Inmetric geometry, themetric envelope ortight span of ametric spaceM is aninjective metric space into whichM can be embedded. In some sense it consists of all points "between" the points ofM, analogous to theconvex hull of a point set in aEuclidean space. The tight span is also sometimes known as theinjective envelope orhyperconvex hull ofM. It has also been called theinjective hull, but should not be confused with theinjective hull of amodule inalgebra, a concept with a similar description relative to thecategory ofR-modules rather than metric spaces.
The tight span of a metric space can be defined as follows. Let (X,d) be a metric space, and letT(X) be the set ofextremal functions onX, where we say anextremal function onX to mean a functionf fromX toR such that
For anyx,y inX,d(x,y) ≤f(x) +f(y), and
For eachx inX,f(x) = sup{d(x,y) - f(y):y inX}.[1]: 124
In particular (takingx =y in property 1 above)f(x) ≥ 0 for allx. One way to interpret the first requirement above is thatf defines a set of possible distances from some new point to the points inX that must satisfy thetriangle inequality together with the distances in (X,d). The second requirement states that none of these distances can be reduced without violating the triangle inequality.
Thetight span of(X,d) is the metric space(T(X),δ), whereis analogous to the metric induced by theℓ∞ norm. (Ifd is bounded, then δ is the subspace metric induced by the metric induced by theℓ∞ norm. Ifd is not bounded, then every extremal function onX is unbounded and so Regardless, it will be true that for anyf,g inT(X), the difference belongs to, i.e., is bounded.)
For a functionf fromX toR satisfying the first requirement, the following versions of the second requirement are equivalent:
For eachx inX,f(x) = sup{d(x,y) - f(y):y inX}.
f is pointwise minimal with respect to the aforementioned first requirement, i.e., for any functiong fromX toR such thatd(x,y) ≤ g(x) + g(y) for allx,y inX, ifg≤f pointwise, thenf=g.[2]: 93, Proposition 4.6.2 [Note 1][Note 2][3]: Lemma 5.1
IfX is finite, then for any functionf fromX toR that satisfies the first requirement, the second requirement is equivalent to the condition that for eachx inX, there existsy inX such thatf(x) +f(y) =d(x,y). (If then both conditions are true. If then the supremum is achieved, and the first requirement implies the equivalence.)
Say|X|=2, and choose distincta, b such thatX={a,b}. Then is the convex hull of{{(a,1),(b,0)},{(a,0),(b,1)}}. [Add a picture. Caption: IfX={0,1}, then is the convex hull of{(0,1),(1,0)}.][4]: 124
Every extremal functionf onX isKatetov:[5][6]: Section 2 f satisfies the first requirement and or equivalently,f satisfies the first requirement and (is 1-Lipschitz), or equivalently,f satisfies the first requirement and[2]: Proof of Proposition 4.6.1 [Note 4]
Not every Katetov function onX is extremal. For example, leta,b be distinct, letX = {a,b}, letd = ([x≠y])x,y inX be thediscrete metric onX, and letf = {(a,1),(b,2)}. Thenf is Katetov but not extremal. (It is almost immediate thatf is Katetov.f is not extremal because it fails the property in the third bullet of this section.)
Ifd is bounded, then everyf inT(X) is bounded. In fact, for everyf inT(X), (Note) (Follows from the third equivalent property in the above section.)
Ifd is unbounded, then everyf inT(X) is unbounded. (Follows from the first requirement.)
is closed under pointwise limits. For any pointwise convergent
If(X,d) is compact, then(T(X),δ) is compact.[7][2]: Proposition 4.6.3 (Proof: Theextreme-value theorem implies thatd, being continuous as a function is bounded, so (see previous bullet) is a bounded subset ofC(X). We have shownT(X) is equicontinuous, so theArzelà–Ascoli theorem implies thatT(X) isrelatively compact. However, the previous bullet impliesT(X) is closed under the norm, since convergence implies pointwise convergence. ThusT(X) is compact.)
For any functiong fromX toR that satisfies the first requirement, there existsf inT(X) such thatf≤g pointwise.[2]: Lemma 4.4
For any extremal functionf onX,[2]: Proposition 4.6.1 [Note 5]
For anyf,g inT(X), the difference belongs to, i.e., is bounded. (Use the above bullet.)
Letf inT(X). For anya inX, iff(a)=0, thenf=e(a).[3]: Lemma 5.1 (For everyx inX we have From minimality (second equivalent characterization in above section) off and the fact that satisfies the first requirement it follows that)
(X,d) ishyperbolic if and only if(T(X),δ) is hyperbolic.[3]: Theorem 5.3
For anyY such that is not hyperconvex.[2]: Proposition 4.7.2 ("(T(X),δ) is a hyperconvex hull of(X,d).")
Let be a hyperconvex metric space with and. If for allI with is not hyperconvex, then and(T(X),δ) areisometric.[2]: Proposition 4.7.1 ("Every hyperconvex hull of(X,d) is isometric with(T(X),δ).")
Say|X|=3, choose distincta, b, c such thatX={a,b,c}, and leti=d(a,b), j=d(a,c), k=d(b,c). Then where [Add a picture. Caption: IfX={0,1,2}, thenT(X)=conv{(,,),(,,)} u conv{(,,),(,,)} u conv{(,,),(,,)} is shaped like the letter Y.] (Cf.[4]: 124 )
If a set of points in the plane, with theManhattan metric, has a connectedorthogonal convex hull, then that hull coincides with the tight span of the points.
The figure shows a setX of 16 points in the plane; to form a finite metric space from these points, we use theManhattan distance (ℓ1 distance).[8] The blue region shown in the figure is theorthogonal convex hull, the set of pointsz such that each of the four closed quadrants withz as apex contains a point ofX. Any such pointz corresponds to a point of the tight span: the functionf(x) corresponding to a pointz isf(x) =d(z,x). A function of this form satisfies property 1 of the tight span for anyz in the Manhattan-metric plane, by the triangle inequality for the Manhattan metric. To show property 2 of the tight span, consider some pointx inX; we must findy inX such thatf(x)+f(y)=d(x,y). But ifx is in one of the four quadrants havingz as apex,y can be taken as any point in the opposite quadrant, so property 2 is satisfied as well. Conversely it can be shown that every point of the tight span corresponds in this way to a point in the orthogonal convex hull of these points. However, for point sets with the Manhattan metric in higher dimensions, and for planar point sets with disconnected orthogonal hulls, the tight span differs from the orthogonal convex hull.
The definition above embeds the tight spanT(X) of a set ofn () points intoRX, a real vector space of dimensionn. On the other hand, if we consider the dimension ofT(X) as apolyhedral complex,Develin (2006) showed that, with a suitable general position assumption on the metric, this definition leads to a space with dimension betweenn/3 andn/2.
An alternative definition based on the notion of ametric space aimed at its subspace was described byHolsztyński (1968), who proved that the injective envelope of a Banach space, in the category of Banach spaces, coincides (after forgetting the linear structure) with the tight span. This theorem allows to reduce certain problems from arbitrary Banach spaces to Banach spaces of the form C(X), where X is a compact space.
Develin & Sturmfels (2004) attempted to provide an alternative definition of the tight span of a finite metric space as thetropical convex hull of the vectors of distances from each point to each other point in the space. However, later the same year they acknowledged in anErratumDevelin & Sturmfels (2004a) that, while the tropical convex hull always contains the tight span, it may not coincide with it.
^In two dimensions, the Manhattan distance is isometric after rotation and scaling to theℓ∞ distance, so with this metric the plane is itself injective, but this equivalence betweenℓ1 andℓ∞ does not hold in higher dimensions.