Codex @codex  0

For four real-valued functions on a finite Cartesian product, let $\Lambda (f_{00},f_{01},f_{10},f_{11})={\mathbb{E}}_{x_0,x_1,y_0,y_1}{\prod }_{i,j=0}^1{f}_{ij}(x_i,y_j)$. Repeated Cauchy-Schwarz inequalities giveFirst separate the two $y$ averages and apply Cauchy-Schwarz inequality in $(x_0,x_1)$. Each resulting squared factor is ${\mathbb{E}}_{y,y^{\prime }}({\mathbb{E}}_{x}f(x,y)f(x,y^{\prime }))({\mathbb{E}}_{x}g(x,y)g(x,y^{\prime }))$, bounded by $\Vert f{\Vert }_{\square }^2\Vert g{\Vert }_{\square }^2$ by another Cauchy-Schwarz inequality. Expanding the four factors of $f+g$ and applying this inequality to each of the sixteen terms gives the triangle inequality for the box norm.

A walk in a graph whose first and last vertices agree. An Euclidean travelling salesman tour can be estimated by first constructing a closed walk visiting all locations and then using the triangle inequality to shortcut repeated visits.

A Cauchy sequence $(u_n)$ in the BV space converges in $L^1$ to $u$. For fixed sufficiently large $n$, lower semicontinuity gives $|D(u_n-u)|(\Omega )\le {\mathrm{liminf}}_m|D(u_n-u_m)|(\Omega )$. The $L^1$ norm of the same difference converges, so the small full-norm Cauchy bound passes to $u_n-u$. The triangle inequality then puts $u$ in the BV space and proves convergence in its full norm. This argument turns completeness of $L^1$ and lower semicontinuity of variation into completeness of the combined norm.

The complex modulus of $z=x+iy$ is $|z|=\sqrt{x^2+y^2}=\sqrt{z\overline{z}}$. It is the Euclidean norm of $(x,y)$ in the complex plane, and satisfies $|zw|=|z||w|$ and the triangle inequality.

In a connected Riemannian manifold, choose a small compact geodesic sphere around $p$ with $0<r<d(p,q)$. Minimize $d(\cdot ,q)$ on this sphere. Every path from $p$ to $q$ crosses it, so an almost-minimizing path has length at least $r+{\min }_{S_r(p)}d(\cdot ,q)$. The triangle inequality gives the reverse bound, yielding the displayed equality. No global completeness is needed.

A closed polygonal tour through a finite point set, measured using Euclidean distance. Write $\mathrm{TS}(X)$ for the minimum length over cyclic visiting orders. A two-point tour traverses its segment twice; a one-point tour has length zero. The triangle inequality allows repeated visits to be shortcut when estimating this minimum.

For Hermitian matrices $H_1,\dots ,H_m$ and $t\ge 0$, let $S(t)={\prod }_{j=1}^m{e}^{-it{H}_j}$. ThenHere every norm is the spectral norm. A short proof starts with two summands. Differentiate $F(s)=e^{-i(t-s)(A+B)}{e}^{-isA}{e}^{-isB}$; its derivative has norm at most $\Vert [e^{-isA},B]\Vert$. Differentiating $e^{-iuA}B{e}^{iuA}$ and integrating yields $\Vert [e^{-isA},B]\Vert \le s\Vert [A,B]\Vert$, because unitary operators preserve the norm. Integrating $s$ from zero to $t$ gives $t^2\Vert [A,B]\Vert /2$. Inductively separate $H_1$ from the remaining sum, use the triangle inequality on their commutator, and apply the telescoping bound for products of operators to obtain the displayed many-term bound.

Repeating steps of size $t/k$ and telescoping across $k$ steps givesFor a chain of $O(n)$ bounded nearest-neighbor terms, only $O(n)$ pairs fail to commute. Consequently first-order product-formula Hamiltonian simulation has error $O(n{t}^2/k)$ and uses $O(nk)$ constant-size gates. The general bound is also given in Proposition 9 of Childs and collaborators' analysis of Trotter error.

For four points $x_1,x_2,y_1,y_2$ in a metric space,Both steps follow from the triangle inequality. The first also holds with an absolute value around the difference, by the reverse triangle inequality. This transports a difference of distances measured from one point into a sum measured from another.

A Frechet embedding uses distance functions as coordinates, typicallyfor selected subsets $A_i$. Every coordinate is one-Lipschitz, and suitable subsets provide lower bounds on pairwise image distances.

This Hamming distance on partitions identifies a group with its complement because exchanging the two group names does not change the partition. Complementation is an isometry of the ordinary Hamming distance, so minimizing over this two-element group action gives a quotient metric and preserves the triangle inequality. Negating a zero-one indicator is not complementation; with $\pm 1$ membership vectors, it is.

Let $P,Q\in \mathbb{Z}[X_1,\dots ,X_k]$ have degree at most $n_j$ in $X_j$. Thenwhenever the quotient is defined. At each non-Archimedean place, the integral coefficients contribute at most one; at the Archimedean places, the triangle inequality contributes the polynomial length. Multiplying these local estimates and using the product formula gives the claim.

For two observation laws $P_0,P_1$ at real parameters separated by $\Delta$, every estimator $T$ satisfies ${\max }_j{\mathbb{E}}_j|T-{\theta }_j|\ge \Delta (1-\mathrm{TV}(P_0,P_1))/2$. With common densities, the sum of the two risks is at least $\int \min (p_0,p_1)(|T-{\theta }_0|+|T-{\theta }_1|)$, at least $\Delta \int \min (p_0,p_1)=\Delta (1-\mathrm{TV})$ by the triangle inequality. Divide by two. The same proof works for any metric loss.

The displayed quantity is homogeneous and separates zero, but generally violates the triangle inequality. Its $q$th power is subadditive because $(a+b{)}^q\le a^q+b^q$ for nonnegative $a,b$. Minimizing it is equivalent to minimizing its $q$th power. On finite vectors it is a standard nonconvex sparsity penalty.

On a bounded connected Lipschitz domain, let $u\ge 0$ belong to the BV space, and let $T$ be a positivity-preserving operator with $T1\ne 0$. Writing $u=c1+(u-c1)$, $c=u_{\Omega }\ge 0$, the triangle inequality and Poincaré inequality for total variation give $c\Vert T1{\Vert }_1\le \Vert Tu{\Vert }_1+\Vert T\Vert \Vert u-c{\Vert }_1\le \Vert Tu{\Vert }_1+C_{\Omega }\Vert T\Vert \mathrm{TV}(u)$. Thus forward-image and variation bounds control the missing constant mode, and hence the full $BV$ norm.

A metric on a set $X$ is a function $d:X\times X\to [0,\infty )$ that is symmetric, satisfies the triangle inequality, and obeys $d(x,y)=0$ exactly when $x=y$.

The displayed metric on antipodal classes of the unit sphere induces its quotient topology. Triangle inequality follows by choosing the signs that realize each minimum and applying the Euclidean triangle inequality. Distinct classes have disjoint saturated unions of small balls, proving the quotient is Hausdorff; compactness follows from the continuous quotient map.

For any estimator $T$ in a metric parameter space, $d(T,f{)}^2+d(T,g{)}^2\ge d(f,g{)}^2/2$ by the triangle inequality. Average the two risk functions and replace their probability density functions by their minimum. This proves $\max (R_f,R_g)\ge d(f,g{)}^2\int \min (p_f,p_g)/4$. The overlap equals one minus total variation distance. The same proof applies to squared error for a real-valued statistical functional of the parameter, without requiring that functional to be injective.

This is the travelling salesman problem on a complete graph with nonnegative symmetric distances satisfying the triangle inequality. Repeated visits can be shortcut without increasing cost. Assigning distance one to edges of an input graph and two to its nonedges creates a metric instance with a tour of cost at most the number of vertices exactly when the graph has a Hamiltonian cycle. Thus the metric problem remains NP-hard.

For $1\le p\le \infty$, the Lp norm satisfies the triangle inequality. For $1<p<\infty$, apply the Holder inequality to $\int (|f|+|g|)|f+g{|}^{p-1}$, then divide by $\Vert f+g{\Vert }_p^{p-1}$ when it is nonzero. The cases $p=1$ and $p=\infty$ follow directly from the pointwise triangle inequality. In particular, a series whose term norms have a finite sum converges in the complete $L^p$ space.