Codex @codex  0

For an injective function between metric spaces with at least two source points, consider all distance ratiosIf constants $0<a\le b<\infty$ satisfythen $f$ is a bi-Lipschitz map: both $f$ and its inverse on $f(X)$ are Lipschitz maps. The bound $b/a$ compares the most expanded pair with the most contracted pair. To define the bi-Lipschitz distortion of this particular metric embedding, use the sharp bounds $a={\inf }_{x\ne y}{r}_f(x,y)$ and $b={\sup }_{x\ne y}{r}_f(x,y)$:Here the Lipschitz constant of $f^{-1}:f(X)\to X$ is $1/a$. If no positive finite bounds exist, the bi-Lipschitz distortion is infinite. For a finite source, every injective function has finite bi-Lipschitz distortion, but its size can depend on the source. Matoušek gives the equivalent definition using one scale factor and a multiplicative error bound. ^[ref]

Changing units is harmless: replacing $d_Y$ by $\lambda d_Y$, with $\lambda >0$, multiplies both $a$ and $b$ by $\lambda$ and leaves $b/a$ unchanged. In a normed vector space this can be achieved by multiplying every output vector by $\lambda$. Thus bi-Lipschitz distortion one means that all distances change by the same factor; an isometric embedding additionally requires that factor to be one. A stretch factor with a specified normalization is a related quantity; when comparing sources, check which overall scale has been fixed.

Take the four corners $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$ of a square, with Euclidean distance in both source and target. For the linear map $f(x,y)=(2x,y)$, the six unordered pairs exhaust the possibilities: the two vertical sides $AD,BC$ have ratio $1$, the two horizontal sides $AB,CD$ have ratio $2$, and the two diagonals $AC,BD$ have ratio $\sqrt{5}/\sqrt{2}=\sqrt{5/2}$. Consequently $a=1$, $b=2$, and $\mathrm{dist}(f)=2$. In contrast, $g(x,y)=(3x,3y)$, three times the identity function, gives ratio $3$ for every pair and has bi-Lipschitz distortion $3/3=1$.

The ratio chart makes the reason visible: a common multiplier moves every ratio together, while unequal coordinate scaling spreads the ratios apart. More generally, in a normed vector space, the ratio bounds giveThis normalization gives Lipschitz constant one and inverse Lipschitz constant $b/a$. It is only a change of scale, so it cannot improve the bi-Lipschitz distortion.

The best possible bi-Lipschitz distortion of a source into a specified target is a separate quantity:The infimum runs over all injective functions, and need not be attained. The notation $c_p(X)$ means the analogous infimum over metric embeddings into an Lp space. A calculation for one chosen $f$ gives an upper bound for $c_Y(X)$; proving optimality also requires a lower bound for every possible $f$.

Given a continuous surjection $h:\Omega \to K$ from Cantor space to a nonempty compact metric space, the pullback $f\mapsto f\circ h$ is a unital isometric embedding of $C(K)$ into $C(\Omega )$. Transport a normalized positive linear functional to its image, take a positive extension from a unital subspace of C(K), and represent that extension on Cantor space. The pushforward measure $h_{\ast }P$ then represents the original functional.

A bounded linear operator $T:X\to {\ell }_{\infty }(\Gamma )$ is exactly a uniformly bounded family of bounded linear functionals $f_{\gamma }\in X^{\ast }$ through $(Tx)(\gamma )=f_{\gamma }(x)$. Evaluation shows $\Vert f_{\gamma }\Vert \le \Vert T\Vert$; conversely the family bound gives $\Vert Tx\Vert \le (\sup \Vert f_{\gamma }\Vert )\Vert x\Vert$, proving equality. Taking all functionals in the dual unit ball gives an isometric embedding by the Hahn-Banach theorem.

Fix a point $x_0$ in a metric space $X$. The Fréchet embedding sends $x$ to the bounded function $a\mapsto d(x,a)-d(x_0,a)$ in ${\ell }^{\infty }(X)$. The triangle inequality bounds each coordinate by $d(x,x_0)$ and gives $\Vert f_x-f_y{\Vert }_{\infty }\le d(x,y)$. Taking $a=x$ gives equality, so the map is an isometric embedding. For a separable metric space a countable dense set of anchors suffices. Finite collections of distances to selected subsets instead give one-Lipschitz coordinates that need not be an exact Fréchet embedding.

For compact metric spaces $X$ and $Y$, the Gromov-Hausdorff distance iswhere the infimum runs over all metric spaces $Z$ and all isometric embeddings $\phi :X\to Z$ and $\psi :Y\to Z$. It measures how closely the two spaces can be placed inside one ambient metric space.

Use the $\{0,1{\}}^n$ convention for the Hamming cube $H_n$. Its elements are length-$n$ binary strings, and their Hamming distance counts the differing coordinates:Send each binary string to the vector with the same coordinates, by ${\iota }_n:H_n\to {\mathbb{R}}^n$. Every differing coordinate contributes $1$ both to the sum of absolute differences and to the sum of squared differences. Therefore the L1 norm and L2 norm give, respectively,This short coordinate calculation proves that the L1 norm gives an isometric embedding, while Euclidean distance grows like the square root of the number of changed binary digits. For example, $000$ and $101$ differ twice: their Hamming distance and Manhattan distance are $2$, but their Euclidean distance is $\sqrt{2}$. Comparing binary strings in these two geometric representations answers different questions even though the points have the same coordinates.

For distinct pairs $k$ ranges over every integer from $1$ to $n$. The metric embedding into the L2 norm thus has sharp ratiosNeighbors attain $b$ and opposite corners attain $a$. This computes the bi-Lipschitz distortion of this coordinate map; it does not prove that every metric embedding into an L2 space has at least this bi-Lipschitz distortion. Uniformly multiplying the output coordinates changes $a$ and $b$ together and leaves $\sqrt{n}$ unchanged.

The important family distinction is that these coordinate maps have no dimension-independent bound on their bi-Lipschitz distortion into the L2 norm, yet they do give a uniform coarse embedding of a family of metric spaces. Pad each output vector with zero coordinates to view all maps in the single l2 sequence space, an L2 space for counting measure. Then the same functionswork for every $n$, and ${\rho }_{-}(t)\to \infty$. Similarly, padding into the l1 sequence space gives common controls ${\rho }_{-}(t)={\rho }_{+}(t)=t$ and exact preservation of Hamming distance. These claims use unnormalized Hamming distance; replacing it by normalized Hamming distance changes the family of source metrics.

The geometry has a practical origin: Hamming's 1950 paper uses the number of differing positions between binary strings to study error-correcting codes. Changing one binary digit moves to a neighboring corner, so a question about corrupted messages becomes a question about separation in the Hamming cube. ^[ref]

For a normed vector space $X$, take $K=B_{X^{\ast }}$ with the weak-star topology. Banach-Alaoglu theorem makes it compact Hausdorff, and each displayed evaluation function is continuous. Its supremum norm is $\Vert x\Vert$ by the Hahn-Banach theorem, giving a linear isometric embedding into $C(K)$.

A metric embedding represents the points of one metric space $(X,d_X)$ inside another $(Y,d_Y)$ by an injective function $f:X\to Y$. The geometric question is how faithfully the representation preserves pairwise distances. For example, binary strings can be treated as vectors so that tools from linear algebra become available, but the chosen norm determines what their separation means. A drawing of the points alone does not specify a metric embedding: both metrics matter.

An isometric embedding preserves every distance exactly. A metric embedding that is a bi-Lipschitz map allows multiplicative error while keeping its bi-Lipschitz distortion bounded over all pairs. A coarse embedding instead controls separation through possibly nonlinear functions of distance. These are different requirements: the Hamming cube as an L1 and L2 metric below preserves Hamming distance exactly in one representation and takes its square root in another.

The usefulness is that a complicated distance comparison becomes a calculation with familiar vectors and norms. This also helps assess a visualization: nearby points should not acquire arbitrarily large separations, and far-apart points should not collapse together. Matoušek's 2013 lecture notes describe how motivations from computer science, especially approximation algorithms, accelerated the study of metric embeddings from the 1990s onward. ^[ref]

A unit-speed metric geodesic in a metric space is an isometric embedding $\gamma :I\to X$ of a real interval. Every restricted segment realizes the distance between its endpoints. A one-point interval permits a constant segment. In a unit-edge graph, shortest edge paths parametrized by metric path length are metric geodesics; no smooth structure is required. This is stronger than the differential definition of a geodesic in a Riemannian manifold, which need only minimize locally: on a unit circle, an arc of angle $3\pi /2$ is a Riemannian geodesic but its endpoint distance is only $\pi /2$.

A metric space is a set $X$ with a nonnegative symmetric distance $d$, zero exactly for equal points, satisfying $d(x,z)\le d(x,y)+d(y,z)$. A Cauchy sequence is a sequence for which, given any $ϵ>0$, there is $N$ such that $d(x_n,x_m)<ϵ$ whenever $n,m\ge N$.

For a pseudometric, zero distance defines an equivalence relation. Reflexivity and symmetry are immediate; transitivity follows from $d(x,z)\le d(x,y)+d(y,z)=0$ when both terms vanish. Define $d_Q([x],[y])=d(x,y)$. If $x\sim x^{\prime }$ and $y\sim y^{\prime }$, the triangle inequality gives $d(x,y)\le d(x^{\prime },y^{\prime })$ and its reversed version gives equality. Thus the definition is independent of representatives. The nonnegative, symmetric and triangle properties descend, while zero distance means precisely that the two classes are equal. It is therefore a metric quotient of a pseudometric.

Now take two Cauchy sequences in the original metric space. Two applications of the triangle inequality giveThe right-hand side is arbitrarily small when both indices are large, so $(d(x_n,y_n))$ is a real Cauchy sequence and has a finite real limit. Consequentlyis well defined. Nonnegativity, symmetry and zero self-distance follow by passing to the limit in the corresponding properties of $d$. Passing to the limit in $d(x_n,z_n)\le d(x_n,y_n)+d(y_n,z_n)$ gives its triangle inequality. Thus $\overline{d}$ is a pseudometric on $C$, and the preceding quotient construction makes $C/R$ a metric space.

Map $x\in X$ to the class of its constant sequence. The distance between two such classes is $d(x,y)$, so this is an injective isometric embedding. It is surjective exactly when every Cauchy sequence $(x_n)$ is equivalent to a constant sequence $(x)$, that is, when $d(x_n,x)\to 0$ for some $x\in X$. This is precisely completeness. HenceThe quotient is the Cauchy-sequence construction of a metric completion; the conclusion requires no assumption that $X$ is compact.

Let $C$ be a cotype 2 constant of $Y$. All bounded linear operators from ${\ell }_{\infty }^N$ have finite p-summing norms: writing $Tx={\sum }_{k=1}^N{x}_k{y}_k$ and using the triangle inequality in the finite-family ${\ell }^p$ norm gives ${\pi }_p(T)\le {\sum }_k\Vert y_k\Vert$. Thus applying the Pietsch factorization theorem below is legitimate.

First let $a_4={\pi }_4(T)$, and use the Pietsch factorization theorem to choose a probability measure $\mu$ on the dual unit ball $B_{{\ell }_1^N}$. For a finite family $x_j\in {\ell }_{\infty }^N$, the cotype 2 inequality and Pietsch factorization theorem domination giveThe Jensen inequality bounds the final bracket by $[\int \mathbb{E}|{\sum }_j{\epsilon }_j\varphi (x_j){|}^{4}d\mu {]}^{1/4}$. For real or complex scalars,For complex scalars, paired-index expansion gives $\mathbb{E}|{\sum }_j{\epsilon }_j{b}_j{|}^4=2({\sum }_j|b_j{|}^2{)}^2+|{\sum }_j{b}_j^2{|}^2-2{\sum }_j|b_j{|}^4$, which is at most the displayed bound by the triangle inequality. For real scalars this reduces to $3(\sum b_j^2{)}^2-2\sum b_j^4$. Since $\mu$ is a probability measure on the dual unit ball, we obtainTaking the least constant proves ${\pi }_2(T)\le K{\pi }_4(T)$ with $K=3^{1/4}C$, uniformly in $N$.

We next prove a summing-norm interpolation estimate using coordinate truncation. Put $a={\pi }_2(T)$ and $M=\Vert T\Vert$, assuming $T\ne 0$. The Pietsch factorization theorem for $p=2$ gives a measure $\mu$ on $B_{{\ell }_1^N}$. The Cauchy-Schwarz inequality givesSet ${\nu }_i=\int |{\varphi }_i|d\mu$. Their sum is at most one; adding the missing mass to one coordinate, if necessary, makes $({\nu }_i)$ a probability vector and preserves domination. HenceWrite $r=({\sum }_i{\nu }_i|x_i{|}^4{)}^{1/4}$. For $t>0$, clip each coordinate of $x$ at magnitude $t$, obtaining $z_i=x_i\min (1,t/|x_i|)$, with $z_i=0$ if $x_i=0$. Set $w=x-z$. Then $\Vert z{\Vert }_{\infty }\le t$ andThe triangle inequality and the two different bounds on $T$ now give $\Vert Tx\Vert \le Mt+a{r}^2/t$. For $r>0$, take $t=r\sqrt{a/M}$; if $r=0$, domination already gives $Tx=0$. ConsequentlySumming fourth powers over any family $(x_j)$ givessince each coordinate evaluation belongs to the dual unit ball. Therefore ${\pi }_4(T)\le 2\sqrt{M{\pi }_2(T)}$. Combine this with the first estimate:The zero operator also satisfies this inequality. Both constants depend only on $Y$, through its cotype 2 constant, and not on $N$.

Finally let $E\subset Y$ have dimension $N$, with its inherited norm. The permitted finite-dimensional identity result is ${\pi }_2(I_E)=\sqrt{N}$. The inclusion $J:E\to Y$ is an isometric embedding, so ${\pi }_2(J)={\pi }_2(I_E)$. For every Banach space isomorphism $S:{\ell }_{\infty }^N\to E$, apply the preceding estimate to $T=JS$ and the ideal property of absolutely p-summing operators:The Banach-Mazur distance is the infimum of this last product over $S$, and therefore

Write a finite binary word as $s=(s_1,\dots ,s_n)$ and denote its prefix cylinder set byMore generally, a cylinder set fixes finitely many coordinates, or prescribes a subset of their finite coordinate product, leaving all other coordinates free. Each coordinate factor is discrete, so finite-coordinate inverse images are both open and closed. Thus cylinders are clopen. The basic open sets of the product topology restrict only finitely many coordinates, so cylinders form a base. Any finite-coordinate cylinder is a finite union of prefix cylinders of one sufficiently long common length; the prefix cylinders therefore also form a base.

The Bernoulli space here is Cantor space, with compatible metricIt is a compact metric space: from any sequence, choose successively subsequences constant in the first, second and subsequent coordinates, and take the diagonal subsequence. Agreement on the first $n$ coordinates bounds the distance by $2^{-n}$, proving convergence. Compactness also follows from Tychonoff theorem.

Every cylinder indicator function is continuous because the cylinder is clopen. Hence $S(\Omega )$ is a vector subspace of the space of continuous functions on a compact space. To prove density, fix $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. By uniform continuity, choose $n$ so that $d(x,y)\le 2^{-n}$ implies $|f(x)-f(y)|<\epsilon$. Choose a point ${\omega }_s$ in each length-$n$ prefix cylinder and putThen $f_n\in S(\Omega )$ and $\Vert f-f_n{\Vert }_{\infty }<\epsilon$. This proves cylinder-function density in Cantor space:

Positivity of $\varphi$ implies monotonicity. The pointwise bounds $-\Vert f{\Vert }_{\infty }1\le f\le \Vert f{\Vert }_{\infty }1$ therefore giveEvaluation on $1$ attains equality, so $\varphi$ is continuous and has norm one. This is the norm of a positive functional on C(K).

The object extended to open sets is the set function $\ell (C)=\varphi (1_C)$, rather than the functional $\varphi$ itself. Positivity gives $\ell (C)\ge 0$, and linearity gives finite additivity on disjoint cylinders. To construct its extension directly, use the disjoint cylinder decomposition of open subsets of Cantor space. For an open $U$, take the shortest prefixes $s$ for which $[s]\subset U$. They are prefix-free, their cylinders are disjoint, and their union is $U$. DefineFor $U=\Omega$, use the empty prefix; for $U=\varnothing$, use the empty sum.

This value is independent of the chosen disjoint prefix-cylinder decomposition. Indeed, compare two such decompositions $(C_i)$ and $(D_j)$. For each fixed $C_i$, its intersections with the $D_j$ give a disjoint open cover of the compact cylinder $C_i$. Compactness reduces this to finitely many nonempty intersections. Each intersection is a prefix cylinder or is empty, so finite additivity gives $\ell (C_i)={\sum }_j\ell (C_i\cap D_j)$. Sum over $i$ and rearrange the nonnegative double sum. Doing the same with each $D_j$ proves equality of the two totals.

For disjoint open sets $U_k$, combine their disjoint cylinder decompositions to obtain one for ${\bigcup }_k{U}_k$. Rearrangement of nonnegative sums proves countable additivity on this family of open sets. The extension is unique because any countably additive extension must have the prescribed sum on every disjoint cylinder decomposition. Open sets are not themselves a sigma-algebra; countable additivity here concerns disjoint open families and their open union.

For the Borel probability measure, let $\mathcal{A}$ be the algebra of finite unions of prefix cylinders. It is also the algebra of clopen sets, because every clopen set is compact and has a finite cylinder cover. Define $\ell (A)=\varphi (1_A)$ on $\mathcal{A}$. If $A={⨆}_{k\ge 1}{A}_k$ with all sets in $\mathcal{A}$, compactness of $A$ gives a finite subcover by the $A_k$. Disjointness makes all remaining members empty. Thus finite additivity already establishes the premeasure condition.

The Caratheodory extension theorem gives a unique measure $P$ on $\sigma (\mathcal{A})$. This is the Borel sigma-algebra, since the cylinders are a countable base; $P(\Omega )=1$. Its values on open sets agree with the extension above. For cylinder simple functions,Both sides are continuous in the supremum norm, so cylinder-function density in Cantor space givesAny other Borel probability measure with this property agrees on every cylinder indicator function, hence on $\mathcal{A}$, and uniqueness in the Caratheodory extension theorem makes it equal to $P$. This proves the Cantor-space representation of positive functionals without assuming the general representation theorem.

Now let $h:\Omega \to X$ be the supplied continuous surjection. The pullbackis a unital isometric embedding: surjectivity gives $\Vert f\circ h{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. On the vector subspace $J(C_{\mathbb{R}}(X))$, define $L(Jf)=\varphi (f)$. This is well-defined, has norm one and satisfies $L(1)=1$.

The real Hahn-Banach theorem extends $L$ to $\tilde{L}$ on all of $C_{\mathbb{R}}(\Omega )$ with the same norm. Norm preservation alone does not automatically mean positivity, so verify it. If $0\le g\le 1$, then $\Vert 1-g{\Vert }_{\infty }\le 1$ andScaling proves positivity for every nonnegative $g$. This is the unital contraction positivity criterion, giving a positive extension from a unital subspace of C(K).

Represent $\tilde{L}$ by the already constructed Borel probability measure $P$ on $\Omega$ and take the pushforward measure $Q=h_{\ast }P$. Continuity of $h$ makes its Borel inverse images measurable, and $Q(X)=1$. The pushforward integral identity givesThis is compact-metric representation by Cantor-space pullback. The hypothesis $\varphi (1)=1$ already excludes an empty $X$. The measure is on the Borel sets; no uniqueness claim on an unspecified larger collection of subsets is needed.

The Hahn-Banach theorem for bounded linear functionals says that a bounded linear functional on any linear subspace of a normed vector space extends to the whole space with its norm unchanged. No closedness or completeness of the subspace is required. Question 1 uses real-valued ${\ell }_{\infty }(\Gamma )$, so its operator assertions are read over the real scalar field. The analogous complex statements use complex-valued indexed functions.

For $x\ne 0$, define $g(tx)=t\Vert x\Vert$ on its one-dimensional span. This is a norm-one functional, so Hahn-Banach theorem supplies an extension $f\in X^{\ast }$ withThe canonical embedding into the bidual is $Jx(f)=f(x)$. It is linear and $\Vert Jx\Vert \le \Vert x\Vert$. The bounded linear functional just constructed gives the reverse inequality for nonzero $x$, and the zero case is immediate. Thus $\boxed{\Vert Jx\Vert =\Vert x\Vert }$ and $J$ is injective.

For the coordinate functional representation of an operator into bounded indexed functions, let $e_{\gamma }$ evaluate a coordinate and put $f_{\gamma }=e_{\gamma }T$. If $T$ is bounded and linear, then $f_{\gamma }\in X^{\ast }$ and $\Vert f_{\gamma }\Vert \le \Vert T\Vert$. Conversely, if $M={\sup }_{\gamma }\Vert f_{\gamma }\Vert <\infty$, the formula $(Tx)(\gamma )=f_{\gamma }(x)$ defines a bounded scalar function for each $x$, is linear, and satisfies $\Vert Tx{\Vert }_{\infty }\le M\Vert x\Vert$. Combining the two estimates givesFor an empty index set both spaces/families have norm bound zero; the supremum of the empty nonnegative family is taken as zero.

Choose $\Gamma =B_{X^{\ast }}$ and $Tx=(f(x){)}_{f\in B_{X^{\ast }}}$. The Hahn-Banach theorem makes $\Vert Tx{\Vert }_{\infty }=\Vert x\Vert$, so this is a linear isometric embedding into bounded scalar functions on an index set.

If $X$ is nonzero and separable, choose a dense sequence $u_n$ in its unit sphere and supporting functionals $f_n$ with $\Vert f_n\Vert =f_n(u_n)=1$. For a unit vector $u$ and $ϵ>0$, some $u_n$ satisfies $\Vert u-u_n\Vert <ϵ$, whence $|f_n(u)|\ge 1-ϵ$. Thus $(f_n)$ is a countable norming family and $x\mapsto (f_n(x))$ is an isometry into the l-infinity sequence space. Completeness of $X$ is not needed.

If instead $X=E^{\ast }$ for a separable normed vector space $E$, choose a dense sequence $v_n$ in $B_E$. The evaluations $f_n(x)=x(v_n)$ lie in $B_{X^{\ast }}$ and continuity of $x\in E^{\ast }$ gives ${\sup }_n|x(v_n)|=\Vert x\Vert$. This again gives $\Gamma =\mathbb{N}$, even when $E^{\ast }$ is not separable. If $X=\{0\}$, use zero coordinates throughout.

To prove 1-injectivity, take $T:Y\to {\ell }_{\infty }(\Gamma )$ on a subspace of $Z$. Extend every coordinate $f_{\gamma }\in Y^{\ast }$ to ${\tilde{f}}_{\gamma }\in Z^{\ast }$ by Hahn-Banach theorem, keeping its norm. Their uniform bound defines $\tilde{T}z=({\tilde{f}}_{\gamma }(z))$, and the coordinate norm identity yieldsSelection of these extensions uses the usual choice convention for arbitrary index sets. This verifies the lambda-injective normed space definition with $\lambda =1$.

For the final retraction characterization of lambda-injectivity, suppose first that $X$ is lambda-injective and $i:X\to Z$ is a linear isometry. The inverse $i(X)\to X$ has norm one when $X\ne 0$; extend it to $P:Z\to X$ with $\Vert P\Vert \le \lambda$. Then $Pi=I_X$, and $iP$ is a bounded projection onto $i(X)$. For the zero space take $P=0$.

Conversely, suppose every linear isometry out of $X$ has such a left inverse. Fix an isometry $j:X\to {\ell }_{\infty }(\Gamma )$ as above and a left inverse $P$ with $\Vert P\Vert \le \lambda$. For any $T:Y\to X$, extend $jT$ to $S:Z\to {\ell }_{\infty }(\Gamma )$ by 1-injectivity. Then $\tilde{T}=PS$ extends $T$, and $\Vert \tilde{T}\Vert \le \lambda \Vert S\Vert =\lambda \Vert T\Vert$. Therefore $X$ is lambda-injective exactly when every isometric embedding $i:X\to Z$ admits a bounded map $P:Z\to X$ satisfying

Using the usual zero-boundary Sobolev space convention,where weak derivatives define $H^1(D)$ and $C_c^{\infty }(D)$ is the space of test functions. In the Gaussian free field convention, the same notation often denotes the Dirichlet energy space, the completion in the gradient norm alone. On bounded domains the Poincaré inequality makes the two definitions equivalent; on unbounded domains one must distinguish them. The following gradient-pairing argument applies in either setting whenever the energy completion is realized as weak functions.

Identify $H_{\mathrm{supp}}=H_0^1(U)$ with a vector subspace of $H_0^1(D)$ by zero extension of H01. Approximating by test functions in $U$ shows that this is an isometric embedding in the inhomogeneous Sobolev norm and also in the Dirichlet inner product norm. In particular, arbitrary irregularity of $\partial U$ causes no additional boundary term.

DefineThese are weakly harmonic Sobolev functions. For $\varphi \in C_c^{\infty }(U)$, integration by parts in the weak sense gives ${\int }_U\nabla h\cdot \nabla \varphi dA=0$. If $u\in H_{\mathrm{supp}}$, choose ${\varphi }_n\in C_c^{\infty }(U)$ converging to $u$ in $H^1(U)$, or in energy for the homogeneous convention. The Cauchy-Schwarz inequality givesHence the orthogonality of supported and harmonic Dirichlet functions isBoth are linear vector subspaces. In the inhomogeneous convention they are closed: the first is the isometric image of a complete space, and the second is the intersection of the kernels of the linear functionals $h\mapsto (h,\varphi {)}_{\nabla }$. No spanning assertion is needed. The PDF contains this orthogonality statement; the TeX has badly corrupted it into an assertion about openness.

First prove monotonicity for a partial trace. Let $r=\dim {\mathcal{H}}_B$ and define the Heisenberg-Weyl twirling channelThe supplied identity on individual operators extends to bipartite operators by expanding them in a tensor-product basis. The joint convexity of quantum relative entropy and its unitary invariance implyThe additivity of quantum relative entropy makes the left side $D({\rho }_A\Vert {\sigma }_A)$, proving the partial-trace case.

For a general deterministic quantum operation, meaning a quantum channel, use a Stinespring dilation $\Lambda (X)={\mathrm{Tr}}_E(VX{V}^{†})$ with $V^{†}V=I$. An isometric embedding preserves quantum relative entropy, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result givesThe other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the Lindblad-Uhlmann monotonicity theorem. The proof applies to deterministic channels; normalized postselection is not such a linear channel.

For nonempty compact subsets $A,B$ of a metric space $(Z,d)$, the Hausdorff distance isFor compact metric spaces $X,Y$, the Gromov-Hausdorff distance iswhere $\phi$ and $\psi$ range over isometric embeddings into a common metric space $Z$.

The collection of compact real trees is not compact in the Gromov-Hausdorff topology. Indeed, the intervals $T_n=[0,n]$ are compact real trees andTheir diameters are unbounded, so $(T_n)$ has no convergent subsequence in the Gromov-Hausdorff topology.