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Apply the triangle inequality to the path from $x_1$ through $y_2$ to $y_1$:This is one direction of the reverse triangle inequality. A second application of the triangle inequality, this time through $x_2$, givesThe resulting four-point triangle inequality follows without any assumptions on coincidences among the four points.

The Cauchy-Schwarz inequality in ${\mathbb{R}}^n$ isHere the Euclidean norm is induced by the dot product. If $b=0$, the result is immediate. If $b\ne 0$, the squared Euclidean norm of the component of $a$ perpendicular to $b$ is nonnegative:Rearrangement proves the Cauchy-Schwarz inequality. Equality holds precisely when $a$ and $b$ are linearly dependent, including the cases of a zero vector.

Expanding the squared Euclidean norm and using the Cauchy-Schwarz inequality givesTaking nonnegative square roots gives the triangle inequality. Applying the triangle inequality once more gives

Since $a$ is an indicator function of density of a finite subset $\alpha$, the triangle inequality gives $|\hat{a}(r)|\le {\mathbb{E}}_{x}a(x)=\alpha$ for every frequency. Also, character orthogonality and the Parseval identity on a finite group giveFor completeness, the character orthogonality used here iswhich follows by summing a finite geometric series. Expanding the squared Fourier coefficients on a finite abelian group and using this identity proves the displayed Parseval identity on a finite group directly.

Combining the uniform bound with that identity gives the fourth-moment boundIn particular, the Lp norm on the frequency side here is a sum, not a normalized average. This is the fourth Fourier moment bound for an indicator function.

First, convergence in probability at $t=0$ gives $X_0=0$ almost surely, since every $X_0^n=0$. For any finite set of times, the corresponding vectors converge in probability: the union bound controls the probability that any coordinate differs by more than a fixed tolerance. The same holds for their increment vectors, hence also for their convergence in distribution.

For $0=t_0<t_1<\cdots <t_m$, set $D_j^n=X_{t_j}^n-X_{t_{j-1}}^n$ and $D_j=X_{t_j}-X_{t_{j-1}}$. The characteristic function of a random vector factors for the independent increments of each $X^n$:Pass to the limit using convergence in distribution and bounded continuous functions. The resulting factorization of the characteristic function of a random vector, with the uniqueness theorem for characteristic functions, proves independence of the $D_j$. Similarly $X_{s+t}^n-X_s^n\stackrel{d}{=}{X}_t^n$ passes to the limit, proving stationary increments for $X$.

It remains to prove stochastic continuity. For $\epsilon >0$, the triangle inequality and the union bound imply, for every fixed $n$,The second term vanishes by stochastic continuity of $X^n$. Now let $n\to \infty$ and use the additional near-zero approximation hypothesis. We obtain $X_t\to 0$ in probability as $t\downarrow 0$. The stationary increments transfer this to every time: both $X_{s+h}-X_s$ and $X_s-X_{s-h}$ have the law of $X_h$ for $h>0$ when defined. Thus $X$ has all the intrinsic Lévy process properties, proving closure of Lévy processes under locally controlled convergence in probability.

If one requires the supplied process itself to be càdlàg, the hypotheses justify a càdlàg modification, rather than that stronger pathwise assertion. To see the distinction, take $X^n\equiv 0$, let $U$ have uniform distribution on $(0,1)$, and put $X_t=1_{\{t=U\}}$. At each fixed time $t$, $X_t=0$ almost surely, so both approximation hypotheses hold with zero error probability. Nevertheless every path has an isolated spike at $U$ and is not right-continuous there. Its identically zero modification of a stochastic process is a Lévy process with càdlàg paths. The conclusion is exact under the intrinsic definition, and exact up to modification under the convention requiring càdlàg paths.

First justify the dependence structure rather than assume that adjacent squares behave like an ordinary lag-one ARCH model. Write $Z_t=X_t^2$. Iterating its nonnegative recurrence givesThe product coefficient has expected value ${\alpha }_2^m$, so it tends to zero in probability. The stationary $Z_{t-2m}$ are a tight family; consequently the remainder tends to zero in probability, without needing independence between that remainder's factors. The increasing partial sums therefore give the representationThis proves the parity decomposition of a lag-two ARCH process: even and odd observations are functions of disjoint noise families. It also shows that the positive scale ${\sigma }_t$ is a function of past noise, independent of ${\epsilon }_t$. The conditional expectation of $X_t$ given past noise is zero, and the second-moment recursion is $m_2={\alpha }_0+{\alpha }_2{m}_2$. independence of the parity families gives the adjacent-square covariance. ThusTo establish finiteness of the fourth moment before using its recursion, note that the $L^2$ norm of the $k$th term in the positive series for $Z_t$ is ${\alpha }_0\sqrt{3}({\alpha }_2\sqrt{3}{)}^k$. The triangle inequality in L2 space makes the series square-integrable when $3{\alpha }_2^2<1$. independence of the current noise and past scale then yieldsand henceIf $3{\alpha }_2^2\ge 1$, a finite $m_4$ would make $(1-3{\alpha }_2^2)m_4=3{\alpha }_0^2+6{\alpha }_0{\alpha }_2{m}_2>0$, which is impossible. Its fourth moment is then infinite. The adjacent-square product remains integrable by parity independence, even in that case.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Fix the degree bound $n$. Existence can be proved without a general uniqueness theorem: a minimizing sequence in ${\mathcal{P}}_n$ is uniformly bounded because its distances from $f$ are bounded. Values at any fixed $n+1$ distinct points determine its coefficients through an invertible Vandermonde matrix, so those coefficients are bounded. A convergent subsequence supplies a polynomial attaining the minimum.

For uniqueness of best uniform polynomial approximation, suppose $p$ and $q$ attain the same minimum $d$, and let $r=(p+q)/2$. The triangle inequality makes $r$ another minimizer. If $d=0$, both polynomials equal $f$, so assume $d>0$. Put $e=f-r$ and $E=\{|e|=d\}$. At each $x\in E$, the two real errors $f-p$ and $f-q$ lie in $[-d,d]$ and their average equals an endpoint. Hence they are equal there, and $p(x)=q(x)$.

The set $E$ must contain at least $n+2$ points. Otherwise interpolate the values $\mathrm{sign}e(x)$ on its at most $n+1$ points by a polynomial $v\in {\mathcal{P}}_n$. Then $ev>0$ on $E$. Continuity gives this positivity on a neighbourhood of $E$, while the error has a strict gap below $d$ on the compact complement. A sufficiently small positive multiple of $v$ decreases the maximum error, contradicting optimality of $r$. This is an elementary perturbation argument, not an appeal to Haar's theorem.

Thus $p-q$ has at least $n+2$ distinct zeros. A polynomial of degree at most $n$ with that many zeros is identically zero. The best uniform polynomial is unique for every fixed degree bound: $\boxed{p=q}$.

For a block code, the minimum distance of a code is the minimum Hamming distance between distinct codewords, where Hamming distance counts positions in which two equal-length binary strings differ. We use equal-length codewords, so the fact that the codomain allows arbitrary finite strings creates no ambiguity. Minimum distance $\delta$ means every two distinct codewords differ in at least $\delta$ positions and some pair differs in exactly $\delta$.

Changing at most $\delta -1$ bits cannot turn a codeword into another codeword. Testing membership therefore detects every such nonzero error. For $t=\lfloor (\delta -1)/2\rfloor$, suppose a received word lies within distance $t$ of two codewords. The triangle inequality would put their mutual distance at most $2t<\delta$, a contradiction. Hence the radius-$t$ Hamming balls are disjoint and nearest-codeword decoding corrects all errors of weight at most $t$.

For an alphabet of size $m\ge 2$, start with $0$ and the $m-1$ standard unit vectors in $\{0,1{\}}^{m-1}$. Their minimum distance of a code is one. Repeat every coordinate $\delta$ times; every distance is multiplied by $\delta$, so the resulting code has minimum distance of a code exactly $\delta$. A one-letter alphabet has no pair of distinct codewords and therefore no finite minimum distance of a code under this definition; the construction concerns the usual nontrivial coding situation.

A linear map $T:X\to Y$ between normed spaces is a bounded linear operator if $\Vert Tx\Vert \le C\Vert x\Vert$ for one finite $C$ and every $x$. This implies $\Vert Tx-Ty\Vert \le C\Vert x-y\Vert$, hence continuity. Conversely, continuity at zero supplies $\delta >0$ such that $\Vert x\Vert <\delta$ implies $\Vert Tx\Vert <1$. Apply this to $\delta x/(2\Vert x\Vert )$ to obtain $\Vert Tx\Vert \le 2\Vert x\Vert /\delta$ for nonzero $x$. Thus boundedness and continuity are equivalent for linear maps.

The operator norm is $\Vert T\Vert ={\sup }_{\Vert x\Vert \le 1}\Vert Tx\Vert$, equivalently ${\sup }_{x\ne 0}\Vert Tx\Vert /\Vert x\Vert$. The bounded operators form a vector space, since $\Vert (S+T)x\Vert \le (\Vert S\Vert +\Vert T\Vert )\Vert x\Vert$ and scalar multiples remain bounded. Taking suprema proves the triangle inequality and absolute homogeneity for the operator norm; $\Vert T\Vert =0$ forces $Tx=0$ for every $x$. Thus $B(X,Y)$ is a normed space, whether or not $X,Y$ are complete.

For nonempty compact sets, define the distance from a point to a closed set by $d(x,B)={\min }_{b\in B}|x-b|$ and the Hausdorff distance byThe minima and suprema exist: the sets are nonempty and compact, and the distance function is Lipschitz continuous. The Hausdorff distance is finite, nonnegative and symmetric. If it is zero, every point of $A$ belongs to the closed set $B$ and conversely, so $A=B$; the converse is immediate.

For $a\in A$, choose $b\in B$ attaining $d(a,B)$ and then $c\in C$ attaining $d(b,C)$. The triangle inequality gives $d(a,C)\le d(a,B)+d_H(B,C)$. Taking the supremum over $a$, and doing the same with $A,C$ exchanged, proves $d_H(A,C)\le d_H(A,B)+d_H(B,C)$. Hence this is a metric.

The nested compact sets have the finite intersection property inside $K_1$, so $K=\bigcap K_n$ is nonempty. It is a closed subset of $K_1$ and therefore compact. Because $K\subseteq K_n$, only ${\sup }_{x\in K_n}d(x,K)$ contributes to the Hausdorff distance. If this did not tend to zero, some increasing indices $n_j$ and $x_j\in K_{n_j}$ would satisfy $d(x_j,K)\ge ϵ>0$. A convergent subsequence in $K_1$ has limit $x$. For each fixed $m$, its tail belongs to the closed set $K_m$, so $x\in K$. But then $d(x_j,K)\le |x_j-x|\to 0$, a contradiction. Thus $\boxed{d_H(K_n,K)\to 0}$.

The two preliminary sequence limit laws follow directly from the triangle inequality. For the sum, choose indices so that $|x_n-x|<\epsilon /2$ and $|y_n-y|<\epsilon /2$ beyond their respective cutoffs. Beyond the larger cutoff,Thus

For the reciprocal, convergence to nonzero $x$ gives $|x_n-x|<|x|/2$ eventually. The reverse triangle inequality then gives $|x_n|\ge |x|/2$, and consequentlyChoosing the eventual error smaller than both $|x|/2$ and $\epsilon |x{|}^2/2$ provesThe nonzero limit gives the denominator a positive lower bound; nonzero individual terms alone would not do so. The assumption $x_n\ne 0$ ensures the reciprocal sequence is defined even before this eventual bound applies.

The stated bound is Lipschitz continuity with constant $1$. For every $x$ and every $\epsilon >0$, taking $\delta =\epsilon$ ensures $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and uniformly continuous.

For an integer $N\ge 1$, partition $[0,1]$ into $N$ equal intervals and define a step functionAt the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The Lipschitz bound givesChoosing $N\ge 1/\epsilon$ proves the requested uniform step approximation on a compact interval:A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.

The continuous function $f$ and every finite step function $g_N$ are Riemann integrable. Put $c_j=f(j/N)$. Integrating each constant piece explicitly givesSince the sine differences have absolute value at most $2$, the triangle inequality and uniform approximation giveTo prove convergence, fix $\epsilon >0$ and first choose $N$ with $1/N<\epsilon /2$. With this $N$ held fixed, the finite number ${\sum }_j|c_j|$ is independent of $n$, so the second term is also less than $\epsilon /2$ for sufficiently large $n$. ThereforeFirst make the approximation error small, then let the oscillation frequency grow. This is the step-function proof of the Riemann-Lebesgue lemma; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.

A sequence $(x_n)$ is a convergent sequence with limit $x$ ifThe index $N$ may depend on $\epsilon$, but must work for every later term.

Fix $\epsilon >0$. Choose $N$ such that $|x_i-x|<\epsilon /2$ for $i\ge N$, and putThis is a fixed finite number. For $n\ge N$, the triangle inequality givesChoose $n$ sufficiently large that $C/n<\epsilon /2$ as well. The error is then less than $\epsilon$, provingTaking successive arithmetic averages preserves the limit. This is the Cesaro theorem for convergent sequences: any troublesome initial terms contribute only a fixed numerator divided by $n$.

A Cauchy sequence in a metric space satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. Suppose a subsequence $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\epsilon /2$, and then choose $k$ with $n_k\ge N$ and $d(x_{n_k},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole Cauchy sequence converges to the subsequential limit.

Apply the triangle inequality with a third point $x_k$:Letting $k\to \infty$ gives the desired bounds. Interchanging $m,n$ also gives the sharper symmetric lower bound:

A norm on a real or complex vector space $V$ is a function $\Vert \cdot \Vert :V\to [0,\infty )$ with $\Vert v\Vert =0$ exactly when $v=0$, $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$, and the triangle inequality $\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$. These are positivity, absolute homogeneity and the triangle inequality.