On hypersurfaces of positive reach, alternating Steiner formulae and Hadwiger's Problem
aa r X i v : . [ m a t h . DG ] A p r On hypersurfaces of positive reach, alternating Steiner formulæand Hadwiger’s Problem
Sebastian ScholtesSeptember 10, 2018
Abstract
We give new characterisations of sets of positive reach and show that a closed hypersurfacehas positive reach if and only if it is of class C , . These results are then used to prove newalternating Steiner formulæ for hypersurfaces of positive reach. Furthermore, it will turnout that every hypersurface that satisfies an alternating Steiner formula has positive reach.Finally, we provide a new solution to a problem by Hadwiger on convex sets and prove longtime existence for the gradient flow of mean breadth.Mathematics Subject Classification (2000): 53A07, 52A20 In his seminal paper [Fed59] Federer introduced the notion of sets of positive reach. Roughlyspeaking, the reach of a closed set A is the largest s ≥ such that all points whose distance to A is smaller than s possess a unique nearest point in A . Sets of positive reach share many of theproperties that make convex sets so interesting and important, but it is a much broader class. Allclosed convex sets as well as all closed C submanifolds of R n have positive reach in particular.One of Federer’s main results is a Steiner formula for sets of positive reach. In the simplest casethis means that for A ⊂ R n closed and ≤ s < reach( A ) the volume V ( A s ) := H n ( A s ) of theparallel set is a polynomial of degree at most n . More precisely, there are real numbers W k ( A ) , k = 0 , . . . , n , such that V ( A s ) = n X k =0 (cid:18) nk (cid:19) W k ( A ) s k (1)for ≤ s < reach( A ) [Fed59, 5.8 Theorem]. Here, the parallel set of a non-empty set A ⊂ R n isdefined by A s := ( { x ∈ R n | dist( x, A ) ≤ s } , s ≥ , { x ∈ A | dist( x, ∂A ) ≥ − s } , s < . In case of convex sets the W k are called quermaßintegrals and in the more general context ofsets with positive reach total curvatures (although the total curvatures differ from the W k by amultiplicative constant depending on n and k and are usually numbered in reverse order). Theseare important geometric quantities that characterise the sets involved. For example, for a non-empty compact set A with positive reach we have W ( A ) = H n ( A ) , W n ( A ) = χ ( A ) H n ( B (0)) (see [Fed59, 5.19 Theorem]); for n ≥ holds W ( A ) = n − SM ( A ) and if additionally A isconvex and has non-empty interior we even have W ( A ) = n − H n − ( ∂A ) . Here, χ ( A ) is the For an example of a compact set A ⊂ R of positive reach with − H ( ∂A ) < W ( A ) see [ACV08, Example1]. A and SM ( A ) is the outer Minkowski content of A , for a defi-nition see [ACV08]. In case of sets A ⊂ R n of positive reach whose boundaries are of class C , the quermaßintegrals can also be written as mean curvature integrals, that is, as an integral over ∂A of certain combinations of the classical principal curvatures that exist a.e. (see Lemma A.1);this is what the title of Federer’s paper alludes to.There are different characterisations of the reach of a set. For example, it can be defined as thelargest t such that two normals do not intersect in A s for all s < t (see Lemma 2.3 and for thedefinition of normals in this context (5)). In Theorem 1.1 we give two new characterisations ofsets of positive reach. The first tells us that a set has positive reach if and only if the set andits outer parallel sets satisfy an alternating Steiner formula. By alternating we mean that theSteiner formula not only gives the volume of the outer parallel sets (in our case ( A s ) t for t ≥ ),as in Federer’s case, but the same polynomial also describes the volume of the inner parallel sets( t < is admissible). The second characterisation says that a set has positive reach if and onlyif the parallel sets exhibit a semigroup-like structure. Theorem 1.1 (Characterisation of sets of positive reach).
Let A ⊂ R n closed, A
6∈ {∅ , R n } and r > . Then the following are equivalent • for all s ∈ (0 , r ) there are W k ( A s ) ∈ R such that for < s + t < r holds V (( A s ) t ) = n X k =0 (cid:18) nk (cid:19) W k ( A s ) t k , • ( A s ) t = A s + t for all for all s ∈ (0 , r ) and < s + t < r , • reach( A ) ≥ r . By means of the example A := [ − b, b ] \ [ − a, a ] for < a < b , where V ( A s ) = 4( b − a ) + 8( b + a ) s + ( π − s for ≤ s ≤ a, see Figure 1, we find that it is essential to have the Steiner formula for the outer parallel sets, inorder to characterise sets of positive reach.As we have seen before, a set of positive reach possesses a Steiner formula (1) for ≤ s < reach( A ) . Now, it is an obvious question to ask wether or not this formula can also be extendedto the inside of the set, i.e. if there is u < such that (1) also holds for u < s < reach( A ) .Disappointingly, the answer is, in general and even for convex bodies: No! This can easily be seenby A := [ − , , because V ( A s ) = 4 + 8 s + πs for s ≥ but V ( A s ) = (2 + 2 s ) = 4 + 8 s + 4 s for s ∈ ( − , , or by the example of the semi-circle, where the formula for the volume of the innerparallel bodies is not even a polynomial (see [KR12, Example 2]). In [HCS10c] a conjecture byMatheron, that the volume of the inner parallel bodies of a convex set is bounded below by theSteiner polynomial, is disproven and conditions for different bounds on the volume of the innerparallel bodies are given. This line of research was continued in [HCS10b]. Furthermore [KR12]showed that the volume of the inner parallel bodies of a polytope in R n is, what the authorscalled, a degree n pluriphase Steiner-like function , which basically allows the quermaßintegralsto change their values at a finite number of points. In Theorem 1.2 we characterise closed setswhose inner and outer parallel sets posses an alternating Steiner formula as those sets of thisclass whose boundaries have positive reach. Theorem 1.2 (Alternating Steiner formula and reach of the boundary).
Let A ⊂ R n be closed and bounded by a closed hypersurface, r > . Then the following areequivalent A := [ − b, b ] \ [ − a, a ] with outer parallel set. • for all s ∈ ( − r, r ) there are W k ( A s ) ∈ R such that for − r < s + t < r holds V (( A s ) t ) = n X k =0 (cid:18) nk (cid:19) W k ( A s ) t k , • ( A s ) t = A s + t for all for all s ∈ ( − r, r ) and − r < s + t < r , • reach( ∂A ) ≥ r , • ∂A is a closed C , hypersurface with reach( ∂A ) ≥ r . To prove this theorem we need a characterization of closed hypersurfaces of positive reach. Bya closed hypersurface in R n we mean a topological sphere , that is, the homeomorphic image of S n − . Theorem 1.3 (Closed hypersurfaces have positive reach iff C , ). Let A be a closed hypersurface in R n . Then A has positive reach if and only if A is a C , manifold. This result was already featured in [Luc57, §4 Theorem 1], a reference that is not easily accessibleand which does not seem to be widely known. Clearly, the result was stated in a slightly differentform, as Federer had not coined the term reach yet and is also proven by different methods. Inresources more readily available, we find the direction reach( A ) > implies C , in [Lyt05,Proposition 1.4] and [HG10, Theorem 1.2]. The other direction can, other than [Luc57, §4Theorem 1], only be found as a remark without proof, for example in [Fu89, below 2.1 Definitions]or [Lyt04, under Theorem 1.1]. Another hint to this result may be found in [Fed59, 4.20 Remark].Considering that Theorem 1.3 is mostly folklore and a uniform proof of both directions togetheris not available it seems to be worth to give a detailed proof of this result. To show this, we use acharacterisation of C ,α loc functions, Proposition 2.12, which states that a function is of class C ,α loc if and only if | f ( x − h ) − f ( x ) + f ( x + h ) | ≤ C | h | α . Hadwiger’s problem [Had55]. To be more precise, denote by K n the class of non-empty compactconvex sets in R n and by R p ( r ) , for r ≥ and ≤ p ≤ n − , the class of all K ∈ K n such that ϕ i : ( − r, ∞ ) → R , s W i ( K s ) for i = 0 , . . . , p are differentiable with W ′ i ( s ) = ( n − i ) W i +1 ( s ) ,where we abbreviate W i ( s ) = W i ( K s ) . In [HCS10a, Theorem 1.1] the class R n − of convexsets K whose quermaßintegrals are differentiable on ( − r ( K ) , ∞ ) , where r ( K ) is the inradius , isidentified as the set of outer parallel bodies of lower dimensional convex sets, i.e. R n − = { L s | L ∈ K n , dim( L ) ≤ n − , s ≥ } , (2)and [HCS11] gives a characterisation of R n − of a more complicated nature. Using our resultsof the present paper we can give the following new characterisation of the class R n − ( r ) . Theorem 1.4 (Characterisation of R n − ( r ) ). Let K ∈ K n , r > . Then the following are equivalent • K ∈ R n − ( r ) , • there is a convex L with K = L r , • K = ( K − r ) r , • reach( ∂K ) ≥ r , • ∂K is a closed C , hypersurface with reach( ∂K ) ≥ r . Additionally, these results give us a long time existence result for the energy dissipation equality(EDE) gradient flow of the mean breadth W n − on the space K , , of all sets in K n with non-empty interior and C , boundary, equipped with the Hausdorff distance d H . For the essentialnotation see the beginning of Section 3.2 and for more detailed information on gradient flows onmetric spaces we refer to [AGS05]. Proposition 1.5 (Gradient flow of the mean breath W n − on ( K , , d H ) ). Let K ∈ K , and T := ω − n reach( ∂K ) then x : [0 , T ) → K , , t K − ω n t is a gradient flow in the (EDE) sense for W n − on ( K , , d H ) , i.e. W n − ( x ( t )) + 12 Z ts | ˙ x ( u ) | d u + 12 Z ts |∇ W n − | ( x ( u )) d u = W n − ( x ( s )) (3) for all ≤ s ≤ t < T and x is an absolutely continuous curve. Additionally, x ( t ) → x ( T ) in d H for t → T , where x ( T ) := K − reach( ∂K ) , and x ( T ) is either a convex set contained in an affine n − dimensional space or a convex set with non-empty interior with reach( ∂x ( T )) = 0 . By ω n we denote the n -dimensional volume of the unit ball in R n , i.e. ω n := H n ( B (0)) . Actually, these charaterisations were done in a more general setting, which not only considers parallel sets,which are Minkowski sums with balls, but also allows for Minkowski sums with a certain class of convex sets. cknowledgements The author wishes to thank his advisor Heiko von der Mosel for his interest and encouragementas well as reading the manuscript and making many helpful suggestions. Additionally, we thankthe participants of the “2 nd Workshop on Geometric Curvature Energies” 2012 in Steinfeld,Germany, for valuable comments and remarks, and finally Rolf Schneider for his detailed answerto a particular question on quermaßintegrals.
As a generalisation of convex sets Federer introduced in his seminal paper [Fed59] the notion ofsets of positive reach. A closed set A ⊂ R n is said to be of reach t at a point a ∈ A , denoted by reach( A, a ) = t , if t is the supremum of all ρ > such that the restriction ˜ ξ A | B ρ ( a ) of the metricprojection map ˜ ξ A : R n → P ( A ) , x
7→ { a ∈ A | | x − a | = dist( x, A ) } is single valued, or to be more precise, singleton valued. Here, P ( A ) denotes the power set of A .The reach of a set A is then defined to be reach( A ) := inf a ∈ A reach( A, a ) . By Unp( A ) we denotethe set of all points that have a unique nearest point in A , that is Unp( A ) := { x ∈ R n | ξ A ( x ) = 1 } . Now, we introduce another metric projection map ξ A , defined on Unp( A ) so that ˜ ξ A ( x ) is alreadya singleton, by ξ A : Unp( A ) → A, x argmin a ∈ A ( | x − a | ) . This is essentially the same mapping as before, but it “extracts” the unique nearest point fromthe singleton.In what follows, we always assume A ⊂ R n , A
6∈ {∅ , R n } , so that we do not have to worry aboutcertain pathologies. Especially, we have ∂A = ∅ , because else we would have A = A ˙ ∪ ∂A , but A = ∅ and A = R n are the only closed and open sets in R n . We also use dist( x, A ) = dist( x, A ) and for x A additionally dist( x, A ) = dist( x, ∂A ) without further notice. Lemma 2.1 (Properties of ˜ ξ A ). Let A ⊂ R n , a ∈ A . Then a ∈ ˜ ξ A ( x ) if and only if ξ A ( x t ) = a for x t := a + t ( x − a ) and t ∈ [0 , .Proof. Step 1
Let a ∈ ˜ ξ A ( x ) . Suppose there is b ∈ A \{ a } with | x t − b | ≤ | x t − a | for a fixed t ∈ [0 , . Then | x − b | < | x − x t | + | x t − b | ≤ | x − x t | + | x t − a | = | x − [ a + t ( x − a )] | + | [ a + t ( x − a )] − a | = (1 − t ) | x − a | + t | x − a | = | x − a | , (4)but this contradicts a ∈ ˜ ξ A ( x ) . The strict inequality in (4) holds, because else we would have b ∈ x + [0 , ∞ )( a − x ) , which is not compatible with | x t − b | ≤ | x t − a | and | a − x | ≤ | b − x | . Step 2
Let ξ A ( x t ) = a for t ∈ [0 , and assume that there is b ∈ A \{ a } , such that | b − x | < | a − x | .Then − t ) | x − a | + | x − b | = 2 | x t − x | + | x − b | < | x − a | − + 2 − | x − b | / | x − a | < t < , so that | x t − b | ≤ | x t − x | + | x − b | < | x − a | − | x t − x | = t | x − a | = | x t − a | , which contradicts our hypothesis.We define the tangent cone of a set A ⊂ R n at a ∈ A , to be Tan a A := n tv | t ≥ , ∃ a k ∈ A \{ a } : v = lim k →∞ a k − a | a k − a | o ∪ { } and the normal cone of A at a to be the dual cone of Tan a A , in other words Nor a A := dual(Tan a A ) = { u ∈ R n | h u, v i ≤ for all v ∈ Tan a A } . (5)The normal cone is always a convex cone, while it may happen that the tangent cone is notconvex. From [Fed59, 4.8 Theorem (2)] we know that ξ A ( x ) = a implies x − a ∈ Nor a A .Another representation of the normal cone Nor a A = { tv | t ≥ , | v | = s, ξ A ( a + v ) = a } (6)for reach( A, a ) > s > can be found in [Fed59, 4.8 Theorem (12)]. Unfortunately, there seemsto be a small gap at the very end of the proof of this item in Federer’s paper. Namely, it has notbeen taken into consideration that the cone S , which is set to be the right-hand side of (6), cana priori be empty. That this is indeed not the case is shown in Lemma 2.2. From (6) we infer x − a ∈ Nor a A, x = a ⇒ ξ A ( x s ) = a for s < reach( A, a ) and x s = a + s x − a | x − a | , (7)as s x − a | x − a | ∈ Nor a A , so that v from (6) must be equal to s x − a | x − a | . Lemma 2.2 (If reach(
A, a ) > then there is v ∈ Unp( A ) \{ a } with ξ A ( v ) = a ). Let A ⊂ R n be a closed set, a ∈ ∂A and reach( A, a ) > . Then there is v ∈ Unp( A ) \{ a } with ξ A ( v ) = a .Proof. Step 1
We adapt the proof of [Fed59, 4.8 Theorem (11)] to our situation. Let a ∈ ∂A , < r < reach( A, a ) , < ε < reach( A, a ) − r . Without loss of generality we might assume that a = 0 . Then there is a sequence u k ∈ Unp( A ) \ A , with u k → a , | u k | < ε/ . For ρ ∈ [0 , r ] , k ∈ N and δ ( x ) := dist( x, A ) set η ( u k , ρ ) : = ξ A ( u k ) + δ ( u k ) + ρδ ( u k ) ( u k − ξ A ( u k )) . Then | η ( u k , ρ ) | ≤ | u k | + δ ( u k ) + ρ ≤ | u k | + r ≤ ε + r < reach( A, a ) , hence η ( u k , ρ ) ∈ Unp( A ) . Step 2
Now, we want to show that ξ A ( η ( u k , ρ )) = ξ A ( u k ) . Assume that this is not the case.Then ≤ τ := sup { t > | ξ A [ ξ A ( u k ) + t ( u k − ξ A ( u k ))] = ξ A ( u k ) } ≤ δ ( u k ) + ρδ ( u k ) < ∞ , by Lemma 2.1. Now, ξ A ( u k ) + τ ( u k − ξ A ( u k )) Unp( A ) ◦ , by [Fed59, 4.8 Theorem (6)], but | ξ A ( u k ) + τ ( u k − ξ A ( u k )) | ≤ | ξ A ( u k ) | + τ δ ( u k ) ≤ | u k | + δ ( u k ) + ρ ≤ | u k | + r ≤ ε + r < reach( A, a ) . A at two different points. Contradiction . Step 3 As | η ( u k , r ) | ≤ ε + r there must be a convergent subsequence, i.e. there is v ∈ R n with v = lim l →∞ η ( u k l , r ) and | v | = lim l →∞ | η ( u k l , r ) | = r, hence v ∈ Unp( A ) \{ a } and according to Step 2 we have ξ A ( v ) = lim l →∞ ξ A ( η ( u k l , r )) = lim l →∞ ξ A ( u k l ) = ξ A ( a ) = a, since ξ A is continuous on Unp( A ) , see [Fed59, 4.8 Theorem (4)].Note that any closed hypersurface A is compact and by the Jordan–Brouwer Separation Theoremit has a well-defined inside int( A ) and outside ext( A ) . From the definitions it is immediatelyclear that reach( A ) = min { reach(int( A )) , reach(ext( A )) } . (8) Lemma 2.3 (Alternative characterisation of reach
I).
Let A ⊂ R n closed, A
6∈ {∅ , R n } and reach( A ) > . Then reach( A ) = sup { t | ∀ a, b ∈ A, a = b : ( a + Nor a A ) ∩ ( b + Nor b A ) ∩ B t ( A ) = ∅} . (9) Proof.
Let a, b ∈ A , a = b and u ∈ Nor a A , v ∈ Nor b A with a + u = b + v . Then by (7) wemust have either | u | ≥ reach( A ) or | v | ≥ reach( A ) , because else ξ A ( a + u ) = a and ξ A ( b + v ) = b contradicts a = b . Hence reach( A ) is not larger than the right-hand side of (9). This means,for reach( A ) = ∞ we have proven the proposition. Let reach( A ) < ∞ . Clearly, for ε > theremust be a ε ∈ A and u ε ∈ S n − with x ε = a ε + (reach( A ) + ε ) u ε Unp( A ) . Hence, there7re two different points b ε = c ε such that b ε , c ε ∈ ˜ ξ A ( x ε ) . Therefore x ε − b ε ∈ Nor b ε A and x ε − c ε ∈ Nor c ε A , see [Fed59, 4.8 Theorem (2)], i.e. x ε ∈ ( b ε + Nor b ε A ) ∩ ( c ε + Nor b ε A ) and | x ε − b ε | = | x ε − c ε | ≤ | x ε − a ε | = reach( A ) + ε . Consequently, the right-hand side of (9) cannotbe larger than reach( A ) . Lemma 2.4 (Properties of parallel sets).
Let A ⊂ R n , A
6∈ {∅ , R n } .(a) For s > holds ∂ [ A s ] ⊂ { x ∈ R n \ A | dist( x, ∂A ) = s } .(b) For s, t ≥ holds ( A s ) t = A s + t .(c) For s ≥ and − s ≤ t ≤ holds A s + t ⊂ ( A s ) t .(d) For s < holds ∂ [ A s ] = { x ∈ A | dist( x, ∂A ) = | s |} .(e) For s, t ≤ holds ( A s ) t = A s + t .(f ) For s ≤ and ≤ t ≤ − s holds ( A s ) t ⊂ A s + t .Proof. (a) Let s > and x ∈ ∂ [ A s ] . As dist( · , A ) is continuous, the set A s is closed and dist( x, A ) ≤ s . Therefore, for every ε > there are points y ∈ B ε ( x ) with dist( y, A ) > s . Hence, x A and dist( x, A ) = dist( x, ∂A ) = s . (b) For s = 0 or t = 0 the equality is evident. Let s, t > . Then A s ⊂ A s + t and for x ∈ ( A s ) t \ A s we have dist( x, A ) ≤ dist( x, ∂ [ A s ]) + dist( ∂ [ A s ] , A ) ≤ t + s and hence x ∈ A s + t . Clearly A s ⊂ ( A s ) t , therefore let x ∈ A s + t \ A s . Then there is y ∈ ˜ ξ ∂A ( x ) and there is t ∈ [0 , , such that | z − y | = s for z = y + t ( x − y ) . Considering Lemma 2.1 weknow that z ∈ A s and additionally we have | x − y | = | x − z | + | z − y | = | x − z | + s ≤ t + s, note that x, y and z are on a straight line with z between x and y . This means | x − z | ≤ t andhence x ∈ ( A s ) t . (c) Let s ≥ , − s ≤ t ≤ and x ∈ A s + t . Then x ∈ A s and − dist( x, ∂ [ A s ]) + s = − dist( x, ∂ [ A s ]) + dist( ∂ [ A s ] , A ) ≤ dist( x, A ) ≤ s + t and hence dist( x, ∂ [ A s ]) ≥ − t , i.e. x ∈ ( A s ) t . (d) Let s < and x ∈ ∂ [ A s ] . As dist( · , ∂A ) is continuous the set A s is closed and dist( x, ∂A ) ≥| s | . Then x ∈ A s and for every ε > there are points y ∈ B ε ( x ) with dist( y, ∂A ) < | s | . Hence dist( x, ∂A ) = | s | . Now, let x ∈ A with dist( x, ∂A ) = | s | . Then x ∈ A s . As ∂A is closed thereexists a ∈ ˜ ξ ∂A ( x ) and according to Lemma 2.1 we have dist( x t , ∂A ) = t | x − a | = t | s | and hence x t ∈ R n \ A s for t ∈ (0 , . Consequently, x ∈ R n \ A s and x ∈ A s , therefore x ∈ ∂ [ A s ] . (e) For s = 0 or t = 0 the equality is evident. Let s, t < and x ∈ ( A s ) t . Then, as ∂A isclosed and non-empty, there is y ∈ ˜ ξ ∂A ( x ) and there is t ∈ [0 , such that | y − z | = | s | for z = y + t ( x − y ) . Considering Lemma 2.1 we have z ∈ A s . From | z − x | ≥ | t | we infer | s + t | = | s | + | t | ≤ | y − z | + | z − x | = | x − y | = dist( x, ∂A ) , note that x, y and z are on a straight line with z between x and y . This means x ∈ A s + t . Nowlet x ∈ A s + t . Then x ∈ A s and | s + t | = | s | + | t | ≤ dist( x, ∂A ) ≤ dist( x, ∂ [ A s ]) + dist( ∂ [ A s ] , ∂A ) = dist( x, ∂ [ A s ]) + | s | ,
8y (d), so that x ∈ ( A s ) t . (f ) Let s ≤ , ≤ t ≤ − s and x ∈ ( A s ) t . Then x ∈ A and − s − t ≤ dist( A s , ∂A ) − dist( x, A s ) ≤ dist( x, ∂A ) , i.e. x ∈ A s + t .The examples ∂B (0) , ∂ [0 , and [0 , suffice to show that the inclusions in (a), (c) and (f),respectively, can be strict. Lemma 2.5 (Alternative characterisation of reach
II).
Let A ⊂ R n closed, A
6∈ {∅ , R n } and r > . Then reach( A ) ≥ r ⇔ ( A s ) t = A s + t for all s ∈ (0 , r ) , t ∈ ( − s, r − s ) . (10) Proof.
Step 1
Let reach( A ) ≥ r . Let s ∈ (0 , r ) . For t = 0 nothing needs to be shown. Let t ∈ (0 , r − s ) . We then always have ( A s ) t = A s + t , see Lemma 2.4 (b). Let s ∈ (0 , r ) , t ∈ ( − s, ,then by Lemma 2.4 (c) we always have A s + t ⊂ ( A s ) t . For x ∈ A we automatically have x ∈ A s + t ,so let x ∈ ( A s ) t \ A . As reach( A ) ≥ r we find a unique y = ξ A ( x ) and by (7) we additionally know dist( x u , A ) = | x u − y | = u for x u := y + u ( x − y ) / | x − y | , u < r . Then x s ∈ ∂ [ A s ] , because x u ∈ A s for ≤ u ≤ s and x u ∈ R n \ A s for s < u < r , so that | x − x s | ≥ − t , dist( x, A ) = | x − y | < s andhence dist( x, A ) = | x − y | = | x s − y | − | x s − x | ≤ s + t, note that y, x and x s are on a straight line with x between y and x s . Hence x ∈ A s + t . Step 2
The other direction is a the contrapositive of Lemma 2.6 if we put s = σ + τ and t = − τ . Lemma 2.6 (If reach( A ) < r then ( A σ + τ ) − τ \ A σ contains an inner point). Let A ⊂ R n be closed, A
6∈ {∅ , R n } and reach( A ) < r . Then there are σ ∈ (0 , r ) , τ ∈ (0 , r − σ ) such that ( A σ + τ ) − τ \ A σ contains an inner point.Proof. Let reach( A ) < r . Then there is x ∈ A u \ A for some u ∈ (0 , r ) and y, z ∈ A , y = z with y, z ∈ ˜ ξ A ( x ) . Let | x − y | = | x − z | =: t then < t < r . Case 1
Let x ∈ A ◦ t . Then dist( x, ∂ [ A t ]) > and B dist( x,∂ [ A t ])+ ε ( x ) ⊂ A t + ε for all ε > .Choose < ε < r − t and < δ < min { − dist( x, ∂ [ A t ]) , − t } . Then B δ ( x ) ⊂ ( A t + ε ) − ( ε + δ ) and for all w ∈ B δ ( x ) holds dist( w, A ) ≥ dist( x, A ) − | x − w | = t − | x − w | > t − δ so that B δ ( x ) ∩ A t − δ = ∅ . Hence, x is an inner point of ( A t + ε ) − ( ε + δ ) \ A t − δ , i.e. the propositionholds for σ = t − δ and τ = ε + δ . Case 2
Let x ∈ ∂ [ A t ] . Without loss of generality we might assume that y = − ae , z = ae and x = be with t = a + b and a > . Let ε ∈ (0 , min { r − t , t } ) . Then B ε ( x ) ⊂ ( B t + ε ( y ) ∩ B t + ε ( z )) and the only elements of ∂B t + ε ( y ) ∩ ∂B ε ( x ) and ∂B t + ε ( z ) ∩ ∂B ε ( x ) are x + ε ( x − y ) /t and x + ε ( x − z ) /t , respectively. If these two points do not belong to ∂ [ A t + ε ] then dist( x, ∂ [ A t + ε ]) > ε . Now, (cid:12)(cid:12) x + ε x − yt − z (cid:12)(cid:12) = (cid:12)(cid:12) (1 + ε/t ) be − (1 − ε/t ) ae (cid:12)(cid:12) = (1 + ε/t ) b + (1 − ε/t ) a = (1 + ε/t ) ( a + b ) − εa /t = ( t + ε ) − εa /t < ( t + ε ) , At first glance it might seem rather strange that dist( x, A ) = t and x ∈ A ◦ t , but it is seen easily that this isindeed possible, for example for A = ∂B (0) , x = 0 and t = 1 . x + ε ( x − y ) /t ∈ B t + ε ( z ) and, by interchanging y and z we obtain x + ε ( x − z ) /t ∈ B t + ε ( y ) . Hence, we have shown that x lies in the interior of ( A t + ε ) − ε . This means that thereis δ > such that B δ ( x ) ⊂ ( A t + ε ) − ε . Now, B δ ( x ) \ A t is open and non-empty, as x ∈ ∂ [ A t ] , so that there must be w ∈ B δ ( x ) \ A t and δ ′ > with B δ ′ ( w ) ⊂ B δ ( x ) \ A t . Therefore w is aninner point of ( A t + ε ) − ε \ A t . That is, we have shown the proposition for σ = t and τ = ε . Lemma 2.7 (If reach( R n \ A ) < r then A − σ \ ( A − ( σ + τ ) ) τ contains an inner point). Let A ⊂ R n be closed, A
6∈ {∅ , R n } and reach( R n \ A ) < r . Then there are σ ∈ (0 , r ) , τ ∈ (0 , r − σ ) such that A − σ \ ( A − ( σ + τ ) ) τ contains an inner point.Proof. Let reach( R n \ A ) < r . Then there is x ∈ ( R n \ A ) u \ R n \ A ⊂ A for some u ∈ (0 , r ) and y, z ∈ ∂A , y = z with y, z ∈ ˜ ξ ∂A ( x ) . Let | x − y | = | x − z | =: t then < t < r . Hence, x is aninner point of A − ( t − δ ) for δ ∈ (0 , t ) and consequently B δ ( x ) ⊂ A − ( t − δ ) , since dist( w, ∂A ) ≥ dist( x, ∂A ) − | x − w | ≥ t − δ holds for all w ∈ B δ ( x ) . In the same manner as in Lemma 2.6 Case 2 we can show that for everysmall enough ε > we have dist( x, A − ( t + ε ) ) > ε . Now, fix δ = min { dist( x,A − ( t ε ) ) − ε , t } , i.e.especially ε + 3 δ ≤ dist( x, A − ( t + ε ) ) . Then dist( w, A − ( t + ε ) ) ≥ dist( x, A − ( t + ε ) ) − | x − w | ≥ ε + 2 δ holds for all w ∈ B δ ( x ) . This means we have w ( A − ( t + ε ) ) δ + ε = ( A − ( t − δ + δ + ε ) ) δ + ε for all w ∈ B δ ( x ) , or in other words x is an inner point of A − ( t − δ ) \ ( A − ( t − δ + δ + ε ) ) δ + ε and thuswe have proven the proposition for σ = t − δ and τ = δ + ε . C , manifolds Proposition 2.8 (Normal cones of closed hypersurfaces of positive reach are lines).
Let A be a closed hypersurface in R n and reach( A ) > . Then for a ∈ A there is a direction s ∈ S n − such that Nor a A = R s and Nor a int( A ) = [0 , ∞ ) s. (11) Proof.
Clearly ξ A ( x ) = a , x = a implies B | x − a | ( x ) ∩ A = ∅ . By Lemma 2.2 and (8) we know thatfor all a ∈ A there are x ∈ int( A ) , x ∈ ext( A ) , such that ξ A ( x i ) = a and hence B | x − a | ( x ) ⊂ int( A ) , B | x − a | ( x ) ⊂ ext( A ) . Then we must have that x , x , a lie on a straight line, with a between x and x , as else | x − x | < | x − a | + | a − x | , so that there would be a point y = x + α x − x | x − x | = x + ( | x − x | − α ) x − x | x − x | ∈ B | x − a | ( x ) ∩ B | x − a | ( x ) with ≤ α < | x − a | and ≤ | x − x |− α < | x − a | . Obviously this contradicts int( A ) ∩ ext( A ) = ∅ . Therefore, R ( x − a ) ⊂ Nor a A , by (6), and with the same argument as above we can alsoshow that Nor a A ⊂ R ( x − a ) .An s ∈ S n − with [0 , ∞ ) s ⊂ Nor a int( A ) is called outer normal of a closed hypersurface A at a and correspondingly − s an inner normal . If the outer normal is unique we denote it by ν ( a ) . Note, that we had to distinguish the different cases, because we need B δ ( x ) \ A t to be non-empty. emma 2.9 (Normals are continuous). Let A be a closed hypersurface in R n , reach( A ) > , a k ∈ A , a k → a and s k ∈ S n − be outernormals for A at a k . Then s k → s and s ∈ S n − is the outer normal of A at a .Proof. Let ( s k l ) l ∈ N be a subsequence. Then, as S n − is compact, there is an u ∈ S n − and afurther subsequence with s k lm → u . Since ξ A is continuous, see [Fed59, 4.8 Theorem (4)], wehave a k lm = ξ A ( a k lm + ts k lm ) → a = ξ A ( a + tu ) for all t < reach( A ) . According to [Fed59, 4.8 Theorem (2)] holds u ∈ Nor a A . By Proposition 2.8 there is a single s ∈ S n − such that u = s for all subsequences and s is outer normal of A at a . By Urysohn’sprinciple we have s k → s .The proof also shows that for any closed set of positive reach the limit of normals is a normal atthe limit point. Lemma 2.10 (Closed hypersurface of positive reach is locally a graph).
Let A ⊂ R n be a closed hypersurface, reach( A ) > , a ∈ A such that Nor a A = R s and s ∈ S n − is an outer normal. Then A is locally a graph over a + (Nor a A ) ⊥ . Put more precisely, this meansthat there is ε > such that after a rotation and translation Φ : R n → R n , transforming a to and s to e n , we can write Ψ : R n − ⊃ B ε (0) → Φ( B ε ( a ) ∩ A ) , v ( v, f ( v )) , with a bijective function Ψ and a scalar function f : R n − → R .Proof. Assume that the proposition is not true. Without loss of generality we might assume a = 0 and s = e n . Then for every ε > there are y = y ( ε ) , z = z ( ε ) ∈ B ε (0) ∩ A , y = z such that y i = z i , for i = 1 , . . . , n − . Without loss of generality let < y n < z n . If s y is theouter normal at y , we know by Lemma 2.9 that α y := ∡ ( s, s y ) → , for ε → . By elementarygeometry we have y + (0 , t ) e n ⊂ B t ( y + ts y ) , if sin( α y / ≤ − . This means that z ∈ B t ( y + ts y ) for | y − z | = t < reach( A ) , if ε is small enough. But as we have seen in the proof of Proposition2.8, we have B t ( y + ts y ) ∩ A = ∅ . Contradiction .The subdifferential of a function f : Ω → R , Ω ⊂ R n at x ∈ Ω is the set ∂f ( x ) := n v ∈ R n | lim inf y → x f ( y ) − f ( x ) − h v, y − x i| y − x | ≥ o , see [RW98, Definition 8.3, (a) and 8(4), p.301].The next lemma is a special case of [Lyt05, Proposition 1.4]. Lemma 2.11 (Closed hypersurface of positive reach are C , ). Let A ⊂ R n be a closed hypersurface, reach( A ) > . Then A is a C , hypersurface.Proof. Step 1
From Lemma 2.10 we know that we can write A locally as the graph of a realfunction f . Let a ∈ A . Without loss of generality we assume that s = − e n is the, thanksto Lemma 2.8, unique outer normal of a at A and a = ( x, f ( x )) . By the characterisationof subdifferentials in terms of normal vectors [RW98, 8.9 Theorem, p.304f,] it is clear that ∂f ( x ) = { v } , where ( v, − ∈ Nor ( x,f ( x )) epi( f ) = [0 , ∞ ) s , corresponding to the normal of int( A ) .Likewise ∂ ( − f )( x ) = {− v } , where ( − v, − ∈ Nor ( x, − f ( x )) epi( − f ) = [0 , ∞ ) s , corresponding tothe normal of ext( A ) . This means that lim y → x f ( y ) − f ( x ) − h v, y − x i| y − x | = 0 , lim inf y → x [ − g ( x )] = − lim sup y → x g ( x ) . Hence, f is differentiable at x and ∇ f ( x ) = v . Let x k → x and ( ∇ f ( x k ) , − ∈ Nor ( x k ,f ( x k )) epi( f ) = [0 , ∞ ) s k , | s k | = 1 , k ∈ N . Then, as wehave seen in Lemma 2.9, s k → s and there are t k ∈ [0 , ∞ ) , such that ( ∇ f ( x k ) , −
1) = t k s k .Additionally, t s n = − t k s nk ⇒ t k = t s n s nk → t , and therefore ∇ f ( x k ) → ∇ f ( x ) . This means the Jacobian matrix ∇ f is continuous and hence f is locally Lipschitz. Step 2
Using [Fed59, 4.18 Theorem] and the abbreviations a := ( x, f ( x )) , b ± := ( x ± h, f ( x ± h )) we can estimate (cid:12)(cid:12)(cid:12) f ( x − h ) − f ( x ) + f ( x + h ) q |∇ f ( x ) | (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) − [ f ( x ) − f ( x − h ) + h∇ f ( x ) , ( x − h ) − x i ] − [ f ( x ) − f ( x + h ) + h∇ f ( x ) , ( x + h ) − x i ] q |∇ f ( x ) | (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)D (cid:20) x − hf ( x − h ) (cid:21) − (cid:20) xf ( x ) (cid:21) , q |∇ f ( x ) | (cid:20) ∇ f ( x ) − (cid:21) E(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)D (cid:20) x + hf ( x + h ) (cid:21) − (cid:20) xf ( x ) (cid:21) , q |∇ f ( x ) | (cid:20) ∇ f ( x ) − (cid:21) E(cid:12)(cid:12)(cid:12) = π Nor a A ( b − − a ) + π Nor a A ( b + − a ) = dist( b − − a, Tan a A ) + dist( b + − a, Tan a A ) ≤ | b − − a | t + | b + − a | t = | ( x − h ) − x | + | f ( x − h ) − f ( x ) | t + | ( x + h ) − x | + | f ( x + h ) − f ( x ) | t ≤ c t | h | , for t ≤ reach( A ) . Now, Proposition 2.12 implies that f is of class C , .The interesting direction of the next very useful proposition can be found in [CH70, Lemma 2.1]for a more general modulus of continuity; but as we are only interested in Hölder continuousderivatives, we use this specialised version. We also found the proposition formulated in [Lyt05,Lemma 2.1] in a form very close to the way we present it here. The idea of smoothing thefunction came from [LTR05, Theorem 2.1]. Proposition 2.12 (Characterisation of C ,α loc functions). Let Ω ⊂ R n be open, f : Ω → R m bounded and < α ≤ . Then the following are equivalent • there are ρ > and L > such that for all x ∈ Ω holds f ∈ C ,α ( B ρ x ( x ) , R m ) and [ Df | B ρx ( x ) ] C ,α ≤ L , where ρ x = min { dist( x, ∂ Ω) , ρ } , • there is C > and δ > such that for all x ∈ Ω and all | h | < δ x = min { dist( x, ∂ Ω) , δ } holds | f ( x − h ) − f ( x ) + f ( x + h ) | ≤ C | h | α . Proof.
Step 1
Let f be as requested in the first item. Obviously, it is enough to prove theproposition for m = 1 . Using Taylor’s Theorem for Lipschitz functions, Theorem 2.15, we know f ( x ± h ) − f ( x ) = Z h∇ f ( x ± (1 − t ) h ) , ± h i d t | h | < ρ x , and we obtain | f ( x − h ) − f ( x ) + f ( x + h ) | = (cid:12)(cid:12)(cid:12) Z h∇ f ( x + (1 − t ) h ) , h i + h∇ f ( x − (1 − t ) h ) , − h i d t (cid:12)(cid:12)(cid:12) ≤ Z |h∇ f ( x + (1 − t ) h ) − ∇ f ( x − (1 − t ) h ) , h i| d t ≤ Z L | [ x + (1 − t ) h ] − [ x − (1 − t ) h ] | α | h | d t ≤ α L | h | α . Step 2
Now let f be as specified in the second item. We estimate (cid:12)(cid:12)(cid:12) ∞ X k =0 k ( f ( x ) − f ( x + 2 − ( k +1) h ) + f ( x + 2 − k h )) (cid:12)(cid:12)(cid:12) ≤ C ∞ X k =0 k (2 − ( k +1) | h | ) α = C − (1+ α ) | h | α ∞ X k =0 (2 − α ) k < ∞ , (12)so that the series converges uniformly in ( x, h ) on U := S x ∈ Ω { x }× B δ x (0) by Weierstraß’ M -Test.As the l th partial sum is a telescoping sum, we easily compute S l ( x, h ) := l X k =0 k ( f ( x ) − f ( x + 2 − ( k +1) h ) + f ( x + 2 − k h ))= l X k =0 k f ( x ) − l +1 X k =1 k f ( x + 2 − k h ) + l X k =0 k f ( x + 2 − k h )= (2 l +1 − f ( x ) − l +1 f ( x + 2 − ( l +1) h ) + f ( x + h )= f ( x + h ) − f ( x ) − | h | f ( x + 2 − ( l +1) h ) − f ( x )2 − ( l +1) | h | . (13)Therefore for all ( x, h ) ∈ U , h = 0 the following limit exists (but might depend not only on thedirection, but also on the absolute value of h ) lim l →∞ f ( x + 2 − ( l +1) h ) − f ( x )2 − ( l +1) | h | . Step 3
Let x ∈ Ω and y, z ∈ B δ x / ( x ) , y = z . Clearly B δ x / ( x ) ⊂ B δ z / ( z ) and δ x / ≤ δ z ≤ δ x .Then there is l ∈ N with δ z / ≤ l +1 | y − z | < δ z . According to (13) we have | S l ( z, l +1 ( y − z )) | = (cid:12)(cid:12)(cid:12) f ( z + 2 l +1 ( y − z )) − f ( z ) − l +1 | y − z | f ( z + 2 − ( l +1) l +1 ( y − z )) − f ( z )2 − ( l +1) l +1 | y − z | (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) f ( z + 2 l +1 ( y − z )) − f ( z ) − l +1 | y − z | f ( y ) − f ( z ) | y − z | (cid:12)(cid:12)(cid:12) and (12) yields | S l ( z, l +1 ( y − z )) | ≤ C − (1+ α ) | l +1 ( y − z ) | α ∞ X k =0 (2 − α ) k ≤ (cid:16) C − (1+ α ) | l +2 δ x | α ∞ X k =0 (2 − α ) k (cid:17) l +1 | y − z | =: c l +1 | y − z | . f , i.e. | f ( x ) | ≤ M for all x ∈ Ω , to obtain (cid:12)(cid:12)(cid:12) f ( y ) − f ( z ) | y − z | (cid:12)(cid:12)(cid:12) ≤ c + (cid:12)(cid:12)(cid:12) f ( z + 2 l +1 ( y − z )) − f ( z )2 l +1 | y − z | (cid:12)(cid:12)(cid:12) ≤ c + 2 M δ z ≤ c + 8 Mδ x , so that f is locally Lipschitz. Step 4
In retrospect of Step 2 and Step 3 we know that the mapping g i ( x, λ ) := lim l →∞ f ( x + 2 − ( l +1) λe i ) − f ( x )2 − ( l +1) λ , i = 1 , . . . , n is continuous on [ x ∈ Ω { x } × ([ − δ x , δ x ] \{ } ) , thanks to the uniform limit theorem. Let f ε be the mollification of f , i.e. the convolution withstandard mollifiers η ε . Fix x ∈ Ω and < | λ | < δ x / . We now want to show that there is asequence ε k ↓ such that for all < | λ | < δ x / we have g i ( x, λ ) = lim k →∞ ∂ i f ε k ( x ) , regardlessof the value of λ . Since g i ( x, λ ) equals ∂ i f ( x ) at every point x ∈ Ω where f is differentiable,which is almost every point of Ω , we know by elementary properties of mollifications on Sobolevspaces, note C , ⊂ W , ∞ , that ∂ i f ε ( x ) = (cid:16) η ε ∗ lim l →∞ (cid:16) f ( · + 2 − l λe i ) − f ( · )2 − l λ (cid:17)(cid:17) ( x ) = ( η ε ∗ g i ( · , λ ))( x ) , for all < | λ | < δ x / and ε small enough. As ∂ i f ε ( x ) is bounded in ε , because f is Lipschitzcontinuous, there is a sequence ε k ↓ such that lim k →∞ ∂ i f ε k ( x ) = a i , or in other words, forevery ˜ ε > there is N = N (˜ ε ) , with | a i − ∂ i f ε k ( x ) | ≤ − ˜ ε for all k ≥ N . On the other handwe find N = N (˜ ε ) , such that | ∂ i f ε k ( x ) − g i ( x, λ ) | = | η ε k ( x ) ∗ g i ( x, λ ) − g i ( x, λ ) | ≤ − ˜ ε for all k ≥ N , because g i ( x, λ ) is continuous. Putting the inequalities together we obtain | a i − g i ( x, λ ) | ≤ | a i − ∂ i f ε k ( x ) | + | ∂ i f ε k ( x ) − g i ( x, λ ) | ≤ ˜ ε for all k ≥ max { N , N } , i.e. g i ( x, λ ) = a i . By (12) and (13) this means | f ( x + λe i ) − f ( x ) − a i λ | ≤ C | λ | α , so that f is partially differentiable at x with ∂ i f ( x ) = a i = g i ( x, λ ) with continuouspartial derivatives. Therefore f is differentiable. Step 5
Let x ∈ Ω and y, z ∈ B δ x / ( x ) , y = z as in Step 3. Then | lim l →∞ S l ( z, y − z ) + lim l →∞ S l ( y, z − y ) | = | f ( y ) − f ( z ) − ( y − z ) ∇ f ( z ) + f ( z ) − f ( y ) − ( z − y ) ∇ f ( y ) | = | y − z ||∇ f ( y ) − ∇ f ( z ) | and (12) yields | lim l →∞ S l ( z, y − z ) + lim l →∞ S l ( y, z − y ) | ≤ C − (1+ α ) | y − z | α ∞ X k =0 (2 − α ) k =: ˜ C | y − z | α . .2 Closed C , hypersurfaces have positive reach It is folklore that compact C , submanifolds have positive reach and in fact this can evenbe found in many remarks in the literature, see for example [Fu89, below 2.1 Definitions] or[Lyt04, under Theorem 1.1], but, unfortunately, the author was not able to locate a single proof.Therefore we show the statement in a special case, adapted to our needs. Lemma 2.13 (Closed C , hypersurfaces have positive reach). Let A ⊂ R n be a closed hypersurface of class C , . Then reach( A ) > .Proof. As A is C , it can be locally written as a graph of a C , function. By compactnessof A and Lebesgue’s Number Lemma we find ε, δ > and a finite number N of functions f k ∈ C , ( B ε (0) , R ) , k = 1 , . . . , N , B ε (0) ⊂ R n − such that for every a ∈ A the set A ∩ B δ ( a ) is,after a translation and rotation, covered by the graph of a single f k . Step 1
Let u, v ∈ A with | u − v | ≤ δ . Then both points lie in the graph of a function f = f k and we can write u = ( x, f ( x )) , v = ( y, f ( y )) for x, y ∈ B ε (0) . The distance of v − u to Tan u A is given by the projection of v − u on the normal space Nor u A , i.e. dist( v − u, Tan u A ) = (cid:12)(cid:12)(cid:12)D (cid:20) yf ( y ) (cid:21) − (cid:20) xf ( x ) (cid:21) , q |∇ f ( x ) | (cid:20) ∇ f ( x ) − (cid:21) E(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ( y − x ) ∇ f ( x ) − ( f ( y ) − f ( x )) q |∇ f ( x ) | (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) f ( x ) − f ( y ) − ∇ f ( x )( x − y ) q |∇ f ( x ) | (cid:12)(cid:12)(cid:12) . By Taylor’s Theorem for Lipschitz functions, Theorem 2.15, we can write f ( x ) = f ( a ) + ∇ f ( x ) · ( x − a ) + Z (1 − s )( x − a ) T [Hess f ( a + s ( x − a ))]( x − a ) d s and estimate dist( v − u, Tan u A ) ≤ (cid:12)(cid:12)(cid:12) Z (1 − s )( x − y ) T [Hess f ( y + s ( x − y ))]( x − y ) d s (cid:12)(cid:12)(cid:12) ≤ k Hess f k L ∞ ( B ε (0)) | x − y | ≤ k Hess f k L ∞ ( B ε (0)) | v − u | . Step 2
Let u, v ∈ A with | u − v | > δ . Then dist( v − u, Tan u A ) ≤ diam( A ) < ∞ , so that dist( v − u, Tan u A ) ≤ diam( A ) ≤ diam( A ) δ | u − v | . Step 3
All in all we have shown dist( v − u, Tan u A ) ≤ max n diam( A ) δ , k Hess f k k L ∞ ( B ε (0)) | k = 1 , . . . , N o | u − v | , for all u, v ∈ A . Now the proposition follows with [Fed59, 4.18 Theorem]. Theorem 2.14 (Taylor’s theorem for Sobolev functions).
Let I ⊂ R be a bounded open interval, k ∈ N . Then for all f ∈ W k, ( I ) and x, a ∈ I holds f ( x ) = k − X i =0 f ( i ) ( a ) i ! ( x − a ) i + Z xa f ( k ) ( t )( k − x − t ) k − d t. Proof.
We can follow the usual proof by induction using the fundamental theorem of calculusand integration by parts. This is possible, because the product rule, and therefore integration byparts, also holds for absolutely continuous, and hence W , , functions, see [Hei07, formula (3.4),p.167]. 15 heorem 2.15 (Taylor’s theorem for Lipschitz functions). Let Ω ⊂ R n be open, k ∈ N . Then for all f ∈ C k, (Ω) and x, a ∈ Ω with x + [0 , a − x ) ⊂ Ω holds f ( x ) = k X | α | =0 D α f ( a ) α ! ( x − a ) α + X | β | = k +1 k + 1 β ! Z (1 − t ) k D β f ( a + t ( x − a ))( x − a ) β d t. Proof.
We always have C k, ⊂ W k +1 , ∞ , so that we can use the standard proof that appliesTaylor’s Theorem in dimension one, Theorem 2.14, to g = f ◦ h for h : [0 , → Ω with h ( t ) = a + t ( x − a ) . For this it is important that g ∈ W k, ([0 , , which is clear as f and h are both C k, , hence g ∈ W k +1 , ∞ ([0 , , and that [0 , is bounded. Proof of Theorem 1.2.
The equivalence of the last three items is Theorem 1.3, Lemma 2.4 andLemma 2.5 together with (8).
Step 1
Let V (( A s ) t ) = n X k =0 (cid:18) nk (cid:19) W k ( A s ) t k (14)for all s ∈ ( − r, r ) and − r < s + t < r . We compute V ( A s + t ) = n X k =0 (cid:18) nk (cid:19) W k ( A )( s + t ) k = n X k =0 (cid:18) nk (cid:19) W k ( A ) k X i =0 (cid:18) ki (cid:19) s k − i t i = n X k =0 k X i =0 (cid:18) nk (cid:19)(cid:18) ki (cid:19) W k ( A ) s k − i t i = n X i =0 n X k = i (cid:18) nk (cid:19)(cid:18) ki (cid:19) W k ( A ) s k − i t i = n X i =0 n X k = i (cid:18) ni (cid:19)(cid:18) n − ik − i (cid:19) W k ( A ) s k − i t i = n X i =0 (cid:18) ni (cid:19)(cid:16) n X k = i (cid:18) n − ik − i (cid:19) W k ( A ) s k − i (cid:17) t i . (15)By Lemma 2.4 holds V (( A s ) t ) = V ( A s + t ) for s, t > or s, t < with | s + t | < r , so thatcomparing (14) with (15) yields W i ( A s ) = n X k = i (cid:18) n − ik − i (cid:19) W k ( A ) s k − i . (16)According to Lemma 2.4 we either have A s + t ⊂ ( A s ) t or ( A s ) t ⊂ A s + t for s ∈ ( − r, r ) , − r
Let the last three items hold. Then according to the second item of Lemma 3.1 for B = A , s = t and (8) we have reach( A s ) ≥ reach( ∂A s ) ≥ r − | s | for s ∈ ( − r, r ) . Using Federer’s Steinerformula for sets of positive reach, see [Fed59, 5.6 Theorem], we obtain (14) for all s ∈ ( − r, r ) < t < r − | s | and, obviously, this also holds for t = 0 . In a first part we use this to prove(14) for s ∈ ( − r, r ) and s ≤ s + t < r . These results are then used in a second part to establish(14) for s ∈ ( − r, r ) and − r < s + t < r . Part 1
Making use of Federer’s Steiner formula we can do a computation similar to (15) for V (( A s + t ) u ) = V (( A s ) t + u ) , < t < r − | s | and < u < r − | s | − t , note t + u > , to obtain W i ( A s + t ) = n X k = i (cid:18) n − ik − i (cid:19) W k ( A s ) t k − i . (18)For s ∈ [0 , r ) we already have (14) for all ≤ t < r − s . Let s ∈ ( − r, . Choose u ∈ (0 , r − | s | ) and v ∈ (0 , r − | s + u | ) . Now, again using Federer’s Steiner formula, we can compute V (( A s + u ) v ) and substitute (18), using the same tricks as in (15) and (17), to obtain V (( A s ) u + v ) = V (( A s + u ) v ) = n X k =0 (cid:18) nk (cid:19) W k ( A s + u ) v k = n X k =0 (cid:18) nk (cid:19) n X j = k (cid:18) n − kj − k (cid:19) W j ( A s ) u j − k v k = n X j =0 j X k =0 (cid:18) jk (cid:19)(cid:18) nj (cid:19) W j ( A s ) u j − k v k = n X j =0 (cid:18) nj (cid:19) W j ( A s )( u + v ) j . This means we have shown (14) for all s ∈ ( − r, and u + v = t ∈ (0 , r − | s | + r − | s + u | ) , where r − | s | + r − | s + u | = 2 r − u ≥ r − ( r + s ) = r + s if s + u > and r − | s | + r − | s + u | = 2( r + s ) + u ≥ r + s if s + u ≤ . Iteration yields (14) for all s ∈ ( − r, r ) and s ≤ s + t < r . Part 2
Let s ∈ ( − r, r ) and − r < s + t < r . We want to obtain (14) for this range of parameters.Choose < u with − r < s + t + u < r and < t + u . As in Part 1 we can use the Steinerformula, now with the extended range from Part 1, to compute V (( A s + t ) u ) = V (( A s ) t + u ) , whichyields (18) for s ∈ ( − r, r ) and − r < s + t < r . This time choose < u such that − r < s + t − u .Then by the Steiner formula from Part 1 holds V (( A s ) t ) = V (( A s + t − u ) u ) = n X k =0 (cid:18) nk (cid:19) n X i = k (cid:18) n − ki − k (cid:19) W i ( A s )( t − u ) i − k u k = n X i =0 i X k =0 (cid:18) ik (cid:19)(cid:18) ni (cid:19) W i ( A s )( t − u ) i − k u k = n X i =0 (cid:18) ni (cid:19) W i ( A s ) t i . Proof of Theorem 1.1.
Except for the differences explained below the proof is the same as forTheorem 1.2. For the very last part of the analog of Step 1 in the proof of Theorem 1.2 weassume reach( A ) < r and then obtain a contradiction to to (17) via Lemma 2.6 for s = σ + τ , t = − τ . For the analog of Step 2 it is enough to have reach( A ) > , because we do not have touse Lemma 3.1, as we can simply employ [Fed59, 4.9 Corollary] to obtain reach( A s ) ≥ r − s for s ∈ (0 , r ) . Then we can follow the other steps, skipping the middle part, to obtain the desiredresult. Note that this range could not be covered in the Part 1, because the range of u there is restricted to < u
The equivalence of the first three items is actually shown in [HCS10a,proof of Theorem 1.1] and the equivalence of the last two items is Theorem 1.2.
Step 1
Let K = ( K − r ) r and x ∈ B r ( ∂K ) . If x ∈ ext( ∂K ) we have a unique projection ξ ∂K ( x ) ,so let x ∈ int( ∂K ) . We know that K − r is convex and, as x ∈ ext( ∂ ( K − r )) ∪ ∂ ( K − r ) , we have aunique projection y = ξ ∂ ( K − r ) ( x ) . Let { z } = [0 , ∞ )( x − y ) ∩ ∂K . Then ξ ∂ ( K − r ) ( z ) = y by (7),as K − r is convex and hence reach( K − r ) = ∞ . Then B r ( y ) ⊂ K and | z − y | = r . This means z ∈ ˜ ξ ∂K ( y ) and consequently ξ ∂K ( x ) = z , see Lemma 2.1. Therefore reach( ∂K ) ≥ r . Step 2
Let reach( ∂K ) ≥ r . Then according to Theorem 1.2, we have a Steiner formula for every K s , s ∈ ( − r, r ) . This directly yields (16) and W ′ i ( s ) = ( n − i ) W i +1 ( s ) for the quermaßintegrals.Hence K ∈ R n − ( r ) . Before we start to prove (3) in Proposition 1.5 we should at least, very briefly, explain thenotation that is specific to gradient flows on metric spaces. For a curve x : I → X from aninterval I to a metric space X we define the metric derivative | ˙ x ( t ) | at a point t ∈ I by | ˙ x ( t ) | := lim t → t t ∈ I d ( x ( t ) , x ( t )) | t − t | if this limit exists. The slope |∇ F | ( x ) of map F : X → R at a point x ∈ X is set to be |∇ F | ( x ) := lim sup x → x ( F ( x ) − F ( x )) + d ( x , x ) , where ( a ) + := max { a, } for a ∈ R . A curve x : I → X in a metric space ( X, d ) is called absolutely continuous if there is a function f ∈ L ( I ) such that d ( x ( s ) , x ( t )) ≤ Z ts f ( y ) d y for all s, t ∈ I with s < t. Lemma 3.2 (Computation of the slope |∇ W i | ). For all K ∈ K , we have |∇ W i | ( K ) = ( n − i ) W i +1 ( K ) for i = 0 , . . . , n − . roof. Let t < reach( ∂K ) . According to [Gru07, Theorem 6.13 (iv), p.105] the quermaßintegrals W i are monotonic with regard to inclusion, i.e. for L ⊂ K we have W i ( L ) ≤ W i ( K ) . Hence theset in B t ( K ) ∩ K , with least W i is K − t . Here B t ( K ) is the closed ball about K with regard tothe Hausdorff metric. We compute sup L ∈ B t ( K ) ∩K , ( W i ( K ) − W i ( L )) + = W i ( K ) − W i ( K − t ) and consequently with the help of Theorem 1.4 |∇ W i | ( K ) = lim sup L → KL ∈K , ( W i ( K ) − W i ( L )) + d H ( K, L ) = lim sup t → W i ( K ) − W i ( K − t ) t = W ′ i (0) = ( n − i ) W i +1 (0) = ( n − i ) W i +1 ( K ) . Notice for d H ( K, L ) = t is ( W i ( K ) − W i ( L )) + ≤ W i ( K ) − W i ( K − t ) . Proof of Proposition 1.5.
We have d H ( x ( s ) , x ( t )) = ω n ( t − s ) for s < t , so that x is absolutelycontinuous. For u ∈ (0 , ω − n reach( ∂K )) holds | ˙ x ( u ) | = lim h → d H ( x ( u + h ) , x ( u )) | h | = ω n | h || h | = ω n . By Lemma 3.2 we already know |∇ W n − | ( C ) = W n ( C ) = ω n for all C ∈ K , and together with W n − ( K − t ) = W n − (( K − t ) t ) − W n ( K − t ) t = W n − ( K ) − ω n t, from the usual expansion (16) of W i with ( K − t ) t = K from the proof of Theorem 1.2, we haveproven (3).Clearly x ( t ) → x ( T ) for t → T and x ( T ) is a compact, convex set and hence either contained in alower dimensional affine subspace or it has non-empty interior. Assume that x ( T ) has non-emptyinterior and ∂x ( T ) has positive reach. Then, by Theorem 1.3, ∂x ( T ) is of class C , and we musthave ν ∂K ( a ) = ν ∂K − ωnT ( a − ω n T ν ∂K ( a )) for all a ∈ ∂K . Thus, we obtain a contradiction to ω n T = reach( ∂K ) in the representation of Lemma 2.3, because there must be an ε neighbourhoodof ∂x ( T ) , where the normals cannot intersect, as reach( ∂x ( T )) > . A Quermaßintegrals as mean curvature integrals
Lemma A.1 (Quermaßintegrals as mean curvature integrals).
Let A ⊂ R n , ∂A a closed hypersurface with reach( ∂A ) > . Then W i ( A ) = n − Z ∂A H ( n − i − ( κ , . . . , κ n − ) d H n − , (19) where H ( k ) j is the j th elementary symmetric polynomial in k variables, i.e. H ( k ) j ( x , . . . , x k ) := (cid:18) kj (cid:19) − X ≤ l <... Note that the representation of quermaßintegrals of sets bounded by hypersurfaces of positivereach as mean curvature integrals, Lemma A.1, easily gives us a Gauß-Bonnet Theorem for thesesurfaces Z ∂A K G d σ = nW n ( A ) = nω n χ ( A ) , where K G is the Gauß curvature. Note that here the dimension n does not have to be odd, asin the generalized Gauß-Bonnet Theorem. References [ACV08] Luigi Ambrosio, Andrea Colesanti, and Elena Villa, Outer Minkowski content forsome classes of closed sets , Math. Ann. (2008), no. 4, 727–748.[AGS05] Luigi Ambrosio, Nicola Gigli, and Giuseppe Savaré, Gradient flows in metricspaces and in the space of probability measures , Lectures in Mathematics ETHZürich, Birkhäuser Verlag, Basel, 2005.[CH70] Eugenio Calabi and Philip Hartman, On the smoothness of isometries , Duke Math.J. (1970), 741–750.[CKS02] Jason Cantarella, Robert B. Kusner, and John M. Sullivan, On the minimumropelength of knots and links , Invent. Math. (2002), no. 2, 257–286.[Fed59] Herbert Federer, Curvature measures , Trans. Amer. Math. Soc. (1959), 418–491.[Fed69] , Geometric measure theory , Die Grundlehren der mathematischen Wis-senschaften, Band 153, Springer-Verlag New York Inc., New York, 1969.[Fu89] Joseph H. G. Fu, Curvature measures and generalized Morse theory , J. DifferentialGeom. (1989), no. 3, 619–642.[GMSvdM02] Oscar Gonzalez, John H. Maddocks, Friedemann Schuricht, and Heiko von derMosel, Global curvature and self-contact of nonlinearly elastic curves and rods ,Calc. Var. Partial Differential Equations (2002), no. 1, 29–68.21Gru07] Peter M. Gruber, Convex and discrete geometry , Grundlehren der Mathematis-chen Wissenschaften [Fundamental Principles of Mathematical Sciences], vol. 336,Springer, Berlin, 2007.[Had55] H. Hadwiger, Altes und Neues über konvexe Körper , Birkhäuser Verlag, Basel undStuttgart, 1955.[HCS10a] María A. Hernández Cifre and Eugenia Saorín, On differentiability of quermass-integrals , Forum Math. (2010), no. 1, 115–126.[HCS10b] , On inner parallel bodies and quermassintegrals , Israel J. Math. (2010),29–47.[HCS10c] , On the volume of inner parallel bodies , Adv. Geom. (2010), no. 2,275–286.[HCS11] , Differentiability of quermassintegrals: A classification of convex bodies ,Preprint, to appear in Trans. Amer. Math. Soc., 2011.[Hei07] Juha Heinonen, Nonsmooth calculus , Bull. Amer. Math. Soc. (N.S.) (2007),no. 2, 163–232.[HG10] Ralph Howard and Mohammad Ghomi, Tangent cones and regularity of real hy-persurfaces , Preprint, 2010.[KR12] Şahin Koçak and Andrei V. Ratiu, Inner tube formulas for polytopes , Proc. Amer.Math. Soc. (2012), no. 3, 999–1010.[LTR05] Davide La Torre and Matteo Rocca, C , functions and optimality conditions , J.Concr. Appl. Math. (2005), no. 1, 41–54.[Luc57] Kenneth R. Lucas, Submanifolds of dimension n − in E n with normals satisfying alipschitz condition , Technical Report 18, University of Kansas, May 1957, ContractNonr 58304.[Lyt04] Alexander Lytchak, On the geometry of subsets of positive reach , ManuscriptaMath. (2004), no. 2, 199–205.[Lyt05] , Almost convex subsets , Geom. Dedicata (2005), 201–218.[RW98] R. Tyrrell Rockafellar and Roger J.-B. Wets, Variational analysis , Grundlehren derMathematischen Wissenschaften [Fundamental Principles of Mathematical Sci-ences], vol. 317, Springer-Verlag, Berlin, 1998.[SvdM03] Friedemann Schuricht and Heiko von der Mosel, Global curvature for rectifiableloops , Math. Z. (2003), no. 1, 37–77.[SvdM06] Paweł Strzelecki and Heiko von der Mosel, Global curvature for surfaces and areaminimization under a thickness constraint , Calc. Var. Partial Differential Equa-tions25