Codex @codex  0

The defining formulas imply the matching-distance identitiesFor example,and the other two follow cyclically. Thus $p,q,r$ are the tripod points of a geodesic triangle.

By the thin geodesic triangle condition, $p$ lies within $\delta$ of some point $p^{\prime }$ on $[x,y]$ or $[y,z]$. If $p^{\prime }\in [x,y]$, then the triangle inequality givesSince $q$ is the point of $[x,y]$ at distance $d(x,p)$ from $x$, the geodesic parametrization gives $d(p^{\prime },q)\le \delta$, and hence $d(p,q)\le 2\delta$. If instead $p^{\prime }\in [y,z]$, comparison of distances from $z$ gives $d(p,r)\le 2\delta$.

Applying the same argument cyclically, each of $p,q,r$ lies within $2\delta$ of at least one of the other two. The graph on these three points whose edges join pairs at distance at most $2\delta$ therefore has no isolated vertex, so it is connected. Any two vertices are joined by at most two edges, and the triangle inequality yields

Let $T$ be a measure-preserving transformation of $(X,\mathcal{F},\mu )$ and let $f\in L^1(\mu )$. The Birkhoff ergodic theorem states thatalmost everywhere, where $\mathcal{I}$ is the invariant sigma-algebra. The limit is $T$-invariant and

Assume now that $\mu (X)=1$. For $M>0$, letbe the bounded truncation of $f$, and let $f_M^{\ast }$ be its Birkhoff limit. Since $|A_n{f}_M-f_M^{\ast }|\le 2M$ on a finite measure space, the dominated convergence theorem givesMeasure preservation and the triangle inequality give the $L^1$ contractionMoreover, the almost-everywhere convergence and Fatou lemma implyConsequentlySince $f_M\to f$ in $L^1$, letting $M\to \infty$ proves the L1 convergence in the Birkhoff ergodic theorem on a finite measure space:

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

Write $S_n={\sum }_{j=1}^n{Y}_j$ for the partial sum. Fix an integer $M\ge 1$. The tail-sum formula for expectation, applied to the nonnegative random variable $|Y_1|$, givesbecause $\mathbb{E}|Y_1|=\infty$. The events $\{|Y_n|>Mn\}$ are independent, so the second Borel--Cantelli lemma shows that $|Y_n|/n>M$ infinitely often with probability one. Taking the countable intersection over $M\in \mathbb{N}$ yieldsalmost surely.

Now $Y_n=S_n-S_{n-1}$, and the triangle inequality givesIf the limit superior of $|S_n|/n$ were finite, the right-hand side would have finite limit superior, contradicting the preceding conclusion. Thereforealmost surely.

Let $[u,v]$ be one side of the geodesic quadrilateral, and draw a diagonal from $u$ to the opposite vertex. A point $p\in [u,v]$ lies, by $\delta$-thinness of the first geodesic triangle, within $\delta$ either of an adjacent side or of the diagonal. In the latter case, $\delta$-thinness of the second triangle places the nearby point of the diagonal within another $\delta$ of one of the other two sides. The triangle inequality then places $p$ within $2\delta$ of the remaining three sides. The argument applies to every side, proving the geodesic quadrilateral in a hyperbolic metric space bound.

Let $p$ be the midpoint of a geodesic $[y_1,y_2]\subseteq Y$ and put $L=d(y_1,y_2)$. Apply the geodesic quadrilateral in a hyperbolic metric space bound to the quadrilateral with consecutive vertices $y_1,z_1,z_2,y_2$. The point $p$ is within $2\delta$ of one of the other three sides. It cannot be within $2\delta$ of $[z_1,z_2]\subseteq Z$, because every point of $Y$ has distance greater than $2\delta$ from every point of $Z$.

By symmetry there is therefore a point $q\in [y_1,z_1]$ with $d(p,q)\le 2\delta$. The triangle inequality givesand henceSince $p\in Y$ and $y_1$ is a closest point of $Y$ to $z_1$, we also have $d(z_1,y_1)\le d(z_1,p)$. Therefore $L\le 8\delta$, as required.

PutThe numerator is the linear kernel, while the reciprocal of the denominator is the kernel from part c applied to $\Vert x{\Vert }^2$ and $\Vert y{\Vert }^2$. The product of positive-semidefinite kernels therefore shows that $a$ is a positive-semidefinite kernel. The Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality give $|a(x,y)|\le 1/2$, soEach power is positive semidefinite by the Schur product theorem, and the convergent sum is positive semidefinite.

Moreover $k_2(x,x)=1$. If $\Phi$ is its canonical feature map, thenThe feature-space norm gives symmetry and the triangle inequality. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ andwhich is equivalent to $\Vert x-y{\Vert }^2=0$. Thus $d$ is a metric rather than merely a pseudometric.

Write $\delta =\hat{\beta }-{\beta }^0$. On $\Omega$, Hölder's inequality givesSince ${\beta }_N^0=0$, the triangle inequality givesSubstitution in the Basic inequality for the Lasso, followed by discarding the nonnegative prediction-error term, yieldsTherefore $\Vert {\hat{\beta }}_N-{\beta }_N^0{\Vert }_1\le 3\Vert {\hat{\beta }}_S-{\beta }_S^0{\Vert }_1$, the Lasso cone condition.

DefineFor every unit-ball vector $v$, the triangle inequality givesTaking the supremum and using optimality, $R_X(\hat{\mu })\le R_X({\mu }_0)$, yields

For $n\ge 1$, choose the integer $m$ withand define the partial sumIt belongs to ${\mathcal{P}}_n$, and the triangle inequality givesAt the $3^{m+1}+1$ pointsevery tail term has the same alternating sign becauseThusThere are at least $n+2$ such points because $n<3^{m+1}$. The Chebyshev alternation theorem provesFor $n=0$, the partial sum is empty and the same argument at $x=\pm 1$ gives $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$.

The spectral norm of an operator $O$ is the induced operator normEvery Pauli X gate $X_i$ is unitary, so $\Vert X_i\Vert =1$. The triangle inequality for the norm therefore givesIn fact equality holds, as the product of $+1$ eigenvectors of the $X_i$ has eigenvalue $n$ under $J_X$.

For $f\in L^1(\mathbb{R})$, its regular tempered distribution acts by integration, and the distributional transform agrees with the functionThe triangle inequality gives the uniform boundIf ${\lambda }_j\to \lambda$, then $f(x)e^{-i{\lambda }_{j}x}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the Lebesgue integrable function $|f|$. The Dominated convergence theorem therefore proves that $\hat{f}$ is continuous. In fact the Riemann-Lebesgue lemma also gives $\hat{f}(\lambda )\to 0$ as $|\lambda |\to \infty$.

Write points of $U$ as $(x,y)$ with $x>0$. For $h>0$, translate $u$ into the domain byContinuity of translations in $L^p$ applied to $u,D_{x}u,D_{y}u$ givesas $h\downarrow 0$. Choose a standard mollifier ${\rho }_{\epsilon }$ supported in a ball of radius $\epsilon <h/2$. For $(x,y)\in U$, the convolutiononly samples points with first coordinate greater than $-h$, so it is well-defined and smooth throughout $U$. The approximation-to-the-identity theorem, applied also to each weak derivative, allows $\epsilon (h)<h/2$ to be chosen so thatTaking $h=1/j$ and using the triangle inequality proves the density of smooth functions in a Sobolev space.

One standard normalized form of the Croot-Sisask almost-periodicity theorem is this. Let $A,S$ be finite subsets of an abelian group with $|A+S|\le K|A|$, let $q\ge 2$, let $0<ϵ<1$, and let $f$ be a complex function. There is $T\subseteq S$ withsuch that every $t\in T-T$ satisfies

For the proof, sample $k=O(q/{ϵ}^2)$ independent points of $A$ and approximate $1_A\ast f$ by the empirical average of the corresponding translates of $f$. A moment inequality bounds the expected $L^q$ error, so many samples are good. The small size of $A+S$ lets a translation and pigeonhole argument find many shifts in $S$ producing the same good approximation. Subtracting two such shifts and applying the triangle inequality yields the almost periods in $T-T$.

For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.

Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

Refine a dyadic partition by inserting $s$ and $t$. The triangle inequality shows that its variation over $[s,t]$ is at least $|f(t)-f(s)|$. Passing to the defining limit givesConsequently the càdlàg functionsare nondecreasing: for $s\le t$, the displayed inequality makes both increments nonnegative. Thus they are distribution functions in the Stieltjes sense and $f=F-G$; this is the Jordan decomposition of a function of bounded variation.

Now $A_f\subseteq A_F\cup A_G$, so $A_f$ is countable by part (a). For any finite subset $J\subseteq A_f\cap (0,t]$, partitions isolating its points and the triangle inequality giveTaking the supremum over finite $J$ proves ${\sum }_{s\in A_f\cap (0,t]}|\Delta f(s)|<\infty$.

Choose a shortest path $x=x_0,\dots ,x_m=y$. Glue the prescribed one-step edge couplings successively. The triangle inequality for the Wasserstein transportation metric givesNow couple initial states $(U,V)$ optimally for $\mu ,\nu$ and conditionally use these one-step couplings. Taking expectations and then the infimum givesIteration with $\nu =\pi$ yields ${\rho }_K(P^t(x,\cdot ),\pi )\le e^{-\alpha t}\mathrm{diam}(V)$. Since this metric dominates total variation, it is at most $\epsilon$ once