A Constructive Examination of Rectifiability

We present a Brouwerian example showing that the classical statement ‘Every Lipschitz mapping f : [0, 1] → [0, 1] has rectifiable graph’ is essentially nonconstructive. We turn this Brouwerian example into an explicit recursive example of a Lipschitz function on [0, 1] that is not rectifiable. Then we deal with the connections, if any, between the properties of rectifiability and having a variation. We show that the former property implies the latter, but the statement ‘Every continuous, real-valued function on [0, 1] that has a variation is rectifiable’ is essentially nonconstructive. 2010 Mathematics Subject Classification 03F60, 26A16, 26A99 (primary)


Introduction
Consider a real-valued function f on a closed interval [a, b].
then the corresponding polygonal approximation to f has length We say that f • has bounded length if there exists c > 0 such that l f ,P c for each partition P of [a, b]; • is rectifiable if its length, sup {l f ,P : P is a partition of [a, b]} , exists; Every Lipschitz function has bounded length: for with f , κ, and P as above, The classical least-upper-bound principle ensures that if f has bounded length, then it is rectifiable.But in the constructive context of this paper, that principle implies the law of excluded middle (LEM) and so is inadmissible.The constructive least-upper-bound principle requires the additional hypothesis that the set S ⊂ R whose supremum is sought must be not only inhabited and bounded above, but also upper-order-located, in the sense that whenever α < β , either x β for all x ∈ S or else there exists (we can find) x ∈ S with α < x (see Bishop and Bridges [4, Page 37] or Bridges and Vît ¸ȃ [9,Theorem 2.1.18]).
With the aid of Specker's theorem [17] it is not hard to produce a recursive example of a pointwise, but not uniformly continuous function f : [0, 1] → R that has bounded length but is not rectifiable.The motivation for this paper lies in the question: Is every Lipschitz function f : [0, 1] → R constructively rectifiable?Our first main result (Proposition 2) gives a Brouwerian example showing that the rectifiability of all real-valued Lipschitz functions-and hence of all real-valued, uniformly continuous ones-on [0, 1] implies the essentially nonconstructive limited principle of omniscience: LPO: For each binary sequence (a n ) n 1 , either a n = 0 for all n or else there exists n such that a n = 1.This leads to our second main result (Theorem 7), providing an explicit example of a recursive Lipschitz function that has bounded length but is not rectifiable.The proof of the latter depends on a lemma of interest in its own right (Lemma 5), which enables us to pass from rectifiability over the whole interval [0, 1] to rectifiability over each of its compact subintervals.
In the second part of the paper we consider the possibility of connecting rectifiability with the property of having a variation (which, classically, reduces to that of bounded variation).In particular, we show that rectifiable continuous functions on [0, 1] have a variation, but the converse implies LPO (Corollary 9).
The constructive framework, BISH, of our work is that of Bishop [3,4,9] (see also Troelstra and van Dalen [18]), in which the logic is intuitionistic and we adopt a mathematical foundation such as the set theories CZF (Aczel and Rathjen [1,2]) and CMST (Bridges and Alps [6]), or Martin-Löf's type theory (Martin-Löf [14,15], Nordström, Peterson and Smith [16]).One model (in a purely informal sense) or interpretation of BISH is the recursive one, RUSS, which can be regarded as BISH plus the Church-Markov-Turing thesis and, if desired, Markov's principle of unbounded search (see Kushner [12], Markov [13] or Bridges and Richman [8,Chapter 3]); that model is the setting for Theorem 7.

Lipschitz curves need not be rectifiable
We begin our technical presentation with a lemma.
Lemma 1 Let (a n ) n 1 be an increasing binary sequence with a 1 = 0, and let b > 0. Then there exists a Lipschitz function f : (ii) f has Lipschitz constant 2; (iii) if f is rectifiable, then either a n = 0 for all n or there exists n with a n = 1; and (iv) if f is differentiable at any point of [0, b], then either a n = 0 for all n or there exists n with a n = 1. , Note that if a n = 0 for all n, then f n = 0; whereas if a n = 1 − a n−1 , then the length of the spiked path f n joining 0 to b is (the last inequality following because n 2).Note also that if a n = 1 − a n−1 , then the absolute value of the slope of the spikes of which is less than 2 (and, incidentally, increases to the limit 2 as n → ∞).Hence f n is Lipschitz, with Lipschitz constant 2. Since either f (x) > 0 or f (x) < 2; in the former case, there exists exactly one n such that f (x) = f n (x), so ||f || 2. It then follows that f is Lipschitz, with Lipschitz constant 2, and (see above) that the length of the curve Now suppose that the curve y = f (x) is rectifiable, with length s.Either s > b or else s < 8b/5.In the first case we can find x ∈ [0, 1] with f (x) > 0, and hence n with a n = 1.In the second case we must have a n = 0 for all n.Thus (iii) holds.
To deal with (iv), suppose, initially, that f is differentiable at the point k2 −N b, where N ∈ N and 0 k 2 N .We may further suppose that a N = 0.If there exists n > N such that a n = 1 − a n−1 , then f = f n and k2 −N b = (k2 n−N )2 −n b is a point where two adjacent spikes of f n meet; whence f is not differentiable at k2 −N b, a contradiction.Thus a n = 0 for all n > N and therefore for all n.Now let x be any point of [0, b], and suppose that f (x) exists.Either |f (x)| > 0 or |f (x)| < 1.In the first case there exists h = 0 such that f (x + h) = f (x); so either f (x + h) = 0 or f (x) = 0. Taking, for example, the case where f (x + h) = 0, compute ν such that ν n=1 f n (x + h) = 0. Then there exists n ν such that f n (x + h) = 0 and therefore a n = 1.On the other hand, in the case where |f (x)| < 1, if there exists n with a n = 1 − a n−1 , then, in view of the foregoing observation, x cannot lie on any open segment of a side of any spike of f , and so must be one of the three vertices of a spike.This is absurd, since we have just proved that f is not differentiable at such a vertex.Hence in this case we must have a n = 0 for all n.This completes the proof of (iv).In this context, the following is worth noting.
Proof Let f : [0, 1] → R be sequentially continuous and have bounded length, and let (P n ) n 1 be an enumeration of the partitions of [0, 1] with rational endpoints.Given real numbers α, β with α < β , and using countable choice, construct a binary sequence (λ n ) n 1 such that if λ n = 0, then l f ,Pn < β , and if λ n = 1, then l f ,Pn > α.Applying LPO, we see that either λ n = 0 for all n or else there exists N such that λ N = 1.In the second case, l f l f ,P N > α.In the first case, suppose that there exists a partition P of [0, 1] such that l f ,P > β .Since f is sequentially continuous, we can find such a partition P with rational endpoints.Then P = P ν for some ν , so l f ,Pν > β and therefore λ ν = 1.This contradiction ensures that l f ,P β for all partitions P of [0, 1].Since f has bounded length, its rectifiability now follows from the constructive least-upper-bound principle.
In the next section we convert these Brouwerian examples into a full-blooded counterexample in the recursive setting.

A recursive counterexample
To produce the promised recursive counterexample, we develop a general lemma, whose proof is derived from that of the particular application to functions of bounded variation (Bridges [5,Theorem 3]).
By a pseudoquasimetric on a set X we mean a mapping d : X × X → R such that for all x, y, z in X , • d(x, y) 0 and d(x, x) = 0; A continuous pseudoquasimetric on a metric space (X, ρ) is a pseudoquasimetric that is uniformly continuous as a mapping from X × X , taken with the product metric induced by ρ, into R.
Suppose that s ≡ sup d,P : P is a partition of I exists.Then for each compact subinterval J of I , Since d is uniformly continuous, we can construct a partition P : 0 = x 0 < x 1 < • • • < x N = 1 consisting of distinct rational points of I such that d,P > s − ε.Since a and the x i are rational, there exists p such that x p < a x p+1 .Since d satisfies the triangle inequality, adding a to the partition P does not decrease the value of d,P ; we may therefore assume that a = x m , and likewise that b = x m+k , for some m and k.Letting we have either t > α or t < α + ε.In the latter case, suppose that there exists a partition We have therefore shown that either there exists a partition Q of J with d,Q > α or else d,Q < β for all partitions Q of J .Since α, β are arbitrary positive numbers with α < β , the constructive least-upper-bound principle now ensures that the desired supremum exists.Finally, the continuity of d enables us to remove the restriction that a and b be rational.
Proposition 6 If f : [0; 1] → R is uniformly continuous and rectifiable, then the restriction of f to any compact subinterval of [0, 1] is rectifiable.
Another application of Lemma 5 arises in connection with a uniformly continuous function f : [0, 1] → R of bounded variation.For each partition P : 0 and T f [0, 1] the respective suprema of these quantities as P ranges over all partitions of [0, 1], when the supremum exists; note that if T f [a, b] exists, then we say that f has a variation and that we see that if T + f [0, 1] exists, then so does T + f [a, b] whenever 0 a b 1.Similar properties obtain for T − f and T f .In the case of T f , we recover a special case of Theorem 3 of [5]: if f has a variation on [0, 1], then it has a variation on each compact subinterval of [0, 1].It is then straightforward to prove the additivity of the variation function: This brings us to our recursive example.
Proof Let φ 0 , φ 1 , . . .be an effective enumeration of the computable partial functions in N N , and for each n define an increasing binary sequence (a n,k ) k 1 such that if φ n (n) is computed in exactly K steps, then a n,k = 0 for all k < K and a n,k = 1 for all k K .For each positive integer n let -if g n is rectifiable, then either a n,k = 0 for each k or else there exists k such that a n,k = 1, -if g n is differentiable at any point of [0, 3 −n ], then either a n,k = 0 for each k or else there exists k such that a n,k = 1.

Now construct a uniformly continuous mapping
The function F ≡ ∞ n=0 f n is well defined and uniformly continuous on [0, 1], since ∞ n=0 ||f n || converges by comparison with ∞ n=0 3 −2n .Moreover, since the supports of the functions f n are pairwise apart, the restriction of F to J n is f n , and F has Lipschitz constant 2. Suppose that (the graph of) F is rectifiable.Then by Proposition 6, for each n the restriction of F to J n is rectifiable; whence g n is rectifiable, and therefore either a n,k = 0 for all k or else a n,k = 1 for some k.It follows that the set is recursive, which is known to be false.Hence F is not rectifiable over [0, 1].
Note that if the function F constructed in the proof of Theorem 7 is differentiable at any point of J n , then so is g n , and we can decide whether n ∈ K .It follows that for each n, F cannot be differentiable at any point of J n .In light of this observation, if, in the proof of Theorem 7, we replace J n by 1  2 n , 1 2 n + 2ε 3 n , then we obtain the following: (ii) J is a union of countably many disjoint closed intervals of total length ε; (iv) f is not differentiable at any point of J .

Rectifiability and finite variation
In this section we discuss the question: What, if any, is the connection between rectifiability and having a variation?First we dash any hope of proving that the latter property implies the former.
Proposition 9 For each binary sequence (a n ) n 1 there exists a function g : [0, 1] → R that has a variation but is rectifiable if and only if either a n = 0 for all n or else there exists n with a n = 1.
Proof Given a binary sequence (a n ) n 1 , let f : [0, 1] → R be the function constructed in Lemma 1 above with b = 1, and define g : [0, 1] → R by g(x) = f (x)/2 + x.Then g is increasing: Since g is increasing, it has variation g(1) − g(0) = 1.Now, if a n = 0 for all n, then g(x) = x and l g = √ 2. On the other hand, if a n = 1 for some n, then the slope of g on the intervals k2 and the slope of g on the intervals k Thus the arc length of g over each k2 −n , (k and since there are 2 n such intervals in question, .
Suppose that g is rectifiable.Then either l g < ( √ 34 + √ 10)/6 or else l g > √ 2. In the first case there can be no n with a n = 1, so a n = 0 for all n.In the second case, taking a partition P of [0, 1] such that l g,P > √ 2, we can find x ∈ P such that g( which is absurd.
Corollary 11 If the set of real-valued functions on [0, 1] that have a variation is closed under addition, then LPO is derivable.
Proof With f , g, and (a n ) n 1 as in the proof of Proposition 9, both g and the identity function id on [0, 1] have a variation, but if f = id + g has a variation, then either a n = 0 for all n or there exists n with a n = 1.
In contrast to Corollary 11, we have: The set of real valued, rectifiable functions on [0, 1] is closed under addition.
Proof This is a simple application of the triangle inequality.
Corollary 10 shows that we cannot prove constructively that every Lipschitz, let alone every continuous, function f : [0, 1] → R with a variation is rectifiable.Our final task is to show that, in contrast, we can prove that every rectifiable continuous mapping f : [0, 1] → R has a variation.This will require some preliminaries.
Lemma 13 Let a, b, c be nonnegative numbers.Then Proof First take the case a = 1.Define Then (we omit the details) f 0, which ensures that Next take the case a > 0. By the first case, Finally, if a 0, then for each ε > 0 we have a + ε > 0, so Letting ε → 0 completes the proof.
Proof Denoting by ρ the Euclidean distance function on R 2 , first note that With −1 = 0, an induction now shows that for k < m, k i=0 Note that if x ∈ [x j , x j+1 ] and y ∈ [x j+1 , x j+2 ], then whence g is monotone and so has variation g(1) − g(0).Moreover, for each j, and therefore the variation of g on [x j , x j+1 ] is |f (x j+1 ) − f (x j )|.Since the variation function is additive, it follows that It follows also from (2) that l f ,P = l g,P .Since adding points to a partition cannot decrease the approximations to the length of the curves of f or g, we now see that and hence, via Lemma 13, that We now weaken the assumption (i) by taking P as a strict partition 0 < ξ 1 < ξ 2 < • • • < ξ m = 1.For each i let Note that P i ⊂ P i , that It follows from the constructive least-upper-bound principle that the variation of f on [0, 1] exists.

Concluding remarks
In view of Lemma 1, we might ask for Brouwerian counterexamples to such statements as these: -Every real-valued function on [0, 1] whose derivative exists at each point is rectifiable.
-Every real-valued Lipschitz function on [0, 1] that is differentiable almost everywhere is rectifiable.
There can be no Brouwerian counterexamples for these two statements, since each of them is provable in INT.If f : [0; 1] → R is differentiable everywhere, then intuitionistically its derivative f is not just continuous, but uniformly continuous, so we can rectify the graph of f by the usual calculus formula for arc length.On the other hand, if f has Lipschitz constant c > 0 and is pointwise differentiable almost everywhere, then |f (x)| c at any point x where f is differentiable.A theorem of van Rootselaar (see Heyting [11, page 79] or Bridges and Demuth [7,Theorem 6]) shows that f is intuitionistically measurable, whence by Bishop and Bridges [4,Theorem (7.11), page 263] it is integrable.We can then show that the supremum of the set (1 + f 2 ) 1/2 dx.
It is possible that there is a recursive example of a real-valued Lipschitz function on [0, 1] that is differentiable almost everywhere but not rectifiable; but we do not know of one.

Lemma 1 Proposition 3
immediately provides us with two interesting Brouwerian counterexamples for Lipschitz curves: Proposition 2 The statement 'Every real-valued Lipschitz function on [0, 1] is rectifiable' implies LPO.The statement 'Every Lipschitz function on [0, 1] is differentiable at some point' implies LPO.

Lemma 5
Let I = [0, 1], and let d : I ×I → R be a continuous pseudoquasimetric on I .For each compact interval [a, b] ⊂ I and each partition P

v f ,P − v f ,P = m− 1 i=0( 1 i=0(
v f ,P i − v f ,Pi ),and thatl f ,P − l f ,P = m−l f ,P i − l f ,Pi ), For 0 i < m let ε i = v f ,P i − v f ,Pi 0.Note that since the variation function is additive, Pi = v f ,P − v f ,P > ε.