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Choose a basis of the finite-dimensional vector space and let $|x{|}_{\infty }$ be the maximum absolute coordinate. For any norm $N$, the triangle inequality givesso $N$ is continuous in the coordinate topology. On the compact set $\{x:|x{|}_{\infty }=1\}$, the positive continuous function $N$ attains a positive minimum $m$ and a finite maximum $M$. By absolute homogeneity,This proves the equivalence of norms in finite dimensions: every norm is equivalent to the maximum norm, and transitivity of these inequalities proves that any two norms on a finite-dimensional vector space are Lipschitz equivalent.

Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

A sequence of functions $f_n:A\to \mathbb{R}$ converges uniformly to $f$ when, for every $\epsilon >0$, there is an $N$ such thatfor every $x\in A$. Equivalently, $\Vert f_n-f{\Vert }_{\infty }\to 0$ in the supremum norm.

Yes. Fix $x_0\in A$ and $\epsilon >0$. Uniform convergence gives $N$ such that $|f_N(x)-f(x)|<\epsilon /3$ for every $x\in A$. Since the continuous function $f_N$ is continuous at $x_0$, there is a $\delta >0$ such that $|x-x_0|<\delta$ implies $|f_N(x)-f_N(x_0)|<\epsilon /3$. The triangle inequality then givesThis proves the uniform limit theorem: $f$ is continuous.

Write $Lu=L_{0}u+(a^{ij}-A^{ij})D_{ij}u$, with repeated indices summed. The triangle inequality and the Cauchy-Schwarz inequality over the $n^2$ coefficient pairs giveChooseThen the strict coefficient bound in the question implies

The printed sum contains $Z_n$ although the stated labels end at $n-1$. We use the natural periodic convention $Z_n=Z_0$, with $n\ge 3$. If an open chain was intended, omitting the final term gives the same asymptotic bound.

Set $H_j=X_{j-1}{Z}_j$, with indices modulo $n$, and implement the product-formula Hamiltonian simulationEach factor acts on two qubits, so it is a two-qubit unitary operator. More explicitly, if $a=j-1$, $b=j$, and $C_{a\to b}$ is a controlled-NOT gate,where ${\mathsf{H}}_a$ is a Hadamard gate. This gives a constant number of one-qubit and two-qubit gates per factor; one-qubit gates can also be viewed as two-qubit gates tensored with the identity.

The spectral norm obeys the triangle inequality, submultiplicativity of the operator norm, and invariance under multiplication by unitary operators. In particular, the telescoping bound for products of operators gives $\Vert V^k-W^k\Vert \le k\Vert V-W\Vert$ for unitary $V,W$. The first-order unitary product-formula error bound consequently yieldsOnly neighboring terms can have a nonzero commutator, because all other supports are disjoint. There are $n$ neighboring unordered pairs, and $\Vert [H_j,H_l]\Vert \le 2\Vert H_j\Vert \Vert H_l\Vert =2$. Thus $\Vert U-\tilde{U}\Vert \le n/k$. Choosing $k>n/\epsilon$ givesThe coarser bound that counts all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs also proves the often-used $O(n^3/\epsilon )$ construction. Locality improves that cubic estimate to the quadratic bound above; neither assertion is a lower bound on the best possible circuit.

Let $u^{†}=A^{†}f$. Linearity and the triangle inequality give the noise-bias decomposition for linear regularizationThe first term tends to zero by the assumed noise-amplification bound. The second tends to zero because $\alpha (\delta )\to 0$ and $R_{\alpha }$ is a regularization of an inverse problem. The right side is independent of the particular noisy datum within its allowed ball. Its convergence therefore proves the uniform noisy-data convergence required of a convergent regularization of an inverse problem.

The assumptions prove convergence to the set of J-minimizing solutions, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.

Let $u^{†}$ be any J-minimizing solution, whose existence is granted. It is feasible for every noisy-data constraint because $A{u}^{†}=f$ and $\Vert f-f_{\delta }\Vert \le \delta$. Minimality therefore gives $J(u_{\delta })\le J(u^{†})$. All reconstructions lie in this strongly sequentially compact space sublevel set. For any sequence ${\delta }_k\downarrow 0$, pass to a subsequence with $u_{{\delta }_k}\to u_{\ast }$. Boundedness of $A$ and the triangle inequality implyso $A{u}_{\ast }=f$. By sequential lower semicontinuity,Thus $u_{\ast }$ is a J-minimizing solution. The same compactness argument shows the distance to a set of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full strong convergence.

For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $A=0$, $f=f_{\delta }=0$, and $J(u)=\max \{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\ge 0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1{)}^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is compact-sublevel convergence of the residual method.

Write $L=W_p(\mu ,\nu )$. The assumed equality of the Monge optimal transport problem and Kantorovich optimal transport problem values givesEach interpolated pushforward measure ${\mu }_t=(P_t{)}_{\#}\mu$ has a finite $p$th absolute moment, since $|P_t(x){|}^p\le 2^{p-1}(|x{|}^p+|T^{†}(x){|}^p)$ and $(T^{†}{)}_{\#}\mu =\nu$.

For any $s,t\in [0,1]$, the common-source transport planprovides the upper boundFor the reverse bound assume $0\le s\le t\le 1$. Since ${\mu }_0=\mu$ and ${\mu }_1=\nu$, the triangle inequality for the p-Wasserstein distance and the upper bounds already established giveHence $W_p({\mu }_s,{\mu }_t)\ge (t-s)L$. Combining the bounds and using symmetry provesThis is the constant speed property of displacement interpolation. The given invertibility of $P_s$ also lets one realize the competitor as the map $P_t\circ P_s^{-1}$, assuming its inverse is measurable, but the transport plan argument proves the result without invertibility.

Let $X$ have basis $e_1,\dots ,e_d$, and write $x={\sum }_j{a}_j{e}_j$. On the compact coordinate sphere ${\sum }_j|a_j|=1$, the continuous positive functionhas a positive minimum $m$. Homogeneity and the triangle inequality therefore give constants $m,M>0$ such thatIn particular, each coordinate functional ${\varphi }_j(x)=a_j$ belongs to the continuous dual space $X^{\ast }$.

Every member of $X^{\ast }$ is norm-continuous, so the weak topology is coarser than the norm topology. Conversely, given $x_0\in X$ and $\epsilon >0$, the basic weak neighbourhoodsatisfies $U\subseteq \{x:\Vert x-x_0\Vert <\epsilon \}$. Thus every norm-open set is weakly open, proving that the two topologies coincide. This is the finite-dimensional weak and norm topologies coincide theorem.

Assume $X_n\to X$ in $L^2$. The reverse triangle inequality for the $L^2$ norm givesSquaring the norms yields

Write $h(t)=t/(1+t)$ for $t\ge 0$. Every summand in $d$ is nonnegative and at most $2^{-m}$, so the series converges. Symmetry is immediate. If $d(x,y)=0$, then ${\varphi }_m(x-y)=0$ for every $m$. For arbitrary $\varphi \in ({\ell }^p{)}^{\ast }$, choose a sequence of ${\varphi }_m$ converging to $\varphi$ in operator norm. Since $x-y$ is fixed, $\varphi (x-y)=0$. The continuous dual space separates points, so $x=y$.

The function $h$ is increasing and subadditive:Combining this with the triangle inequality for each $|{\varphi }_m(x-y)|$ and summing proves the triangle inequality for $d$. Hence $d$ is a metric on $B$.

Suppose first that $x^{(n)}$ converges weakly to $x$, meaning thatFor each fixed $m$, the $m$th summand of $d(x^{(n)},x)$ tends to zero. A finite-head and geometric-tail estimate, using that every summand is bounded by $2^{-m}$, then gives $d(x^{(n)},x)\to 0$.

Conversely, if $d(x^{(n)},x)\to 0$, each nonnegative summand tends to zero, and thereforefor every $m$. Given $\varphi \in ({\ell }^p{)}^{\ast }$ and $\epsilon >0$, choose $m$ with $\Vert \varphi -{\varphi }_m\Vert <\epsilon$. Since $x^{(n)},x\in B$,Taking the limit superior and then letting $\epsilon \downarrow 0$ proves weak convergence. ThusThe same approximation argument for neighbourhoods shows that $d$ induces the weak topology on $B$, as in the metrization of the weak topology on a bounded set.

If $q$ is conjugate to $p$, the given duality of sequence spaces applied twice givesso ${\ell }^p$ is a reflexive Banach space. Its closed unit ball is weakly compact by the weak compactness characterization of reflexivity. Since $d$ induces that topology on $B$,

Put $N=p^n$, $\alpha =|A|/N$, and use normalized Fourier analysis on a finite abelian group. If $F=1_A\ast 1_A$ denotes unnormalized convolution and $h=N^{-1}F$, thenby the Parseval identity.

Sample $k=O(m{\epsilon }^{-2})$ characters independently, choosing $\gamma$ with probability $|\hat{h}(\gamma )|/\alpha$, and attach the phase of $\hat{h}(\gamma )$ to the sampled character. The Marcinkiewicz–Zygmund inequality, followed by averaging over $x$, shows that some sampled Fourier sum $g$ satisfiesThis is the sampling argument recorded in the finite-field character approximation principle.

Let $V$ be the intersection of the kernels of the sampled characters. It is a vector subspace of codimension at most $k$, and ${\tau }_{t}g=g$ for every $t\in V$. The triangle inequality therefore givesReturning from normalized convolution and normalized norm to $F$ and the counting norm multiplies the right side by $N^{1+1/(2m)}$. Hencefor every $t\in V$, proving the finite-field convolution almost-periodicity theorem.

A coarse embedding $f:X\to Y$ has nondecreasing control functions ${\rho }_{-},{\rho }_{+}$ such thatA family is a uniform coarse embedding of a family of metric spaces when all its maps use the same controls.

For $H_n=\{0,1{\}}^n$ with the Hamming distance, the coordinate map into ${\ell }_1^n$ satisfiesso the cubes embed isometrically and uniformly into $L^1$. The same coordinate map into ${\ell }_2^n$ satisfiesThus they also uniformly coarsely embed into $L^2$, with ${\rho }_{-}(t)={\rho }_{+}(t)=\sqrt{t}$. This is the Hamming cube as an L1 and L2 metric.

For a finite graph $G=(V,E)$, defineAn expander family is a sequence of graphs with orders tending to infinity, uniformly bounded degrees, and $h(G)$ uniformly bounded below.

Now let $|V|=n$ and let $M$ be a median of $f:V\to \mathbb{R}$. PutBoth supports have at most $n/2$ vertices. The layer-cake representation and the definition of $h$ giveand the same holds for $k$. Since their supports are disjoint,Adding givesThe triangle inequality also givesand henceRepresent an $L^1$-valued map as scalar functions on the underlying measure space and integrate this inequality. The result is the L1 Poincare inequality for an expander graph.

Suppose next that $G$ is $d$-regular and $F:V\to L^1$ is noncontracting with Lipschitz constant $D$. The Poincare inequality givesA ball of radius $r$ in a graph of maximum degree $d$ has at mostvertices. Therefore, forat least $n/2$ vertices lie outside the radius-$r$ ball about each vertex. Integrating the tail count of the distance givesConsequently every such $F$ haswhich proves the L1 distortion lower bound for an expander graph.

Finally, suppose an expander family admitted uniform coarse embeddings $F_n$ into $L^1$. Embedded edges have length at most ${\rho }_{+}(1)$, whereas the preceding Poincare argument and the fact that at least half the ordered pairs have distance comparable to $\log |V_n|$ implyThis contradicts ${\rho }_{-}(t)\to \infty$. Thus expanders do not uniformly coarsely embed into $L^1$. They do not uniformly coarsely embed into $L^2$ either: a Hilbert space embeds isometrically into an $L^1$ space byso an $L^2$ coarse embedding would produce an $L^1$ one. This is the expander graph obstruction to uniform coarse embedding.

Put $q_i=1-{\beta }_i$. Repeating part (v) with radiation from both bodies givesThe two triangular points are intersections of circles with these radii and centre separation one. Iftheir coordinates areFor ordinary outward radiation pressure, $0\le {\beta }_i\le 1$, the non-collinear points exist precisely when the strict triangle inequalityholds. Equality merges the two points on the line of centres.

It is sufficient thatand that the stopping rule obeyFor example, $k(\delta )=\lfloor {\delta }^{-1}\rfloor$ works.

For exact data $f\in \mathcal{D}(A^{†})$, each factor $(1-\tau {\sigma }_i^2{)}^k$ tends to zero. The Picard criterion makes $⟨f,y_i⟩/{\sigma }_i$ square summable, while $|1-(1-\tau {\sigma }_i^2{)}^k|$ is uniformly bounded. The dominated convergence theorem on the resulting series yields $u_k(f)\to A^{†}f$.

For noisy data with $\Vert f^{\delta }-f\Vert \le \delta$, the filter representation givesIndeed, when $\tau {\sigma }^2\le 1$, Bernoulli's inequality gives $1-(1-\tau {\sigma }^2{)}^k\le \min (k\tau {\sigma }^2,1)$; when $\tau {\sigma }^2>1$, the quotient is uniformly bounded because $\tau \Vert A{\Vert }^2<2$. Hence the triangle inequality giveswhich proves that early-stopped Landweber iteration is a convergent regularization of an inverse problem.

Suppose first that $\Vert x_n-x_{\infty }\Vert \to 0$. The reverse triangle inequality givesso (i) implies (ii).

It also implies (iii). Given $ϵ>0$, choose $N$ so thatBy the Parseval identity for a Hilbertian basis, choose $I_0$ so that the basis tail of $x_{\infty }$ beyond $I_0$ is less than $ϵ/8$. For $n\ge N$ and $I\ge I_0$,There are only finitely many $n<N$, so increasing $I$ makes each of their tails less than $ϵ$. Thus one $I$ works for every $n$, proving (iii).

Fix $x$ and $\epsilon >0$. Since $\mathcal{K}$ is compact in the uniform norm, choose a finite $\epsilon /3$-net $f_1,\dots ,f_r$. By continuity, there is a neighbourhood $U$ of $x$ on which $|f_i(y)-f_i(x)|<\epsilon /3$ for every $i$. For $f$ in $\mathcal{K}$, choose $f_i$ with $\Vert f-f_i{\Vert }_{\infty }<\epsilon /3$. The triangle inequality gives $|f(y)-f(x)|<\epsilon$ on $U$, uniformly in $f$. Thus $\mathcal{K}$ is equicontinuous.

Equality in the triangle inequality holds exactly when the two displacement vectors $a-z$ and $z-b$ point in the same direction. Thus $z$ lies on the closed line segment joining $a$ and $b$. Conversely, every point of that segment satisfiesThe solution set is therefore precisely that segment in the complex plane.

The triangle inequality givesHence the left-hand side of the proposed equation is always at least $|a-b|$. There are no solutions when $c<|a-b|$.

SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.