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For a function $f:[0,1]\to \mathbb{R}$ and an integer $n\ge 1$, its Bernstein polynomial isThe Bernstein basis supplies nonnegative weights whose sum is one. Equivalently, the weights are the probabilities of a binomial distribution, and the result is an expected value. At a given $x$, this averages the sampled values $f(k/n)$ with weights concentrated near $k/n=x$.^[ref]

The endpoints are exact: $B_{n}f(0)=f(0)$ and $B_{n}f(1)=f(1)$. Interior sampled points need not be interpolated. For example, binomial distribution moments give $B_n(x)=x$ and $B_n(x^2)=x^2+x(1-x)/n$. Thus every affine function is reproduced exactly, whereas $x^2$ is generally changed. Bernstein polynomial degree preservation is a statement about degree of a polynomial, not equality of the entire polynomial.

For a continuous $f$, the uniform continuity of $f$ and the variance $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$ explain uniform convergence. Split the expected value of the error into $|K/n-x|<\delta$, where uniform continuity makes the change small, and its complement, whose probability is at most $1/(4n{\delta }^2)$ by Chebyshev's inequality. The bounded function $f$ bounds the remaining error. This proves convergence without assuming differentiability.

For $f\in C([0,1{]}^d)$, the tensor-product Bernstein polynomial is the expected value of $f(X_1/n,\dots ,X_d/n)$ for independent random variables $X_j\sim \mathrm{Bin}(n,x_j)$. The same uniform continuity and coordinatewise variance estimates give uniform convergence for each fixed finite $d$.

Cauchy's functional equation asks for functions $f:\mathbb{R}\to \mathbb{R}$ satisfyingThe identity must hold for every pair of real numbers, and a solution is an additive function. It says that combining two inputs before applying $f$ gives the same result as combining their outputs.^[ref]

For example, $f(x)=3x$ satisfies the identity because $3(x+y)=3x+3y$. At $x=2/3$ and $y=5/6$ the two sides giveFor positive widths, a rectangle of height three makes this visible: cutting its width into $x$ and $y$ partitions its area into $3x$ and $3y$. The same identity also holds for negative inputs, where $3x$ is a signed output rather than an ordinary area.

Every function $f(x)=cx$ with a fixed real number $c$ solves Cauchy's functional equation. The surprising question is whether these are all the solutions. Rational homogeneity of an additive function forces $f(q)=qf(1)$ for every rational number $q$, but reaching an irrational number requires an additional argument. Continuity at just one point or being a bounded function on an interval of positive length supplies that argument; without such a condition, discontinuous additive functions exist.^[ref]

This makes Cauchy's functional equation a useful test of the boundary between algebra and analysis: a rule about addition fixes all rational multiples, while a tiny amount of continuity fixes the entire real line. For example, the continuous exponential functional equation reduces to an additive function after taking a logarithm, so this result explains why its nonzero solutions are exponential 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.

An entire function is a holomorphic function whose domain of a function is the whole complex plane. Its derivative exists at every $z\in \mathbb{C}$. Every polynomial, the complex exponential function $e^z$, the product $z{e}^z$, and the complex sine function $\sin z$ are entire functions. The Taylor series of an entire function about any finite centre converges throughout the complex plane.

The complex exponential function is a particularly revealing entire function. Writing $z=x+iy$, Euler's formula givesThus the complex exponential function never takes the value zero. Conversely, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages under the complex exponential function areHere $\ln |w|$ is the real natural logarithm, and $k$ ranges over the integers. Substitution gives $e^{z_k}=|w|e^{i\theta }=w$; equality of moduli and arguments shows these are all the solutions. The complex exponential function therefore has image exactly $\mathbb{C}\setminus \{0\}$, with infinitely many preimages for each nonzero value.

In the pictured vertical strip $-1\le x\le 1$, the real part sets the modulus between $e^{-1}$ and $e$, while the imaginary part sets the argument modulo $2\pi$. The two drawn bands each make one full turn around the same annulus; their corresponding points differ by $2\pi i$ and have identical images. Allowing all real $x$ gives every positive modulus, so the whole complex plane covers every nonzero output.

An entire function has exceptionally strong global constraints. The Liouville theorem says that a bounded function which is also an entire function must be a constant function. Accordingly, the nonconstant entire functions above are unbounded somewhere in the complex plane, even when their restrictions to the real axis are bounded: the complex sine function satisfies $\sin (iy)=i\sinh y$, growing through the hyperbolic sine. The Little Picard theorem adds a different constraint: an unbounded nonconstant entire function still cannot leave two distinct complex values out of its image. These results make entire functions useful when global bounds or restrictions on the image can establish rigidity without solving for the function explicitly.

In the opening of his 1880 memoir, Émile Picard credits Karl Weierstrass with the terminology of entire functions and describes them by everywhere-convergent power series. ^[ref]

For initial data $u_0$ on the real line, Fourier transformation or convolution with the fundamental solution gives

For a bounded function $u_0$, the convolution is well defined at every positive time because the heat kernel has integral one. The error-function diffusion front uses a Heaviside step function initial condition and gives one smoothed interface. The heat equation with interval-indicator initial data instead has two interfaces; subtracting their two error functions produces the finite-interval profile.

Let nonnegative integrable functions $k_n$ on $[0,b]$ satisfy ${\int }_0^b{k}_n\to 1$ and ${\int }_{\delta }^b{k}_n\to 0$ for every $0<\delta <b$. If $f$ is a bounded function and continuous at zero, and the products are integrable, then ${\int }_0^{b}f(x)k_n(x)dx\to f(0)$. To prove this, subtract $f(0)\int k_n$ and split the difference integral at $\delta$. The first piece is bounded by ${\sup }_{[0,\delta ]}|f-f(0)|\int k_n$; the second by $2\Vert f{\Vert }_{\infty }{\int }_{\delta }^b{k}_n$. First use continuity to choose $\delta$, then pass to the limit in $n$. The normalization term $f(0)(\int k_n-1)$ also vanishes.

For $b=1$, the Gaussian integral shows that $k_n(x)=2n{e}^{-n^2{x}^2}/\sqrt{\pi }$ has these properties. The factor two compensates for retaining only the nonnegative half of the full normalized Gaussian profile. Replacing it by $n/\sqrt{\pi }$ would instead give limiting mass $1/2$.

Let $h:Y\to \mathbb{R}$ be continuous, without a boundedness assumption. The homeomorphism $\theta (t)=t/(1+|t|)$ maps the real line onto $(-1,1)$, with inverse ${\theta }^{-1}(s)=s/(1-|s|)$. The bounded function $g=\theta \circ h$ extends by the preceding bounded construction to a continuous real-valued $G$ on $X$. Since $|G|<1$ on $Y$, the strict-range modification gives a continuous $\tilde{G}$ with range in $(-1,1)$ and the same restriction to $Y$. Thereforeis continuous everywhere: its denominator is pointwise positive. On $Y$ it equals ${\theta }^{-1}(\theta (h))=h$. Thus $Y$ is $C$-embedded. It is important to modify the extension before applying the inverse: an extension allowed to attain $1$ or $-1$ outside $Y$ would make this formula undefined there.

Let $f$ be continuous on a closed bounded interval $[a,b]$. First prove that $f$ is a bounded function. If it failed, we could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x_{\ast }$. Because the interval is closed, $x_{\ast }\in [a,b]$. Continuity gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j\to \infty$.

Now let $M=\sup \{f(x):x\in [a,b]\}$, which exists by completeness of the real numbers and the bound just proved. Choose $y_n\in [a,b]$ with $f(y_n)>M-1/n$. A convergent subsequence has limit of a sequence $y_{\ast }\in [a,b]$, and continuity gives $f(y_{\ast })=M$. Applying the same argument to $-f$ gives attainment of the infimum. This proves the extreme value theorem: the maximum and minimum both exist and are attained. The argument also covers a one-point interval, where the conclusions are immediate.

Let $g$ be a bounded function with $P_{f}g=0$. A bounded density tilt realizes this direction:These are nonnegative probability density functions, since $\int f_t=1+t{P}_{f}g=1$. The uniform Taylor expansion of the square root givesThe squared L2 norm of this remainder is $O(t^4{P}_f{g}^4)=o(t^2)$, since $g$ is bounded. Thus the statistical path is differentiable in quadratic mean with score function $g$.

Conversely every score function is centered by the mean-zero score identity under quadratic-mean differentiability. Choosing all these bounded density tilts therefore gives the statistical tangent setThis is a valid choice of statistical tangent set; it does not assert that every possible score function in the unrestricted density model is bounded.

Consider any differentiable-in-quadratic-mean path with score function $g$. Put ${\delta }_t=\sqrt{f_t}-\sqrt{f}=(t/2)g\sqrt{f}+r_t$, where $\Vert r_t{\Vert }_2=o(|t|)$. The quadratic-mean to L1 density derivative follows fromBy the Cauchy-Schwarz inequality,Since $a$ is a bounded function, multiplying this L1 norm bound by $\Vert a{\Vert }_{\infty }$ provesThe last equality uses $P_{f}g=0$. The derivative is a bounded linear functional of $g\in L^2(P_f)$, so the required pathwise differentiability of a statistical functional holds, in particular relative to the statistical tangent set from part (b). Its derivative isFor the explicit bounded density tilts in part (b), this derivative is also obtained by direct integration, with no remainder term.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

For a bounded function, the lower and upper integrals are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.

Here $u(x)=1/x-\lfloor 1/x\rfloor$ for $x>0$, so $0\le u<1$. On $[\delta ,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta ]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta \downarrow 0$ proves integrability on $[0,1]$.

For a partitionputThe upper and lower Darboux sums areThe upper and lower Darboux integrals areThe bounded function is Riemann integrable when these two numbers agree; their common value is its integral.

The Darboux–Riemann criterion says that a bounded function on $[a,b]$ is Riemann integrable exactly when, for every $\epsilon >0$, some partition $P$ satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, every lower sum is at most every upper sum. Taking the supremum of lower sums and infimum of upper sums, the criterion makes their difference smaller than every positive $\epsilon$, so they are equal; this common value is the Riemann integral.

Uniform continuity means that for every $\epsilon >0$ there is a single $\delta >0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\epsilon$ for all $x,y$. A $d^{\prime }$-Cauchy sequence converges uniformly pointwise to a bounded function; passing a uniform-continuity estimate through a sufficiently close member proves that the limit is uniformly continuous. Thus $C_{b,u}(X)$ is complete.

If $f\in C_0({\mathbb{R}}^n)$, decay at infinity and compactness of a ball make it bounded. Uniform continuity on a sufficiently large compact ball, together with small values outside it, proves global uniform continuity. Hence $C_0\subset C_{b,u}$.

A uniform limit of functions vanishing at infinity also vanishes at infinity, so $C_0$ is closed. It is not compact: translate a fixed compactly supported bump of height one so that the supports are disjoint. The resulting sequence has pairwise sup distance one and no convergent subsequence.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.

Write $f_n\to f$ and $g_n\to g$ with uniform convergence. The triangle inequality givesHence $f_n+g_n\to f+g$ with uniform convergence.

If every $f_n,g_n$ is a bounded function, part iii gives constants $M_f,M_g$ bounding the two families. Their pointwise limits satisfy $|f|\le M_f$ and $|g|\le M_g$. ConsequentlyThus both sums and products have uniform convergence under the stated hypotheses. The product estimate is the uniform convergence product rule for bounded functions.

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.

For $a>0$, the fact that sine is a bounded function gives $|f_a(x)-f_a(0)|\le |x{|}^a$. Given $\epsilon >0$, choose $\delta ={\epsilon }^{1/a}$. If $|x|<\delta$, then $|f_a(x)-f_a(0)|<\epsilon$. This proves continuity directly from its definition.

For $a\le 0$, use the first sequence from part (b), for which $f_a(x_n)=x_n^a$. When $a=0$ these values are one; when $a<0$ they tend to $+\infty$. In either case, for $\epsilon =1/2$ every neighbourhood of zero contains a sufficiently late $x_n$ with $|f_a(x_n)-f_a(0)|\ge 1/2$. This contradicts the definition of continuity. Hence