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On the vector space of continuous functions $f:[0,1]\to \mathbb{R}$ of bounded variation with $f(0)=0$, the total variation of a function is a norm. Its triangle inequality and absolute homogeneity of a norm follow by applying the corresponding absolute value identities to every partition of an interval, then taking the supremum. The partition of an interval $(0,t,1)$ gives $|f(t)|\le V_0^1(f)$, so zero total variation of a function forces $f=0$. The same inequality gives $\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_V$, proving that convergence for this norm implies uniform convergence. Without $f(0)=0$, every constant function would have zero total variation of a function.

For a univalent exterior map $G(\zeta )=\zeta +b_0+{\sum }_{n\ge 1}{b}_n{\zeta }^{-n}$, the image of $|\zeta |=R>1$ is a positively oriented Jordan curve. Green's theorem gives its enclosed area as $\pi (R^2-{\sum }_{n\ge 1}n|b_n{|}^2{R}^{-2n})$. Nonnegativity and $R\downarrow 1$ give the displayed bound. Laurent-series differentiation and integration on a circle of radius greater than one are justified by uniform convergence there.

For $m\ge 1$ and $n\ge 2$, the Bernstein polynomial of the monomial $x^m$ satisfiesIt follows by applying $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ and expanding $k=(k-1)+1$ with the binomial theorem. The coefficients sum to one; only the first tends to one as $n\to \infty$. Induction on $m$ proves uniform convergence of these monomial approximants, starting from $B_n(1)=1$.

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 $d$.

Suppose a series of functions $\sum f_n$ has uniform absolute convergence on $E$ and $(g_n)$ is a uniformly bounded family of functions there, with common bound $M$. The triangle inequality givesThe right-hand side tends uniformly to zero as $q>p\to \infty$. The Uniform Cauchy criterion proves uniform convergence of $\sum f_n{g}_n$. Applying the same estimate to $\sum |f_n{g}_n|$ proves uniform absolute convergence too.

The shared bound over both $n$ and $x$ matters: separate bounds for each $g_n$ can grow too fast. For example, the constant functions $f_n(x)=2^{-n}$ and $g_n(x)=2^n$ have a summable original series but products identically one.

A Chebyshev series expands a function in Chebyshev polynomials $T_j$. If $|c_j|\le M{q}^j$ for $0<q<1$, then $|T_j(x)|\le 1$ on $[-1,1]$ gives uniform convergence and the tail bound $\Vert f-{\sum }_{j=0}^d{c}_j{T}_j{\Vert }_{\infty }\le M{q}^{d+1}/(1-q)$ by the geometric series.

Let $K$ be a compact Hausdorff space. A test subspace $H\subset C(K,\mathbb{R})$ containing a strictly positive function is sufficient for uniform convergence of positive linear operators on continuous functions if, for each $x\in K$, the only positive linear functional agreeing with point evaluation on $H$ is that evaluation itself. By the Riesz-Markov-Kakutani representation theorem, this is uniqueness of the positive measure with the prescribed moments. Boundedness from the positive test function and the Banach-Alaoglu theorem prove the assertion by compactness. When $1\in H$, nonnegative tests with a unique zero at each prescribed $x$ give a convenient sufficient condition.

For a fixed bounded measurable uniformly elliptic operator in divergence form, solutions bounded by one have uniform local Caccioppoli inequalities and Hölder continuity estimates. A diagonal subsequence argument yields a local weak $W^{1,2}$ limit, while the Arzelà-Ascoli theorem gives uniform convergence on a fixed smaller closed Euclidean ball. The weak equation passes to the limit because the coefficients are fixed and bounded. A global supremum normalization does not force the interior limit to be nonzero.

If $u_j\to u$ in weak convergence of distributions, then every fixed distributional derivative converges too. For each test function $\varphi$, its derivative $D^{\alpha }\varphi$ is another test function, andThus a series converging as distributions can be differentiated term by term in that topology. For example, uniform convergence on compact sets implies convergence of the associated regular distributions, even when the ordinary series of derivatives does not converge pointwise.

Target: Even-weight principal-value distribution. Parent: Principal-value reciprocal distribution. Intended file: distribution-theory.bigb.

The continuous functions from the p-adic integers to the P-adic numbers form a Banach space over ${\mathbb{Q}}_p$ with supremum norm $\Vert f{\Vert }_{\infty }={\sup }_{x\in {\mathbb{Z}}_p}|f(x){|}_p$. Compactness of the domain bounds the norm, and uniform convergence in the complete codomain proves completeness. Translation and the forward difference operator $\Delta f(x)=f(x+1)-f(x)$ are bounded operators, with $\Vert \Delta f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }$.

Here $C^{\infty }([0,1])$ means the real- or complex-valued functions continuous on $[0,1]$, smooth on $(0,1)$, with every interior derivative having a continuous extension to the endpoints; endpoint values of derivatives are these extensions. On this space of smooth functions on a compact interval, defineThese finite-valued seminorms increase with $k$ and separate points. In fact every $p_k$ is a norm, since it includes $p_0(f)=\Vert f{\Vert }_{\infty }$. The terminology seminorm includes norms; it does not assert that every measurement has a nontrivial kernel of a seminorm. Smooth convergence means $p_k(f_n-f)\to 0$ for every fixed $k$, equivalently uniform convergence of every derivative separately. It does not demand one tolerance or one convergence index that works for all orders simultaneously. The maximum convention is equivalent to the derivative supremum norm convention using a finite sum, since $p_k(f)\le {\sum }_{j=0}^k\Vert f^{(j)}{\Vert }_{\infty }\le (k+1)p_k(f)$.

For a worked example, $f(t)=t^3$ gives $p_0(f)=1$, $p_1(f)=3$, $p_2(f)=6$, and $p_k(f)=6$ for $k\ge 3$. The linear functional $\ell (f)=f^{\prime \prime }(1/2)$ obeys $|\ell (f)|\le p_2(f)$ and is continuous. Similarly, differentiation is a continuous linear operator $D:f\mapsto f^{\prime }$ for smooth convergence, because $p_k(Df)\le p_{k+1}(f)$.

The space of smooth functions on a compact interval is complete for this seminorm topology. A Cauchy sequence has uniform limits $g_j$ for its derivatives of each order. Passing to the limit in the fundamental theorem of calculus identitygives $g_j(t)-g_j(0)={\int }_0^t{g}_{j+1}(s)ds$, so $g_j^{\prime }=g_{j+1}$ and $g_0$ belongs to the same space. This is the completeness of the smooth-function metric argument; the metric there and the increasing-$p_k$ metric above induce the same smooth topology and the same Cauchy sequences.

If the partial sums of $\sum b_n$ are bounded and $a_n$ decreases to zero, then $\sum a_n{b}_n$ converges. The uniform version holds for functions $b_n(x)$ when their partial sums have one bound $M$ independent of $x$. Indeed partial sums starting at index $N$ have absolute value at most $2M$, and summation by parts bounds every weighted tail from $N$ to $K$ by $2M{a}_N$. The uniformly Cauchy sequence criterion proves uniform convergence. This justifies integrating a conditionally convergent Fourier series on intervals staying away from its endpoint jump.

On a finite-measure measurable set, a sequence of measurable functions converging pointwise almost everywhere converges uniformly outside a set of arbitrarily small measure. For tolerance $1/k$, the sets on which every index after $N$ has error at most $1/k$ increase to full measure as $N\to \infty$. Choose $N_k$ with exceptional measure at most $\epsilon 2^{-k-1}$ and intersect the good sets. Countable subadditivity bounds the discarded measure, and the resulting tail estimates prove uniform convergence. Finite measure is essential: $1_{[n,\infty )}$ on $[0,\infty )$ is a counterexample.

The empirical distribution function counts the fraction of observations at or below a threshold. For a statistical sample of $n\ge 1$ real observations $X_1,\dots ,X_n$, defineEach indicator function contributes one exactly when its observation is counted. Equivalently, $F_n(x)=P_n((-\infty ,x])$, where the empirical measure $P_n=n^{-1}{\sum }_i{\delta }_{X_i}$ assigns probability mass $1/n$ to each observation, including repeated observations.^[ref]

Choose the illustrative statistical sample $[1,1,2,4,6]$. Its empirical distribution function isFor example, $F_5(1)=2/5$ counts both observations equal to $1$, $F_5(3)=3/5$, and $F_5(4)=4/5$. There is no observation at $3$, so there is no jump there. The five observations produce only four distinct jump locations.

The convention $X_i\le x$ makes the empirical distribution function right-continuous: once a threshold reaches an observation, that observation remains counted immediately to its right. For a finite statistical sample, the count is constant from any $x$ up to the next larger observed value, proving right continuity. Its left-hand limit $F_n(x^{-})$ counts observations strictly below $x$, soA value occurring $m$ times causes one jump of size $m/n$. At $1$ the jump is $2/5$; at $2$, $4$, and $6$ each jump is $1/5$. All jumps sum to one. Thus the empirical distribution function is a nondecreasing step function with limits zero and one, and is itself a valid cumulative distribution function.

This turns many questions about data into simple cumulative distribution function operations. The observed fraction in $(a,b]$ is $F_n(b)-F_n(a)$: for $(1,4]$ it is $4/5-2/5=2/5$, counting $2$ and $4$ but excluding both $1$s. Including the left endpoint gives $F_n(4)-F_n(1^{-})=4/5$. A sample quantile reads the same staircase in reverse: the smallest threshold reaching $1/2$ is $2$. Unlike a histogram, this representation requires no choice of bins; its jumps retain every observation and its multiplicity.

The exact description of a finite statistical sample must be distinguished from a statistical model for how it was obtained. The empirical distribution function is defined for any finite list, with no independence assumption. Once the list is observed, its counts are fixed. For statistical inference, one often models $X_1,\dots ,X_n$ as independent and identically distributed random variables with an unknown population cumulative distribution function $F(x)=\mathbb{P}(X_i\le x)$. Then $F$ describes the sampling law, while $F_n$ varies across repeated statistical samples. The value $F_5(1)=0.4$ is an observed fraction and an estimate of $F(1)$; it does not establish that the population probability is $0.4$. Even when $F$ is continuous, every finite $F_n$ is discrete.

For any threshold $x$ fixed before sampling, the indicator functions are independent Bernoulli random variables with success probability $F(x)$. Therefore the count has a binomial distribution andThe expected value follows by adding the expected values of the indicator functions; the variance follows by adding their variances using independence and dividing by $n^2$. Thus $F_n(x)$ is an unbiased estimator of $F(x)$ at a fixed threshold. Controlling the error at every threshold at once requires more than a fixed-threshold calculation: the DKW inequality supplies such control, and the Glivenko-Cantelli theorem describes its uniform convergence almost surely as $n$ grows.

Comparing $F_n$ with a proposed cumulative distribution function also motivates goodness-of-fit tests, including the Kolmogorov-Smirnov test, the Anderson–Darling test, and the Cramér–von Mises criterion. A standard Kolmogorov-Smirnov test calibration assumes a fully specified continuous population law; fitting its statistical parameters from the same statistical sample changes that calibration.^[ref]

For a positive integer $k$, multiply the exponential series by $t^{k-1}$ and integrate term by term from zero to $x$. Its uniform convergence on every compact interval justifies the operation and proves the title identity. The resulting power series converges absolutely for every real or complex $x$, by the ratio test. Repeated integration by parts evaluates the right side as an exponential times a polynomial, with its constant chosen to make the integral vanish at zero.

For Picard iteration $y_{n+1}(x)=b+{\int }_a^{x}f(s,y_n(s))ds$, $y_0=b$, assume $|f|\le M$ and a Lipschitz constant $L$ on a rectangle containing every iterate. The first difference is at most $M|x-a|$. Integrating the preceding bound multiplied by $L$ proves the displayed estimate by mathematical induction. Its uniform bounds sum to a convergent exponential series for every finite $Lh$, so the Weierstrass M-test gives uniform convergence without requiring $Lh<1$.

For two arbitrary initial functions $y,z$ in a function set preserved by the Picard iteration operator $T$, the same induction givesThus some iterate is a contraction mapping on any fixed finite interval for which every iterate stays in that set. The iterated contraction argument gives a unique fixed point of $T$ on a nonempty complete metric space, without requiring $Lh<1$. For a rectangle with field bound $M$ and vertical radius $k$, $Mh\le k$ supplies the required self-mapping property.

The normalized Fejér kernel satisfies $0\le F_n(t)\le \min \{n/(2\pi ),\pi /(2n{t}^2)\}$ on $0<|t|\le \pi$. Splitting its first absolute moment at $1/n$ gives $\int |t|F_n(t)dt\le 1/(2\pi n)+(\pi /n)\log (\pi n)$. The chaining inequality for the modulus of continuity, $\omega (f,t)\le (1+t/\delta )\omega (f,\delta )$, then bounds the Fejér sum error by $\omega (f,\delta )[1+{\delta }^{-1}\int |t|F_n]$. Taking $\delta =\log n/n$ proves the stated bound for $n\ge 2$ and hence uniform convergence for continuous periodic functions. The logarithm vanishes at $n=1$, so the displayed estimate is not intended at that index.

The period-two extension of $1-|x|$ on $[-1,1]$ is even. Its Fourier cosine series has $a_0=1$ and $a_n=2{\int }_0^1(1-x)\cos (n\pi x)dx=2[1-(-1{)}^n]/(n^2{\pi }^2)$ by integration by parts. This proves the displayed series, with only odd nonconstant harmonics. Continuity and piecewise smoothness give pointwise equality through the Dirichlet-Jordan convergence theorem; the summable coefficient magnitudes also imply uniform convergence by the Weierstrass M-test. At $x=0$, it gives ${\sum }_{m\ge 0}(2m+1{)}^{-2}={\pi }^2/8$.

For $y^{\prime }+ay=g$ with real $a>0$ and $2\pi$-periodic forcing, comparing Fourier coefficients gives $(a+in){\hat{y}}_n={\hat{g}}_n$. There is no vanishing denominator, including the constant mode. With real cosine and sine coefficients $a_n,b_n$ for $g$ and $A_n,B_n$ for $y$, the equivalent equations giveWhenever termwise differentiation is justified, these formulas construct a periodic particular solution. For a forcing series with coefficients $O(n^{-2})$, the differentiated response coefficients are also summable, which supplies such a justification by uniform convergence. All other solutions add the homogeneous solution $C{e}^{-ax}$. Since this exponential cannot be periodic unless $C=0$, the periodic solution is unique. This is also a simple example of how a linear equation changes the amplitude and phase of each harmonic independently.

On $0\le x\le L$, with $L>0$, the tent $f(x)=\min (x,L-x)$ has Fourier sine series coefficients $b_n=4L\sin (n\pi /2)/({\pi }^2{n}^2)$. Split the coefficient integral at $L/2$ and integrate its two linear pieces by parts; their cosine endpoint terms cancel and their sine terms add. Reflection about the midpoint makes every even-index coefficient zero, and the remaining odd-index coefficients alternate in sign. The $n^{-2}$ bound gives absolute convergence and uniform convergence; the series equals the continuous tent, including its zero endpoints.