local minima. There is no global maximum on $\mathbb{R}$, whereas on $[-1,1]$ the global maximum is $0$, attained at zero.

Finding critical points supplies candidates rather than a global certificate. The squared expression above is beautiful because one inequality compares infinitely many inputs at once. A global minimum is always a local minimum under the relative-neighbourhood convention; the converse fails. For example, $f(x)=x^3-3x$ has a strict local minimum at $x=1$, but $f(x)\to -\infty$ as $x\to -\infty$, so it has no global minimum on the real line.

Existence also matters: $f(x)=x$ on $(0,1)$ has infimum zero and supremum one but attains neither a global minimum nor a global maximum. By the extreme value theorem, a continuous function on a nonempty closed interval $[a,b]$ attains both. To find those values, compare all interior critical numbers, including points where the derivative does not exist, and both endpoints. Checking only stationary points can miss a corner or an endpoint. ^[ref]

If $f$ is a real-valued function and an integrable function, its Fourier transform has Hermitian symmetry: $F(-u)=\overline{F(u)}$. The value at negative frequency is the complex conjugate of the value at positive frequency.

On an open convex set, a twice-differentiable real-valued function is convex if and only if its Hessian matrix is a positive semidefinite matrix at every point. In particular continuous second partial derivatives suffice for the differentiability hypothesis. Restrict $g$ to any line segment: $\varphi (t)=g(x+tv)$ has ${\varphi }^{\prime \prime }(t)=v^{\mathsf{T}}H(g)(x+tv)v$. The second-derivative criterion for convexity proves both implications, because convexity is equivalent to convexity of all such restrictions.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

For integrable real-valued functions $f,g$ on a measure space $(X,\mu )$ and fixed real numbers $a,b$, their linear combination $af+bg$ is integrable, and the Lebesgue integral satisfiesThe same rule applies to any finite summation: ${\int }_X{\sum }_{j=1}^n{a}_j{f}_{j}d\mu ={\sum }_{j=1}^n{a}_j{\int }_X{f}_{j}d\mu$ for integrable functions $f_j$ and fixed scalars $a_j$. Finite signed values matter: integrability ensures all these integrals are finite, so no undefined $\infty -\infty$ occurs.

This is the distributive property of finite summation carried into integration: a fixed factor can be pulled out, and a finite sum can be integrated term by term. For simple functions, distribute the finite sums of value times measure; approximation extends additivity to nonnegative measurable functions, and splitting into a positive part and negative part gives the integrable signed case. See Lecture 12: Lebesgue Integrable Functions, the Lebesgue Integral and the Dominated Convergence Theorem (Proposition 122: linearity for integrable functions). For example, ${\int }_0^1(2x-3)dx=2{\int }_0^{1}xdx-3{\int }_0^{1}1dx=-2$. With a probability measure, linearity of integration gives linearity of expectation.