Continuous valuations on the space of Lipschitz functions on the sphere
Andrea Colesanti, Daniele Pagnini, Pedro Tradacete, Ignacio Villanueva
aa r X i v : . [ m a t h . M G ] M a y CONTINUOUS VALUATIONS ON THE SPACE OFLIPSCHITZ FUNCTIONS ON THE SPHERE
ANDREA COLESANTI, DANIELE PAGNINI, PEDRO TRADACETE,AND IGNACIO VILLANUEVA
Abstract.
We study real-valued valuations on the space of Lipschitzfunctions over the Euclidean unit sphere S n − . After introducing anappropriate notion of convergence, we show that continuous valuationsare bounded on sets which are bounded with respect to the Lipschitznorm. This fact, in combination with measure theoretical arguments,will yield an integral representation for continuous and rotation invariantvaluations on the space of Lipschitz functions over the 1-dimensionalsphere. Contents
1. Introduction and preliminaries 12. τ -continuous valuations on Lip( S n − ) 43. Boundedness on norm-bounded sets 74. Functions supported on outer parallel bands 105. The control measures µ λ,γ ’s 126. The representing measures ν g ’s on S K ( λ, γ ) 288. Integral representation and final remarks 31Acknowledgements 33References 331. Introduction and preliminaries
A (real-valued) valuation on a class F of sets is a function V : F −→ R such that V ( A ∪ B ) + V ( A ∩ B ) = V ( A ) + V ( B ) , for every A, B ∈ F with A ∪ B, A ∩ B ∈ F . Apparently, the first results con-cerning valuations arise in the context of convex polytopes and M. Dehn’ssolution of Hilbert’s third problem in 1901. Valuations defined on the class K n of convex bodies of R n have been particularly relevant in convex ge-ometry, one of the cornerstones being Hadwiger’s characterization theoremfor continuous and rigid motion invariant valuations (see [11], and also [1]for more recent developments). We refer to the monographs [14, 21] for anup-to-date account of this theory. Mathematics Subject Classification.
Key words and phrases.
Geometric valuation theory, Lipschitz functions, integralrepresentation.
In recent years, Geometric Valuation Theory has seen a considerablegrowth, partly motivated by the study of valuations defined in a functionspace setting: if X is a space of functions, then a (real-valued) valuation on X is a mapping V : X −→ R such that V ( f ∨ g ) + V ( f ∧ g ) = V ( f ) + V ( g ) , for every f, g ∈ X with f ∨ g, f ∧ g ∈ X , where ∨ and ∧ respectively denotethe pointwise maximum and pointwise minimum.In particular, several characterization results have been provided for valu-ations on different function spaces, including, for instance, spaces of convexor quasi-concave functions [4, 6, 7, 20], Lebesgue L p spaces [12, 13, 24, 25,26], spaces of continuous functions [22, 23, 27] and Sobolev spaces [16, 17, 18](see also the survey paper [15] for more results of this type). Among otherthings, what these results are implying is that the connections between con-vex geometry and functional analysis are far from being exhausted. As amatter of fact, a structure theory of valuations on Banach lattices can bedeveloped (the last two named authors have initiated it in [24]).In this paper, our analysis will focus on continuous and rotation invariantvaluations defined on the space of Lipschitz functions over the Euclideansphere S n − .Let Lip( S n − ) be the space of real-valued Lipschitz continuous maps de-fined on S n − , i.e., the set of functions f : S n − −→ R for which the Lipschitzconstant L ( f ) = sup (cid:26) | f ( x ) − f ( y ) |k x − y k : x, y ∈ S n − , x = y (cid:27) is finite, where k·k denotes the Euclidean norm on R n . The space Lip( S n − )endowed with the pointwise ordering and the norm k f k Lip = max {k f k ∞ , L ( f ) } , is a complete normed lattice (cf. [28, Proposition 1.6.2]) satisfying L ( f ∧ g ) , L ( f ∨ g ) ≤ max { L ( f ) , L ( g ) } . However, it follows immediately from the definition that the space Lip( S n − )is not a Banach lattice, as the norm is not monotone on the positive cone,so the results given in [24] for valuations on Banach lattices are not directlyavailable in this context.Recall that every valuation on a lattice of real-valued functions, and inparticular every valuation on Lip( S n − ), satisfies the Inclusion-ExclusionPrinciple, which can be proved by induction. Proposition 1.1 (Inclusion-Exclusion Principle) . Let V : Lip( S n − ) −→ R be a valuation. Then V N _ j =1 f j = X ≤ j ≤ N V ( f j ) − X ≤ j Let V : Lip( S ) −→ R be a τ -continuous and rotation in-variant valuation. Then there exists K : R × R + −→ R such that K ( · , γ ) isa Borel function for every γ ∈ R + and, for every f ∈ L ( S ) , V ( f ) = Z π K ( f ( t ) , | f ′ ( t ) | ) dH ( t ) . A. COLESANTI, D. PAGNINI, P. TRADACETE, AND I. VILLANUEVA In particular, for every f ∈ Lip( S ) and { f i } ⊂ L ( S ) such that f i τ → f , V ( f ) = lim i →∞ Z π K ( f i ( t ) , | f ′ i ( t ) | ) dH ( t ) . Let us briefly sketch the path leading to this result: we will start, inSection 2, by introducing τ -convergence for functions on Lip( S n − ) andjustifying why this is the proper notion of convergence for our purposes. InSection 3, it will be shown that τ -continuous valuations are bounded on setswhich are bounded with respect to the norm k·k Lip . This fact will prove tobe essential for several steps of our construction. In Section 4, we will seethat τ -continuous valuations are small on functions which are supported onouter parallel bands of small width (see Lemma 4.2). Section 5 is devotedto the construction, based on the previous results, of a family of controlmeasures µ λ,γ ’s on S n − . Loosely speaking, if a (Borel) set A ⊂ S n − issuch that µ λ,γ ( A ) is small, then small perturbations (with respect to k · k ∞ and in gradient) of a function f ∈ Lip( S n − ) on the set A will not alter muchthe valuation on f . The most technical part of this section is to show thatthe µ λ,γ ’s are subadditive on open sets (Lemma 5.2): this step is needed inorder to prove that they actually define Borel measures. Using the rotationinvariance of the valuation, it follows that the measures µ λ,γ ’s coincide withthe Hausdorff measure H n − up to a constant (depending on λ and γ ).Up to this point, everything works for S n − with any n ∈ N ; the remainingsteps are still quite technical and, so far, we have only been able to handlethem for the one-dimensional sphere S . In Section 6, we will first observethat τ -continuous, rotation invariant valuations which are zero on constantfunctions “do not see” flat regions of functions (this will be formalized inLemma 6.1). Using this fact, for every piecewise linear function g ∈ L ( S )we will be able to introduce a measure ν g on S associated to it. Severaltechnical arguments (see Lemmas 6.3 and 6.4) will show that the measures ν g ’s are absolutely continuous with respect to the Hausdorff measure H on S . Then, in Section 7, using a particular family of piecewise linearfunctions we will define a pseudo kernel K : R × R + −→ R ; by meansof Radon-Nikodym’s theorem, we will give an integral representation for τ -continuous and rotation invariant valuations on Lip( S ) in Section 8.2. τ -continuous valuations on Lip( S n − )It would seem natural a priori to study valuations V : Lip( S n − ) −→ R which are continuous with respect to the standard Lipschitz norm k·k Lip .However, we show next that they are too large a class and, in general, do notadmit an integral representation or a manageable form. After showing this,we present the stronger continuity that seems to be the correct requirementif one searches for the largest reasonably tractable class of valuations withan integral representation.We start by recalling two well-known facts from the classical theory ofBanach spaces, which can be found in [9].First, the dual of L ∞ ( S n − , H n − ) can be naturally identified with thespace of bounded finitely additive measures on the Borel sets of S n − whichvanish on sets of H n − -measure zero (cf. [9, IV.8.16]). ALUATIONS ON Lip( S n − ) 5 Second, one such measure µ is countably additive if and only if it ad-mits an integral representation with respect to H n − (cf. [9, III.10.2]).That is, if we call ˜ T µ the continuous functional associated to µ , then µ iscountably additive if and only if there exists a Radon-Nikodym derivative g µ ∈ L ( S n − , H n − ) such that, for every f ∈ L ∞ ( S n − , H n − ),˜ T µ ( f ) = Z S n − f g µ dH n − . So, consider now µ to be a bounded finitely additive measure on thedual of L ∞ ( S n − , H n − ) which is not countably additive, and call ˜ T µ itsassociated continuous linear functional. Recall that Rademacher’s theoremimplies that for every f ∈ Lip( S n − ), the gradient ∇ f ( x ) is defined for a.e. x ∈ S n − , and satisfies k∇ f ( x ) k ≤ L ( f ). Thus, ˜ T µ defines a linear functional T µ : Lip( S n − ) −→ R by T µ ( f ) = ˜ T µ ( ∇ f ) . The simple fact that f + g = f ∨ g + f ∧ g implies that linear functionalsare valuations, and the reasonings above show that the valuation T µ will notadmit a reasonable integral representation.Therefore, as we said above, we need to restrict ourselves to valuationswith stronger continuity properties. Let us introduce the following notionof convergence. Definition 2.1. For f, f k ∈ Lip( S n − ) , k ∈ N , we say that f k τ → f when (1) k f k − f k ∞ → k ; (2) there exists C > such that sup k k∇ f k ( x ) k ≤ C , for a.e. x ∈ S n − ; (3) ∇ f k ( x ) → k ∇ f ( x ) for a.e. x ∈ S n − . In this paper we will deal with valuations V : Lip( S n − ) −→ R which are τ -continuous , that is, satisfying that for every ( f k ) k ∈ N , f in Lip( S n − ) suchthat f k τ → f , it holds that V ( f k ) → V ( f ).As a first justification of our choice of continuity, consider µ to be abounded finitely additive measure vanishing on H n − -measure zero sets.Then it is easy to see that the linear functional T µ defined as above is τ -continuous if and only if µ is countably additive, which in turn, usingthe Radon-Nikodym Theorem, is equivalent to T µ admitting an integralrepresentation.Let us recall the definition of support function: for a convex body K ⊂ R n ,namely a compact and convex non-empty subset of R n , its support function is h K : S n − −→ R defined by h K ( x ) = max y ∈ K h x, y i , x ∈ S n − , where h· , ·i is the standard dot product in R n . Let H ( S n − ) denote the spaceof support functions. Clearly, as every h ∈ H ( S n − ) is convex, we have H ( S n − ) ⊂ Lip( S n − ). It was shown in [8, Lemma 2.7] that τ -continuousvaluations on H ( S n − ) correspond to valuations on the convex bodies of R n which are continuous with respect to the Hausdorff metric. A. COLESANTI, D. PAGNINI, P. TRADACETE, AND I. VILLANUEVA Although τ -convergence does not correspond to a metric topology, we candefine the following metric: for f, g ∈ Lip( S n − ) let(2) d τ ( f, g ) = k f − g k ∞ + Z S n − |∇ f ( x ) − ∇ g ( x ) | dH n − ( x ) . Also, given M > 0, letLip M ( S n − ) = { f ∈ Lip( S n − ) : k f k Lip ≤ M } = M B Lip( S n − ) . We have the following lemma. Lemma 2.2. A valuation V : Lip( S n − ) −→ R is τ -continuous if and onlyif for every M > , the restriction V | Lip M ( S n − ) is continuous with respectto the metric d τ .Proof. Suppose that V : Lip( S n − ) −→ R is τ -continuous and for some M, δ > 0, ( f k ) ⊂ Lip( S n − ) and f ∈ Lip( S n − ) are such that k f k k Lip , k f k Lip ≤ M , d τ ( f k , f ) −→ k →∞ | V ( f k ) − V ( f ) | ≥ δ for each k ∈ N . In particular, we have that • k f k − f k ∞ ≤ d τ ( f k , f ) −→ k →∞ • sup k k∇ f k ( x ) k ≤ sup k L ( f k ) ≤ M for a.e. x ∈ S n − , • R S n − |∇ f k ( x ) − ∇ f ( x ) | dH n − ( x ) ≤ d τ ( f k , f ) −→ k →∞ 0, hence there isa subsequence such that ∇ f k j ( x ) −→ j →∞ ∇ f ( x ) for a.e. x ∈ S n − [5,Propositions 3.1.3, 3.1.5].Therefore, we have that f k j τ → f , which implies | V ( f k j ) − V ( f ) | −→ j →∞ V | Lip M ( S n − ) is continuous withrespect to the metric d τ .Conversely, suppose that for every M > V | Lip M ( S n − ) is continuouswith respect to the metric d τ , and for some δ > f k τ → f we have(4) | V ( f k ) − V ( f ) | ≥ δ, for every k ∈ N . Since f k τ → f , there is some C > k k∇ f k ( x ) k ≤ C, for a.e. x ∈ S n − . Therefore, we also have that L ( f k ) ≤ C , so f k , f ∈ Lip M ( S n − ) for some M > k ∈ N . Since ∇ f k ( x ) −→ k →∞ ∇ f ( x )for a.e. x ∈ S n − , there is a subsequence such that ∇ f k j −→ j →∞ ∇ f in H n − -measure [5, Proposition 3.1.2].This means that for every ε > 0, we have H n − ( { x ∈ S n − : |∇ f k j ( x ) − ∇ f ( x ) | ≥ ε/ } ) −→ j →∞ . ALUATIONS ON Lip( S n − ) 7 Therefore, given ε > 0, we can find j ∈ N such that for j ≥ j the set A j,ε = { x ∈ S n − : |∇ f k j ( x ) − ∇ f ( x ) | ≥ ε/ } satisfies H n − ( A j,ε ) < ε/ M . Thus, for j ≥ j we have Z S n − |∇ f k j ( x ) − ∇ f ( x ) | dH n − ( x ) = Z A j,ε |∇ f k j ( x ) − ∇ f ( x ) | dH n − ( x )+ Z S n − \ A j,ε |∇ f k j ( x ) − ∇ f ( x ) | dH n − ( x ) ≤ M H n − ( A j,ε ) + ε H n − ( S n − ) < ε. It follows that d τ ( f k j , f ) −→ j →∞ 0, so we must have V ( f k j ) −→ j →∞ V ( f ), incontradiction with (4). (cid:3) Boundedness on norm-bounded sets We do not know whether k·k Lip -continuous valuations on Lip( S n − ) arebounded on k·k Lip -bounded sets. We will see next that this is the case for τ -continuous valuations.Let us start with a technical lemma (cf. [10, Section 3.1.2, Corollary I]). Lemma 3.1. Let f : S n − −→ R be a Lipschitz function, c ∈ R and let Z c = { x ∈ S n − : f ( x ) = c } . Then ∇ f ( x ) = 0 for a.e. x ∈ Z c .Proof. Every Lipschitz function f : S n − −→ R can be extended to a func-tion ˆ f : R n −→ R which is still Lipschitz continuous with the same Lipschitzconstant (see [19]). It follows from the definition of spherical gradient that,for a.e. x ∈ S n − and v ∈ R n with h v, x i = 0,(5) h∇ f ( x ) , v i = h∇ e ˆ f ( x ) , v i , where ∇ e denotes the Euclidean gradient.For every c ∈ R , the function ¯ f = ˆ f − c : R n −→ R is still Lipschitzcontinuous. Set V c = { x ∈ R n : ¯ f ( x ) = 0 } and note that by [10, Section 3.1.2, Corollary I] we have ∇ e ¯ f ( x ) = 0, for all x ∈ V c outside of a set of Lebesgue measure zero. In particular, ∇ e ˆ f ( x ) = 0for these x ’s, and then ∇ f ( x ) = 0 for H n − -a.e. x ∈ Z c ⊂ V c , using (5). (cid:3) Lemma 3.2. Let V : Lip( S n − ) −→ R be a τ -continuous valuation. Then V is bounded on k·k Lip -bounded sets.Proof. We reason by contradiction. If the result is not true, then there exist L > f i ) i ∈ N ⊂ Lip( S n − ), with k f i k Lip ≤ L for every i ∈ N and such that | V ( f i ) | → + ∞ .Consider the function θ : R −→ R defined by θ ( c ) = V ( c ) , A. COLESANTI, D. PAGNINI, P. TRADACETE, AND I. VILLANUEVA for c ∈ R , where denotes the constant function equal to one. The continu-ity of V implies that θ is continuous. Clearly, θ is uniformly continuous on[ − L, L ] and thus bounded, that is, there exists C > c ∈ [ − L, L ], | V ( c ) | ≤ C. We define inductively two sequences ( a j ) j ∈ N , ( b j ) j ∈ N ⊂ R : first set a = − L , b = L , c = a + b . Note that V ( f i ∨ c ) + V ( f i ∧ c ) = V ( f i ) + V ( c ) . Since | V ( c ) | ≤ C and | V ( f i ) | → + ∞ , we know that there must exist aninfinite set M ⊂ N such that for i ∈ M either | V ( f i ∨ c ) | → + ∞ or | V ( f i ∧ c ) | → + ∞ as i increases to ∞ . In the first case, we set a = c , b = L and f i = f i ∨ c , while in the second case we set a = − L , b = c and f i = f i ∧ c . Note that, in either case, we have k f i k Lip ≤ L for every i ∈ M . Now we define c = a + b and proceed similarly.Inductively, we construct two sequences ( a j ) , ( b j ) ⊂ R , a decreasing se-quence of infinite subsets M j ⊂ N , and sequences ( f ji ) i ∈ M j ⊂ Lip( S n − )such that, for every j ∈ N , | a j − b j | = L j − , and for every i ∈ M j , t ∈ S n − , a j ≤ f ji ( t ) ≤ b j , and with the property that lim i →∞ | V ( f ji ) | = + ∞ . Passing to a further subsequence we may assume without loss of generalitythat lim i →∞ | V ( f ii ) | = + ∞ . Call λ = lim i a i and g i = f ii , for i ∈ N . The sequence ( g i ) i ∈ N ⊂ Lip( S n − )satisfies a i ≤ g i ≤ b i , k g i k Lip ≤ L , k g i − λ k ∞ → | V ( g i ) | → ∞ .We do now a second induction to obtain τ -convergence. To this end, wewill define a double indexed sequence ( m ji ) i,j ∈ N . In the first step, for every i ∈ N , we consider the number m i := m ( g i ) , where m ( g i ) is a median of f i ,defined as a number in [ − L, L ] which satisfies H n − ( { x ∈ S n − : g i ( x ) ≥ m ( g i ) } ) ≥ H n − ( S n − ) = 12 H n − ( { x ∈ S n − : g i ( x ) ≤ m ( g i ) } ) ≥ H n − ( S n − ) = 12 . Note that the continuity of g i and the fact that S n − is connected, implythat, in this step, the median is unique. However, at this point, we do notneed uniqueness in our proof. In the next steps, the median is defined onnot necessarily connected subsets, so uniqueness will not follow.The valuation property implies that V ( g i ∨ m i ) + V ( g i ∧ m i ) = V ( g i ) + V ( m i ) . ALUATIONS ON Lip( S n − ) 9 Since | V ( m i ) | ≤ C and | V ( g i ) | → + ∞ , we know that there must exist aninfinite set which with a little abuse of notation will still be called M ⊂ N ,such that for i ∈ M either | V ( g i ∨ m i ) | → + ∞ or | V ( g i ∧ m i ) | → + ∞ as i increases to ∞ .In the first case, we set g i = g i ∨ m i . In the second case, we set g i = g i ∧ m i . Note that in either case, we have k g i k Lip ≤ L for every i ∈ M .Lemma 3.1 implies that ∇ g i ( x ) = 0 for H n − -a. e. x ∈ ( g i ) − ( { m i } ).Since H n − (( g i ) − ( { m i } ) ≥ , we have that ∇ g i ( x ) = 0 for every x in aset of measure larger than or equal to .For every i ∈ M we consider the set A i = { x ∈ S n − : g i ( x ) = m i } . Clearly, H n − ( A i ) ≤ . Then, for every i ∈ M we consider the “median in A i ”, m i , as the number satisfying H n − ( { x ∈ A i : g i ( x ) ≥ m i } ) ≥ H n − ( A i ) H n − ( { x ∈ A i : g i ( x ) ≤ m i } ) ≥ H n − ( A i ) . Again, this median exists.We proceed as before and note that the valuation property implies V ( g i ∨ m i ) + V ( g i ∧ m i ) = V ( g i ) + V ( m i ) . Since | V ( m i ) | ≤ C and | V ( g i ) | → + ∞ , we know that there must exist aninfinite set M ⊂ M such that for i ∈ M either | V ( g i ∨ m i ) | → + ∞ or | V ( g i ∧ m i ) | → + ∞ as i increases to ∞ .In the first case, we set g i = g i ∨ m i . In the second case, we set g i = g i ∧ m i . Note that in either case, we have k g i k Lip ≤ L for every i ∈ M . Assume g i = g i ∨ m i (the other case is analogous).If m i > m i , H n − (( g i ) − ( { m i , m i } )) = H n − ( { g i = m i } ) + H n − ( { g i ≤ m i } )= H n − ( { g i ∨ m i = m i } ) + H n − ( A i ∩ { g i ≤ m i } ) ≥ H n − ( { g i ∨ m i = m i } ) + 12 H n − ( A i )= H n − (( A i ) c ) + 12 H n − ( A i )= 1 − H n − ( A i ) ≥ . If m i ≤ m i instead, H n − (( g i ) − ( { m i , m i } )) = H n − ( { g i ≤ m i } )= H n − (( A i ) c ) + H n − ( A i ∩ { g i ≤ m i } ) ≥ H n − (( A i ) c ) + 12 H n − ( A i )= 1 − H n − ( A i ) ≥ . In either case, we get H n − (( g i ) − ( { m i , m i } )) ≥ . It follows again from Lemma 3.1 that ∇ g i ( x ) = 0 for H n − -a. e. x ∈ ( g i ) − ( { m i , m i } ). Therefore, we have that ∇ g i ( x ) = 0 for every x in a setof measure larger than or equal to .We proceed inductively and we construct a decreasing sequence of infinitesubsets M j ⊂ N , and sequences ( g ji ) i ∈ M j ⊂ Lip( S n − ) such that, for every j ∈ N , for every i ∈ M j , k∇ g ji k ≤ L , a i ≤ g ji ≤ b i ,lim i →∞ | V ( g ji ) | = + ∞ and H n − (cid:16) { t ∈ S n − : ∇ g ji ( t ) = 0 } (cid:17) ≤ j . Passing to a further subsequence if needed, we may assume that the se-quence ( g ii ) i ∈ N satisfies lim i k g ii − λ k ∞ = 0, lim i | V ( g ii ) | = + ∞ , k∇ g ii k ≤ L and H n − (cid:0) { t ∈ S n − : ∇ g ii ( t ) = 0 } (cid:1) ≤ i , for every i ∈ N . Therefore, g ii τ → λ , but | V ( g ii ) | → + ∞ , a contradiction. (cid:3) Functions supported on outer parallel bands Throughout the paper, we will use the notation V λ ( f ) := V ( f + λ ) − V ( λ ) , for λ ∈ R and f ∈ Lip( S n − ), where λ := λ . Moreover, given a set A ⊂ S n − and f ∈ Lip( S n − ), we will write f ≺ A whenever supp( f ) = { x ∈ S n − : f ( x ) = 0 } satisfies supp( f ) ⊂ A . We also introduce the follow-ing definition. Definition 4.1. Given a set ∅ 6 = A ⊂ S n − and ω > , the outer parallelband around A is the set A ω = { t ∈ S n − : 0 < d ( t, A ) < ω } . For convenience, we set ∅ ω := ∅ . The next lemma allows us to control V on outer parallel bands of smallwidth. Lemma 4.2. Let V : Lip( S n − ) −→ R be a τ -continuous valuation. Let A, B ⊂ S n − be Borel sets and let λ ∈ R , γ ∈ R + .Then lim ω → + sup {| V λ ( f ) | : f ≺ A ω ∪ B ω , L ( f ) ≤ γ } = 0 . To prove this, we are going to need the following well-known result. Lemma 4.3. Let A ⊂ S n − be a Borel set. Then lim ω → + H n − ( A ω ) = 0 . ALUATIONS ON Lip( S n − ) 11 Proof. If this were not the case, we would have a number ε > ω i ց H n − ( A ω i ) > ε , for every i ∈ N .If x ∈ T i ∈ N A ω i , then 0 < d ( x, A ) < ω i , for every i ∈ N ; passing to the limit in the second inequality we have acontradiction.Therefore T i ∈ N A ω i = ∅ , hence0 = H n − \ i ∈ N A ω i ! = lim i →∞ H n − ( A ω i ) ≥ ε, which is false. (cid:3) Proof of Lemma 4.2. We reason by contradiction: if the limit is strictly pos-itive, there exist ε > ω i ց {| V λ ( f ) | : f ≺ A ω i ∪ B ω i , L ( f ) ≤ γ } ≥ ε, for every i ∈ N . By definition of supremum, for every i ∈ N there is aLipschitz function f i with f i ≺ A ω i ∪ B ω i and L ( f i ) ≤ γ such that(6) | V λ ( f i ) | > sup {| V λ ( f ) | : f ≺ A ω i ∪ B ω i , L ( f ) ≤ γ } − ε ≥ ε . Since K i = supp( f i ) is compact, for every i ∈ N we can write k f i k ∞ = | f i ( x i ) | , for some x i ∈ K i . Note that, for every i ∈ N , K i ⊂ A ω i ∪ B ω i ⊂ A ω ∪ B ω . By compactness, there exists { x i j } ⊂ { x i } such that x i j → x as j → ∞ , forsome x ∈ A ω ∪ B ω .We actually have x ∈ ∂A ∪ ∂B . Indeed, if x ∂A ∪ ∂B = \ i ∈ N A ω i ∪ \ i ∈ N B ω i = \ i ∈ N A ω i ∪ B ω i = \ i ∈ N A ω i ∪ B ω i , since x ∈ A ω ∪ B ω there must be a number I ∈ N such that x ∈ A ω I ∪ B ω I \ A ω I +1 ∪ B ω I +1 . Now, { x i } i ≥ I +1 ⊂ A ω I +1 ∪ B ω I +1 , which implies x = lim i →∞ x i ∈ A ω I +1 ∪ B ω I +1 , a contradiction. Therefore, x ∈ ∂A ∪ ∂B . Without loss of generality, assume x ∈ ∂A .Let us prove that there exists J ∈ N such that f i j ( x ) = 0 for every j > J .If this was not the case, there would be a sequence { f i jl } ⊂ { f i j } such that f i jl ( x ) = 0, for every l ∈ N . This would imply x ∈ A ω ijl ∪ B ω ijl , but since x ∈ ∂A we would actually have x ∈ B ω ijl , for every l ∈ N . In particular, d = d ( x, B ) > 0, and then there would exist h ∈ N such that ω i jh < d , hence x B ω ijh , a contradiction.By Lipschitz continuity, we get that for sufficiently large j k f i j k ∞ = | f i j ( x i j ) | = | f i j ( x i j ) − f i j ( x ) | ≤ γ k x i j − x k → . Moreover, k∇ f i j k ≤ γ a.e., and since H n − ( K i j ) ≤ H n − ( A ω ij ∪ B ω ij ) → H n − ( { f i j = 0 } ) → 1. From Lemma 3.1we conclude that ∇ f i j → S n − . Therefore, we have f i j τ → 0, whichis a contradiction with (6). (cid:3) The control measures µ λ,γ ’s We fix, for this and the following sections, a τ -continuous and rotationinvariant valuation V : Lip( S n − ) −→ R . We define its “flat component” V flat : Lip( S n − ) −→ R by setting V flat ( f ) = Z S n − V ( f ( t ) · ) dH n − ( t ) , for f ∈ Lip( S n − ).Since η : R −→ R defined by η ( λ ) = V ( λ ) is continuous, it follows that V flat is still a τ -continuous and rotation invariant valuation (see [27]). Thiscan also be verified directly: the valuation property and the rotational in-variance are quite straightforward, while τ -continuity follows from the Dom-inated Convergence Theorem. In a certain way, V flat can be considered asthe component of V that admits an extension to the space of continuousfunctions on the sphere, C ( S n − ). In this case, an integral representationhas been established in [23].Now we can define the “slope component” of the valuation as follows: V slope := V − V flat . Clearly, this satisfies V slope ( λ · ) = V ( λ · ) − V flat ( λ · ) = 0 , for every λ ∈ R . Since V slope is again a τ -continuous and rotation invariantvaluation, up to replacing V by V slope we can and will assume, for the purposeof this paper, that V is null on constant functions.Given λ ∈ R , γ ∈ R + , we can construct a “control measure”. We willseparately build its positive and negative part.We start by defining an outer measure. To this end, let us begin with thedefinition on open sets: for every open set G ⊂ S n − , we define(7) µ ∗ λ,γ ( G ) = lim l → + sup { V λ ( f ) : f ≺ G, k f k ∞ ≤ l, L ( f ) ≤ γ } . Note that the mapping l sup { V λ ( f ) : f ≺ G, k f k ∞ ≤ l, L ( f ) ≤ γ } is well-defined by Lemma 3.2, decreasing in l and lower bounded by V λ (0) =0. Therefore, the limit exists and µ ∗ λ,γ is well-defined.To prove that µ ∗ λ,γ is finitely subadditive on open sets, we will need avariant of McShane’s extension theorem (for the original statement, see [19]).Here and throughout the paper, we will be using the following notation: for f ∈ Lip( S n − ), let f + = f ∨ f − = f ∧ 0. Note in particular that f = f + + f − . ALUATIONS ON Lip( S n − ) 13 Lemma 5.1 (McShane) . Let A , A ⊂ B ⊂ S n − and L > . Let ˜ f : A ∪ A −→ R , g : B −→ R be Lipschitz functions with L ( ˜ f ) , L ( g ) ≤ L suchthat ˜ f | A = g | A , ˜ f | A = 0 . Then ˜ f can be extended to a Lipschitz function f : B −→ R with Lipschitz constant L ( f ) ≤ L such that g − ≤ f ≤ g + on B and k f k ∞ ≤ k g k ∞ .Proof. Consider the McShane extension of ˜ f , defined byˆ f ( t ) = sup s ∈ A ∪ A h ˜ f ( s ) − L k t − s k i , for t ∈ B . This function is Lipschitz continuous with L ( ˆ f ) ≤ L , andˆ f | A ∪ A = ˜ f .Let f = ( ˆ f ∨ g − ) ∧ g + ; then f is still Lipschitz continuous with L ( f ) ≤ L and f | A ∪ A = ˜ f . Moreover, f ≤ g + and f = ( ˆ f ∧ g + ) ∨ g − ≥ g − . Inparticular, it follows that k f k ∞ ≤ k g k ∞ . (cid:3) Despite the simplicity of the statement in the following result, its proofis probably the most technical part of the paper. Drawing a picture can behelpful for the interested reader. Lemma 5.2. For every λ ∈ R , γ ∈ R + and open sets G , G ⊂ S n − , wehave µ ∗ λ,γ ( G ∪ G ) ≤ µ ∗ λ,γ ( G ) + µ ∗ λ,γ ( G ) . Proof. Let G , G ⊂ S n − be open sets. In the following reasoning, the totalspace will be G ∪ G , so that for every set A , its complementary A c willdenote ( G ∪ G ) \ A , and A ω = { t ∈ G ∪ G : 0 < d ( t, A ) < ω } .Given ω > 0, consider the sets G ( ω ) = { t ∈ G : d ( t, G c ) ≥ ω } ,G ( ω ) = { t ∈ G : d ( t, G c ) ≥ ω } . A moment of thought reveals that for every ω > G ∪ G = G ( ω ) ∪ G ( ω ) ∪ (cid:2) G ( ω ) ω ∩ G ( ω ) ω (cid:3) . Fix now ε > 0. Lemma 4.2 implies the existence of an ω > n | V λ ( f ) | : f ≺ ( G c ) ω ∪ ( G c ) ω , L ( f ) ≤ γ o < ε. Given this ω , there exists 0 < l < ω γ such that(9) (cid:12)(cid:12) µ ∗ λ,γ ( G ∪ G ) − sup { V λ ( f ) : f ≺ G ∪ G , k f k ∞ ≤ l, L ( f ) ≤ γ } (cid:12)(cid:12) < ε , (10) sup { V λ ( f ) : f ≺ G , k f k ∞ ≤ l, L ( f ) ≤ γ } < µ ∗ λ,γ ( G ) + ε, (11) sup { V λ ( f ) : f ≺ G , k f k ∞ ≤ l, L ( f ) ≤ γ } < µ ∗ λ,γ ( G ) + ε, by definition of µ ∗ λ,γ . From (9), there also exists a Lipschitz function h ≺ G ∪ G with k h k ∞ ≤ l , L ( h ) ≤ γ , such that µ ∗ λ,γ ( G ∪ G ) < V λ ( h ) + ε. Let ˜ h i : G i ( ω ) ∪ G i (cid:0) ω (cid:1) c −→ R be defined by˜ h i ( t ) = ( h ( t ) t ∈ G i ( ω ) , t ∈ G i (cid:16) ω (cid:17) c , for i = 1 , 2. Note that ˜ h and ˜ h are Lipschitz continuous on their respectivedomains with Lipschitz constants L (˜ h ) , L (˜ h ) ≤ γ . Indeed, for i = 1 , 2, if t, s ∈ G i (cid:0) ω (cid:1) c then | ˜ h i ( t ) − ˜ h i ( s ) | = 0; if t, s ∈ G i ( ω ) then | ˜ h i ( t ) − ˜ h i ( s ) | ≤ γ k t − s k ;and if t ∈ G i ( ω ), s ∈ G i (cid:0) ω (cid:1) c then | ˜ h i ( t ) − ˜ h i ( s ) |k t − s k ≤ l k t − s k ≤ lω < γ. We can now use Lemma 5.1 to extend ˜ h i , i = 1 , 2, to a Lipschitz function h i : G ∪ G −→ R such that h − ≤ h i ≤ h + , L ( h i ) ≤ γ and k h i k ∞ ≤ l .Define ˜ h : (cid:2) G ( ω ) ω ∩ G ( ω ) ω (cid:3) ∪ G (cid:0) ω (cid:1) ∪ G (cid:0) ω (cid:1) −→ R ,˜ h ( t ) = h ( t ) t ∈ G ( ω ) ω ∩ G ( ω ) ω , t ∈ G (cid:18) ω (cid:19) ∪ G (cid:18) ω (cid:19) , and again use McShane’s extension theorem to extend this to h : G ∪ G −→ R with Lipschitz constant L ( h ) ≤ γ .Write h = h + + h − and note that h + = h +0 ∨ h +1 ∨ h +2 ,h − = h − ∧ h − ∧ h − . From the valuation property and the Inclusion-Exclusion Principle, wenow get V λ ( h ) = V λ ( h + ) + V λ ( h − ) = V λ ( h +0 ∨ h +1 ∨ h +2 ) + V λ ( h − ∧ h − ∧ h − )= V λ ( h +0 ) + V λ ( h +1 ) + V λ ( h +2 ) − V λ ( h +0 ∧ h +1 ) − V λ ( h +1 ∧ h +2 ) − V λ ( h +0 ∧ h +2 ) + V λ ( h +0 ∧ h +1 ∧ h +2 ) + V λ ( h − ) + V λ ( h − ) + V λ ( h − ) − V λ ( h − ∨ h − ) − V λ ( h − ∨ h − ) − V λ ( h − ∨ h − ) + V λ ( h − ∨ h − ∨ h − ) . Note that for every f ∈ { h +0 , h +0 ∧ h +1 , h +0 ∧ h +2 , h +0 ∧ h +1 ∧ h +2 , h − , h − ∨ h − , h − ∨ h − , h − ∨ h − ∨ h − } we have | V λ ( f ) | < ε . Indeed, since h ± = 0 on G (cid:0) ω (cid:1) ∪ G (cid:0) ω (cid:1) , it follows that f ≺ ( G c ) ω ∪ ( G c ) ω ; moreover, L ( f ) ≤ γ ,so from (8) we get | V λ ( f ) | < ε. Therefore,(12) V λ ( h ) < V λ ( h +1 ) + V λ ( h +2 ) − V λ ( h +1 ∧ h +2 )+ V λ ( h − ) + V λ ( h − ) − V λ ( h − ∨ h − ) + 8 ε = V λ ( h ) + V λ ( h ) − V λ ( h +1 ∧ h +2 ) − V λ ( h − ∨ h − ) + 8 ε. ALUATIONS ON Lip( S n − ) 15 Now, for i = 1 , 2, we have that h i ≺ G i , k h i k ∞ ≤ l and L ( h i ) ≤ γ , hence(10) and (11) imply V λ ( h i ) < µ ∗ λ,γ ( G i ) + ε , so that from (12) we get(13) V λ ( h ) < µ ∗ λ,γ ( G ) + µ ∗ λ,γ ( G ) − V λ ( h +1 ∧ h +2 ) − V λ ( h − ∨ h − ) + 10 ε. We claim that V λ ( h +1 ∧ h +2 ) , V λ ( h − ∨ h − ) > − ε. In order to see this, suppose that(14) V λ ( h +1 ∧ h +2 ) ≤ − ε, and define g : G ∪ G −→ R to be the extension given by Lemma 5.1 ofthe function g = ( h +1 on G ( ω ) c , G (cid:0) ω (cid:1) . Similarly, let g : G ∪ G −→ R be the extension of g = ( h +2 on G ( ω ) c , G (cid:0) ω (cid:1) . Note that, for i = 1 , g i ∨ ( h +1 ∧ h +2 ) = h + i . Indeed, for t ∈ G ( ω ) c we have g ( t ) ∨ [ h +1 ( t ) ∧ h +2 ( t )] = h +1 ( t ) ∨ [ h +1 ( t ) ∧ h +2 ( t )] = h +1 ( t ) , and for t ∈ G ( ω ) g ( t ) ∨ [ h +1 ( t ) ∧ h +2 ( t )] = g ( t ) ∨ [ h +1 ( t ) ∧ h + ( t )] = g ( t ) ∨ h +1 ( t ) = h +1 ( t ) . Analogously for i = 2.Let g : G ∪ G −→ R be the Lipschitz function defined by g = g ∨ g .From the valuation property and (15) we get V λ ( g ) = V λ ( g ) + V λ ( g ) − V λ ( g ∧ g ) = V λ ( h +1 ) + V λ ( g ∧ h +1 ∧ h +2 ) −− V λ ( h +1 ∧ h +2 ) + V λ ( h +2 ) + V λ ( g ∧ h +1 ∧ h +2 ) − V λ ( h +1 ∧ h +2 ) − V λ ( g ∧ g ) = V λ ( h +1 ∨ h +2 ) − V λ ( h +1 ∧ h +2 ) + V λ ( g ∧ h +1 ∧ h +2 )+ V λ ( g ∧ h +1 ∧ h +2 ) − V λ ( g ∧ g ) . Now, g ∧ h +1 ∧ h +2 ≺ ( G c ) ω ,g ∧ h +1 ∧ h +2 ≺ ( G c ) ω ,g ∧ g ≺ ( G c ) ω ∪ ( G c ) ω , so that (8) implies, for i = 1 , | V λ ( g i ∧ h +1 ∧ h +2 ) | < ε, | V λ ( g ∧ g ) | < ε. Moreover, V λ ( h +1 ∨ h +2 ) = V λ ( h + ) + V λ ( h +0 ∧ ( h +1 ∨ h +2 )) − V λ ( h +0 ) , where h +0 ∧ ( h +1 ∨ h +2 ) ≺ ( G c ) ω ∪ ( G c ) ω , hence | V λ ( h +0 ∧ ( h +1 ∨ h +2 )) | < ε. Putting everything together, from assumption (14) we obtain V λ ( g ) > V λ ( h + ) + 2 ε. The function ˜ g = g + h − satisfies ˜ g ≺ G ∪ G , k ˜ g k ∞ ≤ l , L (˜ g ) ≤ γ and V λ (˜ g ) = V λ (˜ g + ) + V λ (˜ g − ) = V λ ( g ) + V λ ( h − ) > V λ ( h + ) + V λ ( h − ) + 2 ε = V λ ( h ) + 2 ε> µ ∗ λ,γ ( G ∪ G ) + ε, a contradiction with (9).This proves that V λ ( h +1 ∧ h +2 ) > − ε, as claimed. A similar argument also shows that V λ ( h − ∨ h − ) > − ε. From (13) we conclude that µ ∗ λ,γ ( G ∪ G ) < µ ∗ λ,γ ( G ) + µ ∗ λ,γ ( G ) + 25 ε. Since ε > (cid:3) Now, for every A ⊂ S n − , we define µ ∗ λ,γ ( A ) = inf { µ ∗ λ,γ ( G ) : A ⊂ G, G an open set } . This extends the previous definition (7), i.e., the two definitions coincide onopen sets.The next lemma and proposition follow from standard reasonings. Lemma 5.3. For every λ ∈ R and γ ∈ R + , µ ∗ λ,γ is an outer measure on S n − .Proof. By definition, µ ∗ λ,γ is monotone increasing and satisfies µ ∗ λ,γ ( ∅ ) = 0.To check that it is indeed an outer measure we have to prove countablesubadditivity.Let ( A i ) i ∈ N be a sequence of subsets of S n − . By Lemma 3.2, µ ∗ λ,γ ( A i ) < ∞ for every i ∈ N . Let ǫ > 0. For every i ∈ N , choose an open set G ′ i suchthat A i ⊂ G ′ i and µ ∗ λ,γ ( A i ) > µ ∗ λ,γ ( G ′ i ) − ǫ i . Also, consider an open set G ′ ⊃ S i A i such that µ ∗ λ,γ [ i A i ! > µ ∗ λ,γ ( G ′ ) − ǫ. For every i ∈ N , define now G i = G ′ i ∩ G ′ , and G = S i G i . By monotonicityof µ ∗ λ,γ , these sets still satisfy µ ∗ λ,γ ( A i ) > µ ∗ λ,γ ( G i ) − ǫ i and µ ∗ λ,γ [ i A i ! > µ ∗ λ,γ ( G ) − ǫ. ALUATIONS ON Lip( S n − ) 17 For the chosen ǫ , Lemma 4.2 guarantees the existence of ω > f ≺ ( G c ) ω with L ( f ) ≤ γ , we have | V λ ( f ) | < ǫ . This implies µ ∗ λ,γ (( G c ) ω ) ≤ ǫ. Consider now the closed set G ( ω ) = { t ∈ G : d ( t, G c ) ≥ ω } . Note that G ( ω ) ⊂ G = S i G i , and as G ( ω ) is compact, there exist N ∈ N and i , . . . , i N ∈ N such that G ( ω ) ⊂ N [ j =1 G i j . Let G N = S Nj =1 G i j . Then, G = G N ∪ ( G c ) ω . Now, finite subadditivity on open sets implies that µ ∗ λ,γ ( G ) ≤ µ ∗ λ,γ ( G N ) + µ ∗ λ,γ (( G c ) ω ) ≤ N X j =1 µ ∗ λ,γ ( G i j ) + ǫ ≤ X i ∈ N µ ∗ λ,γ ( G i ) + ǫ. Since S i A i ⊂ G ′ ∩ S i G ′ i = G , it follows that µ ∗ λ,γ [ i ∈ N A i ! ≤ µ ∗ λ,γ ( G ) ≤ X i ∈ N µ ∗ λ,γ ( G i ) + ǫ ≤ X i ∈ N µ ∗ λ,γ ( A i ) + 2 ǫ. The fact that ǫ is arbitrary finishes the proof of countable subadditivity. (cid:3) Given an outer measure µ ∗ on S n − , we recall that a set B ⊂ S n − is µ ∗ -measurable if for every A ⊂ S n − , µ ∗ ( A ) = µ ∗ ( A ∩ B ) + µ ∗ ( A ∩ B c ) . It is well-known (cf. [5, Theorem 1.3.4]) that the set of µ ∗ -measurablesets is a σ -algebra. Moreover, µ ∗ restricted to this σ -algebra is a measure. Proposition 5.4. The Borel σ -algebra of S n − , Σ n , is µ ∗ λ,γ -measurable forevery λ ∈ R and γ ∈ R + . Therefore, if we define µ + λ,γ as the restriction of µ ∗ λ,γ to Σ n , then µ + λ,γ is a measure.Proof. It is clearly enough to show that every open set G ⊂ S n − is µ ∗ λ,γ -measurable. It follows from the subadditivity of µ ∗ λ,γ that it suffices to checkthat, for every A ⊂ S n − , µ ∗ λ,γ ( A ) ≥ µ ∗ λ,γ ( A ∩ G ) + µ ∗ λ,γ ( A ∩ G c ) . We fix A ⊂ S n − and ǫ > 0. From Lemma 3.2, we have that µ ∗ λ,γ ( A ) < ∞ .It follows from the definition of µ ∗ λ,γ that there exists an open set U ⊃ A such that µ ∗ λ,γ ( U ) ≤ µ ∗ λ,γ ( A ) + ǫ. As we mentioned before, for every open set U ⊂ S n − , the mapping l sup { V λ ( f ) : f ≺ U, k f k ∞ ≤ l, L ( f ) ≤ γ } is decreasing in l and lower bounded by 0. Let l be such thatsup { V λ ( f ) : f ≺ U, k f k ∞ ≤ l , L ( f ) ≤ γ } < µ ∗ λ,γ ( U ) + ǫ. Now, U ∩ G is an open set, hence we can choose f ≺ U ∩ G such that k f k ∞ ≤ l , L ( f ) ≤ γ and µ ∗ λ,γ ( U ∩ G ) ≤ V λ ( f ) + ǫ. We consider the compact set K = supp( f ) ⊂ U ∩ G . Clearly, we have U ∩ G c ⊂ U ∩ K c , the latter set being open. Choose now f with f ≺ U ∩ K c , k f k ∞ ≤ l , L ( f ) ≤ γ such that µ ∗ λ,γ ( U ∩ K c ) ≤ V λ ( f ) + ǫ. Note that since supp( f ) ⊂ K c = (supp( f )) c , we have that f and f have disjoint supports, both of them contained in U .Therefore, the function g = f + f satisfies g ≺ U , k g k ∞ ≤ l and L ( g ) ≤ γ .Moreover, f +1 ∧ f +2 = f − ∨ f − = 0and f +1 ∨ f +2 = g + ,f − ∧ f − = g − , hence V λ ( f ) + V λ ( f ) = V λ ( f +1 ) + V λ ( f − ) + V λ ( f +2 ) + V λ ( f − )= V λ ( f +1 ∨ f +2 ) + V λ ( f − ∧ f − )= V λ ( g + ) + V λ ( g − ) = V λ ( g ) . This implies µ ∗ λ,γ ( A ) ≥ µ ∗ λ,γ ( U ) − ǫ ≥ sup { V λ ( f ) : f ≺ U, k f k ∞ ≤ l , L ( f ) ≤ γ } − ǫ ≥ V λ ( g ) − ǫ = V λ ( f ) + V λ ( f ) − ǫ ≥ µ ∗ λ,γ ( U ∩ G ) + µ ∗ λ,γ ( U ∩ K c ) − ǫ ≥ µ ∗ λ,γ ( U ∩ G ) + µ ∗ λ,γ ( U ∩ G c ) − ǫ ≥ µ ∗ λ,γ ( A ∩ G ) + µ ∗ λ,γ ( A ∩ G c ) − ǫ. Since ε is arbitrary, the conclusion follows. (cid:3) We could now use an analogous argument to construct measures µ − λ,γ ’s,whose definition on open sets would be given by µ − λ,γ ( G ) = lim l → + sup {− V λ ( f ) : f ≺ G, k f k ∞ ≤ l, L ( f ) ≤ γ } = lim l → + − inf { V λ ( f ) : f ≺ G, k f k ∞ ≤ l, L ( f ) ≤ γ } . The measures defined by µ λ,γ := µ + λ,γ + µ − λ,γ “control” the absolute value of the valuation on open sets: for every openset G ⊂ S n − we have(16) lim l → + sup {| V λ ( f ) | : f ≺ G, k f k ∞ ≤ l, L ( f ) ≤ γ } ≤ µ λ,γ ( G ) . ALUATIONS ON Lip( S n − ) 19 Indeed, | V λ ( f ) | is either V λ ( f ) or − V λ ( f ), which are respectively controlledby µ + λ,γ ( G ) or µ − λ,γ ( G ). Observation 5.5. It follows directly from Lemma 3.2 that for every λ ∈ R , γ ∈ R + , µ λ,γ is a finite measure. It is also clear, from the rotational invari-ance of the valuation, that the measure µ λ,γ is rotation invariant. Therefore,for every λ ∈ R , γ ∈ R + , there exists a number θ ( λ, γ ) ∈ R + such that µ λ,γ = θ ( λ, γ ) H n − . Lemma 5.6. Let λ , γ ∈ R + . Then sup { θ ( λ, γ ) : | λ | ≤ λ , ≤ γ ≤ γ } < ∞ . Proof. If this is not the case, for every C ∈ R there exist λ C , γ C with | λ C | ≤ λ , 0 ≤ γ C ≤ γ , such that θ ( λ C , γ C ) > C. Remembering our normalization H n − ( S n − ) = 1, this implies µ λ C ,γ C ( S n − ) > C. By definition of µ + λ C ,γ C , µ − λ C ,γ C we have that for every ε > < l < (cid:12)(cid:12) µ + λ C ,γ C ( S n − ) − sup { V λ C ( f ) : k f k ∞ ≤ l, L ( f ) ≤ γ C } (cid:12)(cid:12) < ε , (cid:12)(cid:12) µ − λ C ,γ C ( S n − ) + inf { V λ C ( f ) : k f k ∞ ≤ l, L ( f ) ≤ γ C } (cid:12)(cid:12) < ε . From µ λ C ,γ C ’s definition and the triangular inequality we get (cid:12)(cid:12) µ λ C ,γ C ( S n − ) − sup V λ C ( f ) + inf V λ C ( f ) (cid:12)(cid:12) < ε . In particular, C < µ λ C ,γ C ( S n − ) < sup V λ C ( f ) − inf V λ C ( f ) + ε . Now, there exist f C , g C ∈ Lip( S n − ) such that k f C k ∞ , k g C k ∞ ≤ l and L ( f C ) , L ( g C ) ≤ γ C satisfyingsup V λ C ( f ) < V λ C ( f C ) + ε , inf V λ C ( f ) > V λ C ( g C ) − ε , which in turn implies(17) C < V λ C ( f C ) − V λ C ( g C ) + ε. The functions f C + λ C , g C + λ C ∈ Lip( S n − ) satisfy k f C + λ C k ∞ , k g C + λ C k ∞ ≤ l + | λ C | ≤ λ ,L ( f C + λ C ) = L ( f C ) ≤ γ C ≤ γ ,L ( g C + λ C ) = L ( g C ) ≤ γ C ≤ γ , so that k f C + λ C k Lip , k g C + λ C k Lip ≤ Λ := max { λ , γ } . From Lemma 3.2 and what we have just seen, there exists M > C such that | V λ C ( f C ) | , | V λ C ( g C ) | ≤ M. Inequality (17) then implies C < M + ε ;since C is arbitrary, this is a contradiction. (cid:3) The representing measures ν g ’s on S So far we have been able to construct a family of Borel measures on S n − which control the valuation on functions with bounded Lipschitz constantand uniform norm. In this section, we will define another measure which willallow us to provide an integral representation for the valuation on piecewiselinear functions. However, our techniques only work for the one-dimensionalsphere S . For convenience, throughout this and the forthcoming sections,we will identify functions on S with functions f : [0 , π ] −→ R such that f (0) = f (2 π ).Let us show first that, if a valuation is null on constant functions, thenit does not “see” the flat regions of any function. More precisely, given γ ∈ R + , 0 < d ≤ π , 0 ≤ ℓ ≤ π − d ) and t ∈ [0 , π ], consider the function h γ,d,ℓ,t : [0 , π ] −→ R given by h γ,d,ℓ,t ( t ) = γ ( t − t ) for t ∈ ( t , t + d ] ,γd for t ∈ ( t + d, t + d + ℓ ] ,γ ( t + 2 d + ℓ − t ) for t ∈ ( t + d + ℓ, t + 2 d + ℓ ] , , where the intervals of definition have to be considered modulo 2 π . h γ,d,ℓ,t t t + d t + d + ℓ t +2 d + ℓ π We then have the following result. Lemma 6.1. Suppose W : Lip( S ) −→ R is a τ -continuous, rotation in-variant valuation with W ( λ ) = 0 for every λ ∈ R . Given γ ∈ R + and < d ≤ π , for every ≤ ℓ ≤ π − d ) and every t ∈ [0 , π ] we have that W ( h γ,d,ℓ,t ) = W ( h γ,d, , ) . Proof. Since W is rotation invariant it is clear that W ( h γ,d,ℓ,t ) = W ( h γ,d,ℓ, ) , for every 0 ≤ ℓ ≤ π − d ) and t ∈ [0 , π ], so it is enough to prove that W ( h γ,d,ℓ, ) = W ( h γ,d, , ) , for every 0 ≤ ℓ ≤ π − d ). ALUATIONS ON Lip( S n − ) 21 Let us consider the set L = { ℓ ∈ [0 , π − d )] : W ( h γ,d,ℓ, ) = W ( h γ,d, , ) } . We claim that π/ m ∈ L for every m ∈ N = N ∪ { } . We prove this byinduction: let m = 0 and note that h γ,d,π, ∨ h γ,d,π,π = γd and h γ,d,π, ∧ h γ,d,π,π = h γ,d, , ∨ h γ,d, ,π . Since 0 < d ≤ π , h γ,d, , and h γ,d, ,π have disjoint support. Hence, usingrotation invariance, the valuation property and the fact that W ( λ ) = 0 forevery λ , it follows that2 W ( h γ,d,π, ) = W ( h γ,d,π, ) + W ( h γ,d,π,π )= W ( h γ,d,π, ∨ h γ,d,π,π ) + W ( h γ,d,π, ∧ h γ,d,π,π )= W ( γd ) + W ( h γ,d, , ∨ h γ,d, ,π )= 2 W ( h γ,d, , ) . Therefore, π ∈ L . Suppose now that π/ m ∈ L for some m ∈ N . Observethat h γ,d, π m +1 , ∨ h γ,d, π m +1 , π m +1 = h γ,d, π m , , and h γ,d, π m +1 , ∧ h γ,d, π m +1 , π m +1 = h γ,d, , π m +1 . Using rotation invariance, the valuation property and the induction hypoth-esis, we have that2 W (cid:16) h γ,d, π m +1 , (cid:17) = W (cid:16) h γ,d, π m +1 , (cid:17) + W (cid:16) h γ,d, π m +1 , π m +1 (cid:17) = W (cid:16) h γ,d, π m +1 , ∨ h γ,d, π m +1 , π m +1 (cid:17) + W (cid:16) h γ,d, π m +1 , ∧ h γ,d, π m +1 , π m +1 (cid:17) = W (cid:16) h γ,d, π m , (cid:17) + W (cid:16) h γ,d, , π m +1 (cid:17) = 2 W ( h γ,d, , ) . Thus, π/ m +1 ∈ L as desired.Now, we claim that if ℓ , ℓ ∈ L with ℓ + ℓ ≤ π − d ), then ℓ + ℓ ∈ L .Indeed, note that h γ,d,ℓ , ∨ h γ,d,ℓ ,ℓ = h γ,d,ℓ + ℓ , , and h γ,d,ℓ , ∧ h γ,d,ℓ ,ℓ = h γ,d, ,ℓ . Thus, as above, we obtain W ( h γ,d,ℓ , ) + W ( h γ,d,ℓ , ) = W ( h γ,d,ℓ + ℓ , ) + W ( h γ,d, , ) , and then it follows that ℓ + ℓ ∈ L whenever ℓ , ℓ ∈ L .This proves that L contains every number of the form kπ m , with k, m ∈ N ,belonging to the interval [0 , π − d )]. Finally, since these are dense in L , foran arbitrary ℓ ∈ [0 , π − d )] we can consider sequences of natural numbers( k i ) i ∈ N , ( m i ) i ∈ N such that (cid:12)(cid:12)(cid:12)(cid:12) ℓ − k i π m i (cid:12)(cid:12)(cid:12)(cid:12) −→ i →∞ . It is straightforward to check that h γ,d, kiπ mi , τ −→ h γ,d,ℓ, , as i → ∞ . Therefore, by the continuity of W we have that ℓ ∈ L . Thus, L = [0 , π − d )] and the proof is finished. (cid:3) Consider the algebra A defined by A = m [ j =1 I j : m ∈ N , I j = ( a j , b j ] ⊂ [0 , π ] , I j ∩ I k = ∅ for j = k . This coincides with the algebra generated by the semi-open intervals of [0 , π ].Recall that L ( S ) denotes the set of piecewise linear functions f on [0 , π ]such that f (0) = f (2 π ). Fix g ∈ L ( S ) which is symmetric with respect to x = π (that is, g ( t ) = g (2 π − t )). For every interval ( a, b ] ⊂ [0 , π ] let usdefine ν g (( a, b ]) = V ( g ab ) , where(18) g ab = g ( a ) in [0 , a ] ∪ (2 π − a, π ] ,g in ( a, b ] ∪ (2 π − b, π − a ] ,g ( b ) in ( b, π − b ] . We will sometimes use the symbol g a,b instead of g ab . Lemma 6.2. For every g ∈ L ( S ) which is symmetric with respect to x = π ,the function ν g : A −→ R given by ν g m [ j =1 I j := m X j =1 ν g ( I j ) , for every pairwise disjoint and semi-open intervals I , . . . , I m ⊂ [0 , π ] , iswell-defined and finitely additive.Proof. Clearly, to prove that ν g is well-defined, it is enough to show that forconsecutive intervals ( a, b ], ( b, c ] we have ν g (( a, b ]) + ν g (( b, c ]) = ν g (( a, c ]) . To prove this, note that g ab ∨ g bc = g ac ∨ g ( b ) and g ab ∧ g bc = g ac ∧ g ( b ) . Therefore, we have ν g (( a, b ]) + ν g (( b, c ]) = V ( g ab ) + V ( g bc ) = V ( g ab ∨ g bc ) + V ( g ab ∧ g bc )= V ( g ac ∨ g ( b )) + V ( g ac ∧ g ( b )) = V ( g ac ) + V ( g ( b ))= V ( g ac ) = ν g (( a, c ]) . Hence, ν g is well-defined on A . Moreover, it is finitely additive: indeed,if I = ( a, b ] and J = ( c, d ] with a < c < b < d , from what we have just ALUATIONS ON Lip( S n − ) 23 proved we get that ν g ( I ) + ν g ( J ) = ν g (( a, b ]) + ν g (( c, d ])= ν g (( a, c ]) + ν g (( c, b ]) + ν g (( c, b ]) + ν g (( b, d ])= ν g (( a, d ]) + ν g (( c, b ])= ν g ( I ∪ J ) + ν g ( I ∩ J ) . (cid:3) The next technical lemma will allow us to prove that ν g is absolutelycontinuous with respect to the Hausdorff measure H . Lemma 6.3. Let λ ∈ R , γ ∈ R + and let θ ( λ, γ ) be as in Observation 5.5.Take ε > and G ⊂ [0 , π ] an open interval. From (16) , there exists l G > such that, for every f ∈ Lip( S ) with f ≺ G , k f k ∞ ≤ l G and L ( f ) ≤ γ , | V λ ( f ) | ≤ ( θ ( λ, γ ) + ε ) H ( G ) . For every open interval G ′ ⊂ G such that H ( G ) = kH ( G ′ ) for some k ∈ N ,and every f ∈ Lip( S ) with f ≺ G ′ , k f k ∞ ≤ l G and L ( f ) ≤ γ , we have that | V λ ( f ) | ≤ ( θ ( λ, γ ) + ε ) H ( G ′ ) . Proof. We choose an open interval G ′ ⊂ G = ( a, b ) and reason by contradic-tion: suppose there exists a function f ∈ Lip( S ) with f ≺ G ′ , k f k ∞ ≤ l G and L ( f ) ≤ γ such that | V λ ( f ) | > ( θ ( λ, γ ) + ε ) H ( G ′ ). We can write | V λ ( f ) | = ( θ ( λ, γ ) + ε ) H ( G ′ ) + ρ, for a suitable ρ > G and G ′ coincide. Since H ( G ) = kH ( G ′ ), we can divide G into consecutiveintervals ( G i ) ki =1 of equal length, with G = G ′ . In particular, we can write G = k [ i =1 G i , and, for i ∈ { , . . . , k } , G i = ϕ − i ( G ′ ), where ϕ i is an appropriate rotation.Define the function g : [0 , π ] −→ R , g ( x ) = k X i =1 f ( ϕ i ( x )) , x ∈ S . Since f ◦ ϕ i ≺ G i for every i = 1 , . . . , k , the supports of the f ◦ ϕ i ’s arepairwise disjoint, hence g ∨ k _ i =1 ( f ◦ ϕ i ) ∨ ,g ∧ k ^ i =1 ( f ◦ ϕ i ) ∧ . Now, the f ◦ ϕ i ∨ V λ ( g ∨ 0) = k X i =1 V λ (( f ◦ ϕ i ) ∨ , from the Inclusion-Exclusion Principle. Analogously, V λ ( g ∧ 0) = k X i =1 V λ (( f ◦ ϕ i ) ∧ . Moreover, g ≺ G , k g k ∞ ≤ l G and L ( g ) ≤ γ .From the valuation property and rotational invariance we get | V λ ( g ) | = | V λ ( g ∨ 0) + V λ ( g ∧ | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 V λ (( f ◦ ϕ i ) ∨ 0) + k X i =1 V λ (( f ◦ ϕ i ) ∧ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X i =1 V λ ( f ◦ ϕ i ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = k | V λ ( f ) | = k ( θ ( λ, γ ) + ε ) H ( G ′ ) + kρ = ( θ ( λ, γ ) + ε ) H ( G ) + kρ, in contradiction with the hypothesis. (cid:3) Given g ∈ L ( S ), by Lemma 5.6 we can define the number C g = sup { θ ( λ, γ ) : | λ | ≤ k g k ∞ , ≤ γ ≤ L ( g ) } < ∞ . Lemma 6.4. Let g ∈ L ( S ) . For every pairwise disjoint semi-open intervals I , . . . , I m ⊂ [0 , π ] we have (19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ν g m [ j =1 I j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C g + 1) H m [ j =1 I j . Therefore, ν g is absolutely continuous with respect to the Hausdorff measure H on A , and in particular, it is bounded on A .Proof. We preliminarily prove the result for m = 1, i.e.,(20) | ν g ( I ) | ≤ C g + 1) H ( I ) , for I = ( a, b ] ⊂ [0 , π ].Define t max = sup (cid:8) t ∈ [ a, b ] : | ν g (( a, t ]) | ≤ C g + 1) H (( a, t ]) (cid:9) . Clearly t max ≥ a . To prove the case m = 1 we will show that t max = b and | ν g (( a, t max ]) | ≤ C g + 1) H (( a, t max ]) . We start by verifying that(21) | ν g (( a, t max ]) | ≤ C g + 1) H (( a, t max ]) . To prove this, note that g at τ → g at max , as t → t max . Indeed, if t ≤ t max wehavesup x ∈ [0 , π ] | g at ( x ) − g at max ( x ) | = sup x ∈ [ t,π ] | g at ( x ) − g at max ( x ) |≤ sup x ∈ [ t, t max ] | g ( t ) − g ( x ) | + sup x ∈ ( t max , π ] | g ( t ) − g ( t max ) |≤ L ( g ) | t − t max | . If t > t max the same bound is true, with very similar reasonings. Hence g at → g at max uniformly on [0 , π ], as t → t max . ALUATIONS ON Lip( S n − ) 25 Now, let I t = (min { t, t max } , max { t, t max } ] . Then g ′ at ( x ) = g ′ at max ( x ) for a.e. x ∈ [0 , π ] \ I t , and H ( I t ) → t → t max ,so that g ′ at → g ′ at max a.e. for t → t max . Finally, | g ′ at | ≤ L ( g ) a.e. in [0 , π ].Therefore, g at τ → g at max when t → t max , as claimed.So, letting t → t − max in | ν g (( a, t ]) | ≤ C g + 1) H (( a, t ]), by the continuityof V we get | ν g (( a, t max ]) | ≤ C g + 1) H (( a, t max ]) . This proves (21).To finish the proof in the case m = 1 we have to prove that t max = b . Wereason by contradiction.Suppose that t max < b . First note that, in this case, we would actuallyhave equality in (21). This follows from the fact that, for every b > t > t max ,we have | ν g (( a, t ]) | > C g + 1) H (( a, t ]) . Therefore, we use τ -convergence and the continuity of V as above and weobtain | ν g (( a, t max ]) | ≥ C g + 1) H (( a, t max ]) , hence the claimed equality.Consider the open interval G = ( t max , π − t max ) and let λ max = g ( t max ).Fix 0 < ε < 1. From the definition of µ λ max ,γ and Observation 5.5, thereexists l G > f ∈ Lip( S ) with f ≺ G , k f k ∞ ≤ l G , L ( f ) ≤ γ , we have | V λ max ( f ) | ≤ ( θ ( λ max , γ ) + ε ) H ( G ) . Since t max < b and g is piecewise linear, we can choose t ∈ (0 , π ) suchthat • t max < t < min { b, t max + π } , • g | [ t max , t ] is linear, • | g ( t ) − λ max | ≤ l G for every t ∈ [ t max , t ].We can also choose α > 0, as small as we wish, with t + α < b and suchthat if we set G ′ = ( t max , t + α − t max ) , then H ( G ) = kH ( G ′ ) , for some k ∈ N .Suppose g to be increasing in [ t max , t ]; if g | [ t max ,t ] were decreasing, wewould do similar reasonings adapting Lemma 6.1.Observe that if we set γ = g ′ ( t ) for any t ∈ ( t max , t ), d = t − t max and ℓ = 2( π − t ), then it holds that g t max , t − λ max = h γ,d,ℓ,t max . Hence, because of our choice of t and Lemma 6.1, we have that V λ max ( g t max , t − λ max ) = V λ max ( h γ,d, ,t max ) . On the other hand, it is easy to check that, up to a rotation, h γ,d, ,t max ≺ G ′ , k h γ,d, ,t max k ∞ ≤ l G and L ( h γ,d, ,t max ) ≤ γ ≤ γ . Thus, Lemma 6.3 yieldsthat | V λ max ( h γ,d, ,t max ) | ≤ ( θ ( λ max , γ ) + ε ) H ( G ′ ) . Therefore, we have (cid:12)(cid:12)(cid:12) ν g (( t max , t ]) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) V ( g t max ,t ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) V λ max ( g t max , t − λ max ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) V λ max ( h γ,d, ,t max ) (cid:12)(cid:12)(cid:12) ≤ ( θ ( λ max , γ ) + ε ) H (( t max , t + α − t max )) < C g + 1) H (cid:16)(cid:16) t max , t + α i(cid:17) . Letting now α → + we get that(22) (cid:12)(cid:12)(cid:12) ν g (( t max , t ]) (cid:12)(cid:12)(cid:12) ≤ C g + 1) H (( t max , t ]) . From the finite additivity of ν g , (21), (22) and the finite additivity of H ,we have | ν g (( a, t ]) | = | ν g (( a, t max ]) + ν g (( t max , t ]) |≤ C g + 1) H (( a, t max ]) + | ν g (( t max , t ]) |≤ C g + 1) H (( a, t max ]) + 2( C g + 1) H (( t max , t ])= 2( C g + 1) H (( a, t ]) . This is a contradiction with the definition of t max . Hence, we must have t max = b , and the proof of (20) is finished.For the general case, note that for every pairwise disjoint semi-open in-tervals I , . . . , I m we have, because of (20), (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ν g m [ j =1 I j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m X j =1 ν g ( I j ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C g +1) m X j =1 H ( I j ) = 2( C g +1) H m [ j =1 I j . (cid:3) Let us now consider the algebra A = m [ j =1 I j : m ∈ N , I j = ( a j , b j ] ⊂ [ π, π ] , I j ∩ I k = ∅ for j = k . For a symmetric g ∈ L ( S ), we can analogously define a finitely additivefunction ν g on A such that (19) holds for every pairwise disjoint semi-openintervals I , . . . , I m ⊂ [ π, π ]. Definition 6.5. For an arbitrary g ∈ L ( S ) , we can now define a function ν g on the algebra A = m [ j =1 I j : m ∈ N , I j = ( a j , b j ] ⊂ [0 , π ] , I j ∩ I k = ∅ for j = k by setting ν g m [ j =1 I j := 12 ν g m [ j =1 ( I j ∩ [0 , π ]) + ν g m [ j =1 ( I j ∩ [ π, π ]) , for every pairwise disjoint and semi-open intervals I , . . . , I m ⊂ [0 , π ] ,where g , g are the symmetric extensions to [0 , π ] of g | [0 ,π ] and g | [ π, π ] ,respectively. ALUATIONS ON Lip( S n − ) 27 Clearly, we have that ν g is finitely additive and satisfies(23) | ν g | ≤ C g + 1) H on A . The next lemma follows from standard arguments. Lemma 6.6. For every g ∈ L ( S ) , ν g can be extended to a signed measureon the Borel sigma-algebra Σ = σ ( { ( a, b ] : 0 ≤ a ≤ b ≤ π } ) , and ν g is absolutely continuous with respect to H on Σ .Proof. Fix g ∈ L ( S ). Inequality (23) implies the boundedness of ν g on A .From [3, Theorem 2.5.3, (1)-(9)], if we define ν + g ( A ) = sup { ν g ( B ) : B ⊂ A, B ∈ A} ,ν − g ( A ) = sup {− ν g ( B ) : B ⊂ A, B ∈ A} , for A ∈ A , then ν + g , ν − g are non-negative and bounded charges such that ν g = ν + g − ν − g (for the definition of charge, see [3, Definition 2.1.1]). Notethat ν ± g ( A ) ≤ C g + 1) H ( A ) for every A ∈ A , hence ν ± g ≪ H on A .Let us prove that ν + g and ν − g are countably additive on the algebra A .Let { A i } ⊂ A be pairwise disjoint sets such that A = S i ∈ N A i ∈ A . Let ε > 0. Then there exists δ > H ( B ) < δ implies ν ± g ( B ) < ε , forevery B ∈ A . Since X i ∈ N H ( A i ) = H ( A ) < ∞ , there is a number M ∈ N such that H ∞ [ i = m A i ! = ∞ X i = m H ( A i ) < δ for every m ≥ M . Now, ∞ [ i = m A i = [ i ∈ N A i ! \ m − [ i =1 A i ∈ A , hence ν ± g ∞ [ i = m A i ! < ε, for m ≥ M , that is, lim m →∞ ν ± g ∞ [ i = m A i ! = 0 . From the finite additivity of ν ± g on the algebra, we have ν ± g [ i ∈ N A i ! = m − X i =1 ν ± g ( A i ) + ν ± g ∞ [ i = m A i ! . Letting m → ∞ we conclude.Thus ν + g , ν − g are countably additive on the algebra A . Carath´eodory’sextension theorem [3, Theorem 3.5.2] implies that they can be extended to measures on σ ( A ), hence on Σ; this allows us to extend ν g to a signedmeasure on Σ, which still satisfies ν g = ν + g − ν − g .Let us now prove that ν g ≪ H on Σ. It is enough to show that ν ± g ≪ H .Fix ε > 0. From the absolute continuity of ν ± g on the algebra, we havethat there exists a δ > B ∈ A , if H ( B ) < δ then ν ± g ( B ) < ε/ A ∈ Σ such that H ( A ) < δ := δ/ 2. By regularity of theHausdorff measure, there exists an open set U ⊃ A such that H ( U \ A ) <δ/ 2. Then H ( U ) = H ( A ) + H ( U \ A ) < δ. We can write U = S j ∈ N I j , where the I j ’s are pairwise disjoint openintervals. Note that X j ∈ N ν ± g ( I j ) = ν ± g ( U ) < ∞ . Then there exists m ∈ N such that ∞ X j = m ν ± g ( I j ) < ε . Now, if I j = ( a j , b j ) for every j ∈ N , we have H m − [ j =1 ( a j , b j ] = H m − [ j =1 I j ≤ H ( U ) < δ, with m − [ j =1 ( a j , b j ] ∈ A . By monotonicity and additivity of ν ± g , and using thefact that ν ± g is null at singletons, we get ν ± g ( A ) ≤ ν ± g ( U ) = ν ± g m − [ j =1 ( a j , b j ] + ∞ X j = m ν ± g ( I j ) < ε. (cid:3) Definition of the pseudo kernel K ( λ, γ )Lemma 6.6 allows us to use the (signed version of) the Radon-NikodymTheorem [2, Theorem 2.2.1]: for every g ∈ L ( S ), there exists a function D g = dν g dH ∈ L ( S , H ) such that ν g ( A ) = Z A D g ( t ) dH ( t ) , for every A ∈ Σ.For every λ ∈ R , γ ∈ R + , define the function ψ λ,γ ( t ) = ( λ + γ ( t − π ) if t ∈ [0 , π ] ,λ + γ ( π − t ) if t ∈ [ π, π ] . ALUATIONS ON Lip( S n − ) 29 ψ λ,γ π πλ − γπ λ For fixed λ ∈ R , γ ∈ R + and m ∈ N , let us consider the “saw func-tion” S λ,γ,m obtained by joining m shrinked and translated copies of ψ λ,γ asfollows:(24) S λ,γ,m ( t ) = 1 m m X j =1 ψ λm,γ ( mt − j − π ) χ h j − πm , jπm (cid:17) ( t ) , for t ∈ [0 , π ], with the convention that points on the abscissae axis are tobe identified modulo 2 π , so that S λ,γ,m (2 π ) = S λ,γ,m (0). Here χ [ a,b ) denotesthe characteristic function of the interval [ a, b ). S λ,γ,mπm πm πλ − γπ m λ Note that, for every m ∈ N , | S ′ λ,γ,m ( t ) | = γ for a.e. t ∈ [0 , π ], and k S λ,γ,m k ∞ = 1 m max t ∈ [ , πm ] | ψ λm,γ ( mt ) | = 1 m max {| ψ λm,γ (0) | , | ψ λm,γ ( π ) |} ≤ | λ | + γπ , so that k S λ,γ,m k Lip ≤ | λ | + γπ . Thus, it follows from Lemma 3.2 that sup m ∈ N | V ( S λ,γ,m ) | < ∞ .We can then define(25) K ( λ, γ ) : = C lim sup m →∞ V ( S λ,γ,m ) , where C > H = 14 πC L , L being the Lebesgue measure on R . This function K has the followingconnection with our Radon-Nikodym derivative. Lemma 7.1. Let γ ∈ R + and ≤ a < b ≤ π . If g ∈ L ( S ) is such that | g ′ ( t ) | = γ for a.e. t ∈ [ a, b ] , then (26) K ( g ( t ) , | g ′ ( t ) | ) = D g ( t ) , for a.e. t ∈ [ a, b ] . Proof. Let g ∈ L ( S ) be as in the hypothesis. Consider ( c, d ) ⊆ [ a, b ] suchthat g ′ ( t ) = γ for a.e. t ∈ ( c, d ). By the Lebesgue-Besicovitch DifferentiationTheorem [10, Section 1.7.1], we have that for a.e. t ∈ ( c, d )(27) D g ( t ) = lim ε → H (( t − ε, t + ε ]) Z t + εt − ε D g ( s ) dH ( s ) . Take t ∈ ( c, d ), t = π , such that this holds, and set λ = g ( t ). Bythe Inclusion-Exclusion Principle, remembering that V is null on constantfunctions and using the rotational invariance, for every m ∈ N we get V ( S λ,γ,m ) = V m m X j =1 ψ λm,γ ( m · − j − π ) χ h j − πm , jπm (cid:17) = V m _ j =1 (cid:18) m ψ λm,γ ( m · − j − π ) χ h j − πm , jπm (cid:17) + (cid:16) λ − γπ m (cid:17) χ h , j − πm (cid:17) ∪ [ jπm , π ) (cid:19)(cid:19) = m X j =1 V (cid:18) m ψ λm,γ ( m · − j − π ) χ h j − πm , jπm (cid:17) + (cid:16) λ − γπ m (cid:17) χ h , j − πm (cid:17) ∪ [ jπm , π ) (cid:19) = mV (cid:18) m ψ λm,γ ( m · ) χ [ , πm ) + (cid:16) λ − γπ m (cid:17) χ [ πm , π ) (cid:19) = mV λ − γπ m (cid:16) h γm, πm , , (cid:17) . If t < π , applying Lemma 6.1 to V λ − γπ m , which is still a τ -continuous androtation invariant valuation which is null on constant functions, we havethat V ( S λ,γ,m ) = mV λ − γπ m (cid:16) h γm, πm , π − t − πm ,t − π m (cid:17) , if m is big enough so that t + π m < π . We proceed similarly in the case t > π .From the definition of ν g , we find V ( S λ,γ,m ) = mν g (cid:16)(cid:16) t − π m , t + π m i(cid:17) = 1 C H (cid:0)(cid:0) t − πm , t + πm (cid:3)(cid:1) Z t + πm t − πm D g ( s ) dH ( s ) , since H (cid:18)(cid:18) t − πm , t + 2 πm (cid:21)(cid:19) = 14 πC L (cid:18)(cid:18) t − πm , t + 2 πm (cid:21)(cid:19) = 14 πC · πm = 1 mC . Thus, taking the limit superior for m → ∞ and using (27) we obtain (26) fora.e. t ∈ ( c, d ). An analogous argument can be used when g ′ ( t ) = − γ . (cid:3) This allows us to prove the Borel measurability of K ( · , γ ), for every γ ∈ R + . ALUATIONS ON Lip( S n − ) 31 Remark 7.2. Fix γ ∈ R + . For every m ∈ Z , ψ γπm,γ (0) = γπ (2 m − and ψ γπm,γ ( π ) = γπ (2 m +1)2 . If λ ∈ h γπ (2 m − , γπ (2 m +1)2 (cid:17) , then there exists t ∈ [0 , π ] such that ψ γπm,γ ( t ) = λ . From Lemma 7.1, we can thus write K ( λ, γ ) = X m ∈ Z D ψ γπm,γ (cid:0) ψ − γπm,γ ( λ ) (cid:1) χ h γπ (2 m − , γπ (2 m +1)2 (cid:17) ( λ ) , for every λ ∈ R . As a consequence, we have that for every γ ∈ R + , K ( · , γ ) isa Borel function on R (and it is in fact integrable on every bounded interval). Integral representation and final remarks In this section we complete the proof of our main result. Theorem 1.2. Let V : Lip( S ) −→ R be a τ -continuous and rotation in-variant valuation. Then there exists K : R × R + −→ R such that K ( · , γ ) isa Borel function for every γ ∈ R + and, for every f ∈ L ( S ) , V ( f ) = Z π K ( f ( t ) , | f ′ ( t ) | ) dH ( t ) . In particular, for every f ∈ Lip( S ) and { f i } ⊂ L ( S ) such that f i τ → f , V ( f ) = lim i →∞ Z π K ( f i ( t ) , | f ′ i ( t ) | ) dH ( t ) . Proof. Let V : Lip( S ) −→ R be a τ -continuous and rotation invariantvaluation. Decomposing V into its flat and slope components and reasoningas in the beginning of Section 5, we can assume that V ( λ ) = 0 for every λ ∈ R . Define K : R × R + −→ R as in (25); by Remark 7.2, K ( · , γ ) is aBorel function for every fixed γ ∈ R + .Fix a piecewise linear function g ∈ L ( S ). Let ϕ : [0 , π ] −→ [0 , π ] bethe reflection with respect to π , that is, ϕ ( t ) = 2 π − t , t ∈ [0 , π ]. Usingrotation invariance and the valuation property, we have that V ( g ) = 12 (cid:16) V ( g ) + V ( g ◦ ϕ ) (cid:17) = 12 (cid:16) V ( g ∨ ( g ◦ ϕ )) + V ( g ∧ ( g ◦ ϕ )) (cid:17) . Note that both g ∨ ( g ◦ ϕ ) and g ∧ ( g ◦ ϕ ) are piecewise linear and symmetricwith respect to π . Thus, without loss of generality, we can assume g to besymmetric with respect to π .On the one hand, by definition of ν g (see Definition 6.5 and (18)) andsymmetry of g , we have that ν g ([0 , π ]) = 12 (cid:16) ν g ([0 , π ]) + ν g ([ π, π ]) (cid:17) = V ( g ,π ) = V ( g ) . On the other hand, we can take a partition ( t i ) mi =0 ⊂ [0 , π ] such that t = 0, t m = 2 π and the derivative satisfies | g ′ ( t ) | = γ i , for t ∈ ( t i − , t i ]and i = 1 , . . . , m . If D g denotes the Radon-Nikodym derivative of ν g with respect to the Hausdorff measure H , then by Lemma 7.1 we have ν g ([0 , π ]) = Z π D g ( t ) dH ( t ) = m X i =1 Z t i t i − D g ( t ) dH ( t )= m X i =1 Z t i t i − K ( g ( t ) , | g ′ ( t ) | ) dH ( t ) = Z π K ( g ( t ) , | g ′ ( t ) | ) dH ( t ) . Therefore, for a piecewise linear function g ∈ L ( S ) we have V ( g ) = Z π K ( g ( t ) , | g ′ ( t ) | ) dH ( t ) . In particular, for every f ∈ Lip( S ) and any sequence ( f i ) i ∈ N ⊂ L ( S )such that f i τ → f (which exist by the τ -density of L ( S ) in Lip( S )), by τ -continuity of V it follows that V ( f ) = lim i →∞ V ( f i ) = lim i →∞ Z π K ( f i ( t ) , | f ′ i ( t ) | ) dH ( t ) . (cid:3) The first integral formula given in Theorem 1.2 works on the dense set L ( S ). The possibility of extending this formula to the whole space Lip( S )is related to the properties of the pseudo kernel K : R × R + −→ R . Althoughit is conceivable for K to satisfy a strong Carath´eodory condition (as inthe integral representation for valuations on C ( S n − ) given in [23]), at themoment we have not been able to prove that K is a measurable function or,at least, that K ( f ( · ) , | f ′ ( · ) | ) is integrable, or measurable, on S for arbitrary f ∈ Lip( S ).Assuming stronger continuity conditions on the valuation, we can showthat the associated pseudo kernel does have good continuity properties. Forinstance, let us say that a valuation V : Lip( S ) −→ R is uniformly τ -continuous if for every ε > M > δ > | V ( f ) − V ( f ) | < ε, whenever k f k Lip , k f k Lip ≤ M and d τ ( f , f ) < δ (see (2) for the definitionof the metric d τ ). In this case, it is easy to see that the above argumentscan be streamlined and yield the following characterization result. Corollary 8.1. Let V : Lip( S ) −→ R . Then V is a rotation invariant anduniformly τ -continuous valuation if and only if there exists a continuousfunction K : R × R + −→ R such that V ( f ) = Z π K ( f ( t ) , | f ′ ( t ) | ) dH ( t ) , for every f ∈ Lip( S ) . Although Corollary 8.1 provides an integral representation on the wholespace Lip( S ), we should point out that the hypothesis of uniform continuityis very strong: continuous valuations are not necessarily uniformly continu-ous, as the following simple example shows. Consider the valuation defined ALUATIONS ON Lip( S n − ) 33 by V ( f ) = Z π | f ′ ( t ) | χ [1 , ∞ ) ( f ( t )) dH ( t ) , for f ∈ Lip( S ), where χ [1 , ∞ ) denotes the characteristic function of the set[1 , ∞ ). Clearly, this defines a rotation invariant valuation on Lip( S ), andusing the Dominated Convergence Theorem and Lemma 3.1, it is easy to seethat V is τ -continuous. However, if we consider, for m ∈ N , the functions f m = S π m , ,m and g m = S − π m , ,m (we are using the notation of (24)), then we have that for every m ∈ N , V ( f m ) = H ([0 , π ]) and V ( g m ) = 0, but d τ ( f m , g m ) → . Hence, V is not uniformly τ -continuous.Note that the associated pseudo kernel, for λ ∈ R and γ ∈ R + , is givenby K ( λ, γ ) = χ [1 , ∞ ) ( λ ) · γ = γ if λ ≥ , λ < . Clearly, the function K ( · , γ ) is not continuous at λ = 1, for any γ ∈ R + . Acknowledgements A. Colesanti is supported by the G.N.A.M.P.A. research project Prob-lemi di Analisi Geometrica Collegati alle Equazioni alle Derivate Parziali, alCalcolo delle Variazioni, agli Insiemi e alle Funzioni Convesse . D. Pagniniis supported by the G.N.A.M.P.A. group. P. Tradacete is supported byAgencia Estatal de Investigaci´on (AEI) and Fondo Europeo de DesarrolloRegional (FEDER) through grants MTM2016-76808-P (AEI/FEDER, UE)and MTM2016-75196-P (AEI/FEDER, UE), as well as Grupo UCM 910346.I. Villanueva is supported by MINECO (Spain) Grant MTM2017-88385-P,QUITEMAD+-CM (S2013/ICE-2801). P. Tradacete and I. Villanueva alsoacknowledge financial support from the Spanish Ministry of Science and In-novation, through the “Severo Ochoa Programme for Centres of Excellencein R&D” (SEV-2015-0554) and from the Spanish National Research Council,through the “Ayuda extraordinaria a Centros de Excelencia Severo Ochoa”(20205CEX001). References [1] S. Alesker, Continuous Rotation Invariant Valuations on Convex Sets . Ann. of Math.149 (1999), 977-1005.[2] R. B. Ash, Probability and Measure Theory (Second Edition) . Academic Press, SanDiego, 1999.[3] K. P. S. Bhaskara Rao, M. Bhaskara Rao, Theory of Charges. A Study of FinitelyAdditive Measures . With a foreword by D. M. Stone. Pure and Applied Mathematics,109. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York, 1983.[4] L. Cavallina, A. Colesanti, Monotone Valuations on the Space of Convex Functions .Anal. Geom. Metr. Spaces 3 (2015), 167-211.[5] D. L. Cohn, Measure Theory . Reprint of the 1980 original. Birkh¨auser Boston, Inc.,Boston, MA, 1993. [6] A. Colesanti, N. Lombardi, L. Parapatits, Translation Invariant Valuations on theSpace of Quasi-Concave Functions . Studia Math. 243 (2018), no. 1, 79-99.[7] A. Colesanti, M. Ludwig, F. Mussnig, Valuations on Convex Functions . Int. Math.Res. Not. IMRN 2019, no. 8, 2384-2410.[8] A. Colesanti, D. Pagnini, P. Tradacete, I. Villanueva, A Class of Invariant Valuationson Lip( S n − ). Adv. Math. 366, 37 pages, 2020.[9] N. Dunford, J. T. Schwartz. Linear Operators, part I, General Theory . IntersciencePublishers, 1958.[10] L. C. Evans, R. F. Gariepy, Measure Theory and Fine Properties of Functions . CRCPress, 1992.[11] H. Hadwiger, Vorlesungen ¨uber Inhalt, Oberfl¨ache und Isoperimetrie . Springer, Berlin,1957.[12] D. A. Klain, Star Valuations and Dual Mixed Volumes . Adv. Math. 121 (1996), no.1, 80-101.[13] D. A. 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Villanueva, Radial Continuous Valuations on Star Bodies . J. Math.Anal. Appl. 454 (2017), 995-1018.[23] P. Tradacete, I. Villanueva, Continuity and Representation of Valuations on StarBodies . Adv. Math. 329 (2018), 361-391.[24] P. Tradacete, I. Villanueva, Valuations on Banach Lattices . Int. Math. Res. Not.IMRN 2020, no. 1, 287-319.[25] A. Tsang, Valuations on L p -Spaces . Int. Math. Res. Not. IMRN, 20 (2010), 3993-4023.[26] A. Tsang, Minkowski Valuations on L p -Spaces . Trans. Amer. Math. Soc. 364 (2012),no. 12, 6159-6186.[27] I. Villanueva, Radial Continuous Rotation Invariant Valuations on Star Bodies . Adv.Math. 291 (2016), 961-981.[28] N. Weaver, Lipschitz Algebras . Second Edition. World Scientific Publishing Co. Pte.Ltd., Hackensack, NJ, 2018. ALUATIONS ON Lip( S n − ) 35 Dipartimento di Matematica e Informatica “U. Dini” Universit`a degli studidi Firenze, Viale Morgagni 67/A - 50134, Firenze, Italy E-mail address : [email protected] Dipartimento di Matematica e Informatica “U. Dini” Universit`a degli studidi Firenze, Viale Morgagni 67/A - 50134, Firenze, Italy E-mail address : [email protected] Instituto de Ciencias Matem´aticas (CSIC-UAM-UC3M-UCM), Consejo Su-perior de Investigaciones Cient´ıficas, C/ Nicol´as Cabrera, 13–15, Campus deCantoblanco UAM, 28049 Madrid, Spain. E-mail address : [email protected] Universidad Complutense de Madrid, Departamento de An´alisis Matem´aticoy Matem´atica Aplicada. Instituto de Matem´atica Interdisciplinar-IMI. Insti-tuto de Ciencias Matem´aticas ICMAT, Madrid, Spain. E-mail address ::