Reifenberg’s Topological Disc Theorem

Leon Simon^{∗}

Here B_{ρ} = {x ∈ R^{n} : |x| ≤ ρ} and B_{ρ}(y) = {x ∈ R^{n} : |x − y| ≤ ρ}. First we introduce Reifenberg’s ϵ-approximation property for subsets of R^{n}. Definition: If ϵ > 0 and if S is a closed subset of the ball B_{2}, we say that S, containing 0, has the m-dimensional ϵ-Reifenberg approximation property in B_{1} if for each y ∈ S ∩ B_{1} and for each ρ ∈ (0, 1], there is an m-dimensional subspace L_{y,ρ} such that d_{H}(S ∩ B_{ρ}(y), y + L_{y,ρ} ∩ B_{ρ}(y)) < ϵ.

Here d_{H}(A_{1}, A_{2}) is the Hausdorff distance between A_{1}, A_{2}; thus d_{H}(A_{1}, A_{2}) = inf{ϵ > 0 : A_{1} ⊂ B_{ϵ}(A_{2}) & A_{2} ⊂ B_{ϵ}(A_{1})}.

Now we can state the main theorem.

Theorem (Reifenberg’s disc theorem). There is a constant ϵ = ϵ(n) > 0 such that if S, containing 0, is a closed subset of the ball B_{2} which satisfies the above ϵ-Reifenberg approximation property in B_{1}, then B_{1} ∩ S is homeomorphic to the closed unit ball in R^{m}. In fact, there is a closed subset M ⊂ R^{n} such that M ∩ B_{1} = S ∩ B_{1} and such that is homeomorphic to a subspace T_{0} of R^{n} via a homeomorphism τ : T_{0} → M with |τ(x) − x| ≤ C(n)ϵ for each x ∈ T_{0}, and τ(x) = x for each x ∈ T_{0} \ B_{2}. For any given α ∈ (0, 1) we can additionally arrange that τ and τ^{−1} are Hölder continuous with exponent α provided S satisfies the ϵ-Reifenberg condition with suitable ϵ = ϵ(n, α).

We’ll need the following lemma in the proof of the above theorem.

Lemma 1 (Extension Lemma). Let ϵ, r > 0, let y_{1}, …, y_{Q} be a finite collection of points in R^{n} with |y_{i}−y_{k}| ≥ r for each i ≠ k, and assume that f : {y_{1}, …, y_{Q}} → R^{N} is given such that |f(y_{i}) − f(y_{k})| ≤ ϵ whenever |y_{i} − y_{k}| ≤ 6r. Then there is an extension \overline{f} : ∪_{i}B_{2r}(y_{i}) → R^{N} such that |∇\overline{f}| ≤ C(n)ϵr^{−1} and |\overline{f}(x)−f(y_{i})| ≤ C(n)ϵ for x ∈ B_{2r}(y_{i}), i = 1, …, Q. Furthermore there is ϵ = ϵ(n) > 0 such that if N = n^{2} (where R^{n^{2}} is identified with the set of n × n matrices in the usual way) and if each f(y_{i}) is the matrix of an orthogonal projection of R^{n} onto some m-dimensional subspace L_{i} ⊂ R^{n}, then we can choose the extension \overline{f} such that each \overline{f}(x) is the matrix of an orthogonal projection of R^{n} onto some m-dimensional subspace L_{x}.

∗Expository lecture at Universität Tübingen, May ’96; Research partially supported by NSF grant DMS-9504456 at Stanford University

Proof: The proof uses a partition of unity {ψ_{j}} for ∪_{i}B_{2r}(y_{i}) of special type. Indeed we claim that there is a partition of unity for ∪_{i}B_{2r}(y_{i}) with ψ_{i} ∈ C_{c}^{∞}(R^{n}), ψ_{i} ≡ 0 outside B_{3r}(y_{i}), ψ_{i}(y_{i}) = 1, and sup |∇ψ_{i}| ≤ C(n)r^{−1}. We see this as follows: first let ψ^{0} be a C^{∞}(R^{n}) function with ψ^{0}(x) ≡ 1 for |x| < \frac{1}{3}, 0 < ψ^{0}(x) < 1 for < \frac{1}{3}|x| ≤ \frac{5}{2}, and ψ^{0}(x) ≡ 0 for |x| ≥ \frac{5}{2}. For each i = 1, …, Q let

ψ_{i}^{0}(x) = ψ^{0}(\frac{x−y_{i}}{r}), ψ̃_{i}^{0}(x) = ψ_{i}^{0} Π_{k≠i}(1 − ψ_{k}^{0}(x)), and ψ_{i}(x) = \frac{ψ̃_{i}^{0}(x)}{∑_{k} ψ̃_{k}^{0}(x)}. This evidently gives a partition of unity with the stated properties. It is now straightforward to check that

\overline{f}(x) = ∑_{i=1}^{Q} ψ_{i}(x)f(y_{i}).

is a suitable extension. For the second part of the lemma we recall that the orthogonal projections onto m-dimensional subspaces of R^{n} form a smooth (in fact real-analytic) compact submanifold P of R^{n^{2}}, and hence there is a δ = δ(n) > 0 such that there is a smooth nearest-point projection map Ψ of the δ-neighbourhood N_{δ} of S onto S.

Now by the first part of the lemma we have an extension \overline{f}^{0} such that |f(y_{i})−\overline{f}^{0}(x)| ≤ C(n)ϵ for each x ∈ B_{2r}(y_{i}); but by definition f(y_{i}) ∈ S, so this means that if ϵ is small enough (depending only on n) we have \overline{f}^{0}(x) ∈ N_{δ/2} and hence we can define \overline{f} = Ψ ∘ \overline{f}^{0}. Evidently then \overline{f} has the correct properties.

The second lemma involves a simple observation about the subspaces L_{y,ρ} appearing in the ϵ-Reifenberg condition; in particular it shows that these must vary quite slowly (up to tilts of order ϵ) as y and ρ vary.

Lemma 2. If ϵ > 0 and if S satisfies the ϵ-Reifenberg condition above, then ∥L_{y_{1},σ} − L_{y_{2},ρ}∥ ≤ 32ϵ and dist (y_{1}, y_{2} + L_{y_{2},ρ}) ≤ 32ϵρ whenever y_{1}, y_{2} ∈ S ∩ B_{1} and 0 < \frac{ρ}{8} ≤ σ ≤ ρ ≤ 1.

The proof, which involves only the definition of the ϵ-Reifenberg condition and the triangle inequality for d_{H}, is left as an exercise for the reader.

Finally, we need the following “squash lemma”:

Lemma 3 (“Squash Lemma”). There is a constant ϵ_{0} = ϵ_{0}(n) such that the following holds. If ϵ ∈ (0, ϵ_{0}], ρ > 0, L is an m-dimensional subspace of R^{n},

Φ(x) = p_{L}(x) + e(x), x ∈ B_{3ρ},

where p_{L} is orthogonal projection onto L and ρ^{−1}|e(x)| + |∇e(x)| ≤ ϵ for all x ∈ B_{3ρ}, and if

G = {x + g(x) : x ∈ B_{3ρ} ∩ L}

is the graph of a C^{1} function g : B_{3ρ} ∩ L → L^{⊥} with ρ^{−1}|g(x)| + |∇g(x)| ≤ 1 at each point x of B_{3ρ} ∩ L, then Φ(G ∩ B_{3ρ}) is the graph of a C^{1}-function g̃ : U → L^{⊥} over some domain U with B_{11ρ/4} ∩ L ⊂ U ⊂ L and with ρ^{−1}|g̃| + |∇g̃(x)| ≤ 4ϵ on B_{11ρ/4} ∩ L.

Proof of the squash lemma: All hypotheses are written in “scale invariant” form, so there is no loss of generality in taking ρ = 1, which we do. Now by definition

Φ(x + g(x)) = x + e(x + g(x)) (1)

for x ∈ B_{2} ∩ L, and, if h(x) = e(x + g(x)), by the chain rule we have |d_{x}h| ≤ 2ϵ at each point x of L ∩ B_{2}. Now we can write h = h^{⊥} + h^{T}, where h^{⊥} = p_{L}^{⊥} ∘ h and h^{T} = p_{L} ∘ h. Then (1) says

Φ(x + g(x)) = x + h^{T} (x) + h^{⊥}(x), x ∈ B_{2} ∩ L. (2)

Now let

Q(x) = x + h^{T} (x), x ∈ B_{2} ∩ L,

and observe that

|d Q − id| ≤ 2ϵ, |Q − id| ≤ ϵ on B_{2} ∩ L,

and hence, for small enough ϵ ∈ (0, \frac{1}{6}), by the inverse function theorem Q is a diffeomorphism of B_{2} ∩L onto a subset U where L∩B_{11/4} ⊂ U ⊂ L and |d Q^{−1} −id| ≤ 2ϵ(1 + 2ϵ) ≤ 3ϵ. Thus (2) can be written

Φ(x + g(x)) = Q(x) + g̃(Q(x)), x ∈ B_{11/4} ∩ L,

where g̃ = p_{L}^{⊥} ∘ h ∘ Q^{−1} on U, and, since |d h ∘ Q^{−1}| ≤ 2ϵ(1 + 3ϵ) ≤ 3ϵ, we have |d g̃| ≤ 3ϵ and the proof is complete.

Proof of the Reifenberg disc theorem: The proof is based on an inductive procedure, making successive approximations to S_{∗} = S ∩ B_{1} by C^{∞} embedded submanifolds. Let T_{0} = L_{0, 1} (which without loss of generality we could take to be R^{m} × {0}) be an m-dimensional subspace such that d_{H}(S ∩ B_{1}, T_{0} ∩ B_{1}) < ϵ, and let r_{j} = (\frac{1}{8})^{j}, j = 0, 1, …. The quantity r_{j} is going the be the “scale” used at the j^{th} step of the inductive process. We in fact define maps σ_{j} : R^{n} → R^{n} and subsets M_{j} ⊂ R^{n} for j = 0, 1, …, as follows: For j ≥ 1, let B_{r_{j}/2}(y_{ji}), i = 1, …, Q_{j}, be a maximal pairwise disjoint collection of balls centered in S_{∗} = B_{1}∩S. Then evidently S_{∗} ⊂ ∪_{i=1}^{Q_{j}}B_{r_{j}}(y_{ji}) and also dist (S_{∗}, R^{n}\ (∪_{i=1}^{Q_{j}}B_{3r_{j}/2}(y_{ji}))) ≥ r_{j}/2. When j = 0 we take Q_{0} = 1, y_{0 1} = 0, and M_{0} = T_{0}, σ_{0} = the orthogonal projection of R^{n} onto T_{0}.

For j ≥ 1 and for each i = 1, …, Q_{j} let L_{ji} be one of the m-dimensional subspaces L_{y_{ji},8r_{j}} (corresponding to y = y_{ji} and ρ = 8r_{j} in the ϵ-Reifenberg condition). Thus

d_{H}(S ∩ B_{8r_{j}}(y_{ji}), (y_{ji} + L_{ji}) ∩ B_{8r_{j}}(y_{ji})) < 8ϵ r_{j}, i = 1, …, Q_{j}.

For j ≥ 1 we have by Lemma 2 that

d_{H}((y_{ji} + L_{ji}) ∩ B_{r_{j}}(y_{ji}), (y_{ℓ k} + L_{ℓ k}) ∩ B_{r_{j}}(y_{ji})) ≤ 264ϵr_{j} (1)

for any pair y_{j i}, y_{ℓ k} with |y_{j i} − y_{ℓ k}| ≤ 6r_{j−1}, where either ℓ = j − 1 and k ∈ {1, …, Q_{j−1}} or ℓ = j and k ∈ {1, …, Q_{j}}. Notice of course that (1) implies

|p_{ji} − p_{ℓk}| < 264ϵ, dist (y_{ji}, y_{ℓk} + L_{ℓk}) < 264ϵr_{j} (2)

for such j, ℓ, i, k, where p_{ji} denotes the orthogonal projection of R^{n} onto L_{ji}.

In view of the inequalities (2) (together with the fact that |y_{ji} − y_{jk}| ≥ r_{j} for each i ≠ k), we can apply the extension lemma with r = r_{j}, with y_{ji} in place of y_{i} and with the orthogonal projection p_{ji} in place of f(y_{i}), to give orthogonal projections p_{j,x} of R^{n} onto m-dimensional subspaces L_{j,x} such that p_{j,x} = p_{ji} when x = y_{ji} and

|\frac{∂p_{j,x}}{∂x^{ℓ}} | ≤ \frac{C(n)ϵ}{r_{j}}, x ∈ ∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji}), ℓ = 1, …, n,

|p_{j,x} − p_{ji}| ≤ C(n)ϵ, x ∈ B_{2r_{j}}(y_{ji}), i = 1, …, Q_{j}. (3)

Next let ψ_{ji} be a partition of unity for ∪_{i=1}^{Q_{j}}B_{3r_{j}/2}(y_{ji}) such that |∇ψ_{ji}| ≤ C(n)/r_{j} and support ψ_{ji} ⊂ B_{2r_{j}}(y_{ji}) for each i = 1, …, Q_{j}. (This is constructed in precisely the same way as our partition of unity for the extension lemma, except that we start with a smooth function φ with support in B_{2}(0) rather than in B_{3}(0) as before; actually the construction can be simplified here because we do not need ψ_{ji}(y_{ji}) = 1 and ψ_{jk}(y_{ji}) = 0 for i ≠ k.) Now we can define σ_{j} and M_{j} for j ≥ 1. First we define ^{1}

σ_{j}(x) = x − ∑_{i=1}^{Q_{j}} ψ_{ji}(x)p_{j,x}^{⊥}(x − y_{ji}), x ∈ R^{n}, (4)

and then we take

M_{j} = σ_{j}(M_{j−1}). (5)

First note that, since σ_{j}(x) ≡ x for x ∈ R^{n} \ (∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji})), we have

M_{j} \ (∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji})) = M_{j−1} \ (∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji})) (6) for each j ≥ 1. We claim that each M_{k} is a properly embedded C^{∞} m-dimensional submanifold of R^{n} and that for each k ≥ 1 and each i ∈ {1, …, Q_{k}}

1of course it doesn’t matter that the p_{j,x} are not defined outside ∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji}) because the ψ_{ji} vanish identically there. (If you wish to be pedantic, you can define e.g. p_{j,x} to be the orthogonal projection onto T_{0} for x ∈ R^{n} \ (∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji})).)

M_{k} ∩ B_{2r_{k}}(y_{ki}) = graph g_{ki}
(7)  sup |∇g_{ki}| ≤ γ ϵ,  sup |g_{ki}| ≤ γ ϵ r_{k}.

where γ ≥ 1 is a constant (to be specified as a function of n alone) and where g_{ki} is a C^{∞} function over a domain in the affine space y_{ki} + L_{ki} with values normal to L_{ki}. We want to inductively to check this. Observe that if j ≥ 1 and if M_{j−1} is a smooth embedded submanifold satisfying (7) with k = j − 1, then by the definition (4) we have

σ_{j}(x) − x = −∑_{k=1}^{Q_{j}}ψ_{j}(x)p_{j,x}^{⊥}(x − y_{jk})

= −∑_{k=1}^{Q_{j}}ψ_{j}(x)p_{jk}^{⊥}(x − y_{jk}) + ∑_{k=1}^{Q_{j}}ψ_{j}(x)(p_{jk}^{⊥} − p_{j,x}^{⊥})(x − y_{jk}). (8)

Now for each i ∈ {1, …, Q_{j}}, we can pick an i_{0} ∈ {1, …, Q_{j−1}} such that y_{ji} ∈ B_{r_{j−1}}(y_{j−1 i_{0}}). Then, assuming that (7) holds with k = j − 1 and with some constant γ = γ_{j−1}, for x ∈ B_{2r_{j}}(y_{ji}) ∩ M_{j−1}(⊂ B_{2r_{j−1}}(y_{j−1 i_{0}}) ∩ M_{j−1}) we can write x = ξ + g_{j−1}(ξ), with g_{j−1}(ξ) ∈ L_{j−1 i_{0}}^{⊥}, ξ ∈ (y_{j−1 i_{0}} + L_{j−1 i_{0}}) ∩ B_{2r_{j−1}}(y_{j−1 i_{0}}) and with r_{j−1}^{−1}|g_{j−1}(ξ)| + |∇g_{j−1}(ξ)| ≤ γ_{j−1}ϵ. Then we have, for each k ∈ {1, …, Q_{j}},

p_{jk}^{⊥}(x − y_{jk}) = p_{j−1 i_{0}}^{⊥}(ξ + g_{j−1}(ξ) − y_{j−1 i_{0}})
+ p_{j−1 i_{0}}^{⊥}(y_{jk} − y_{j−1 i_{0}}) + (p_{jk}^{⊥} − p_{j−1 i_{0}}^{⊥})(ξ + g_{j−1}(ξ) − y_{jk}),

and using (2), (3) together with the fact that p_{j−1 i_{0}}^{⊥}(ξ − y_{j−1 i_{0}}) = 0 (because ξ − y_{j−1 i_{0}} ∈ L_{j−1 i_{0}}), we have clearly then that

|p_{jk}^{⊥}(x − y_{jk})| ≤ C(n)ϵ(1 + γ_{j−1})r_{j}, x ∈ B_{2r_{j}}(y_{ji}) ∩ M_{j−1}, |y_{jk} − y_{ji}| ≤ 6r_{j}.

Using this in (8), and keeping in mind that for any i ∈ {1, …, Q_{j}} and for any x ∈ B_{2r_{j}}(y_{ji}), we have that at most C(n) terms in the sums on the right of (8) can be non-zero, and that these terms correspond to the indices k such that |y_{ji} − y_{jk}| ≤ 6r_{j}, hence, using also (3), we again deduce from (8) that

|σ_{j}(x) − x| ≤ C(n)(1 + γ_{j−1})ϵr_{j}, x ∈ ∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji}) ∩ M_{j−1}. (9)

By first differentiating in (8) and using similar considerations on the right side, we also conclude

sup_{x∈M_{j−1}} |∇′(σ_{j}(x) − x)| ≤ C(n)(1 + γ_{j−1})ϵr_{j}, (9)′

where ∇′ denotes gradient taken on the submanifold M_{j−1}.

We refer to (9) and (9)′ subsequently as “the coarse estimates” for |σ_{j}(x)−x|, because, although useful, they are insufficient in themselves to complete that inductive proof that there is a fixed constant γ = γ(n) such that (7) holds for all k; indeed after k applications of this coarse inequality, we will only have established that (7) holds with γ = C(n)^{k}. Now assume that j ≥ 2 and that (7) holds for k = 1, …, j − 1, take an arbitrary i_{0} ∈ {1, …, Q_{j}}, and write y_{0} = y_{j i_{0}}, p_{0} = p_{j i_{0}}, and L_{0} = L_{j i_{0}}. Since ∑_{i=1}^{Q_{j}} ψ_{ji} ≡ 1 in U_{j} ≡ ∪_{i=1}^{Q_{j}}B_{3r_{j}/2}(y_{ji}) we can rearrange the defining expression for σ_{j} to give

σ_{j}(x) = y_{0} + p_{0}(x − y_{0}) + e(x), x ∈ U_{j}, (10)

where e is given by

e(x) ≡ ∑_{i=1}^{Q_{j}} ψ_{ji}(x)p_{0}^{⊥}(y_{ji} − y_{0}) − ∑_{i=1}^{Q_{j}} ψ_{ji}(x)(p_{j,x}^{⊥} − p_{0}^{⊥})(x − y_{ji}), x ∈ R^{n}. (11)

Now observe that by (2) and (3) we have |p_{j,x} −p_{0}| ≤ C(n)ϵ r_{j} for x ∈ B_{6r_{j}}(y_{0}). Using additionally the first inequality in (3) and the fact that |∇ψ_{ji}| ≤ C(n)/r_{j}, it then follows easily that

r_{j}^{−1}|e(x)| + |∇e(x)| ≤ C(n)ϵ, if x ∈ B_{3r_{j}/2}(y_{0}), (12)

where C(n) is a fixed constant determined by n alone (and which is independent of any properties of M_{j−1}; in particular it is independent of whatever constant γ appears in (7)). But now we can apply the Squash Lemma with σ̃_{j}(x) ≡ σ_{j}(x + y_{0}) − y_{0} in place of Φ, 2 r_{j} in place of ρ, and C(n)ϵ in place of ϵ. Assuming that (7) holds with γ, ϵ such that γϵ ≤ \frac{1}{2}, we thus conclude

σ_{j}(M_{j−1} ∩ B_{3r_{j}/2}(y_{0})) = G_{j}, (13)

where G_{j} = {x + g_{j}(x) : x ∈ Ω_{j}} is the graph of a C^{∞} function g_{j} defined over a domain Ω_{j} contained in the affine space y_{0} + L_{0} with B_{11r_{j}/8}(y_{0}) ∩ (y_{0} + L_{0}) ⊂ Ω_{j} and with

r_{j}^{−1}|g_{j}| + |∇g_{j}| ≤ C(n)ϵ, x ∈ B_{11r_{j}/8}(y_{0}) ∩ (y_{0} + L_{0}), (14)

with C(n) not depending on γ. Of course since |σ_{j}(x)−x| < C(n)γϵ (by (8)), we thus have, so long as C(n)γϵ ≤ \frac{1}{32} that σ_{j}(M_{j−1} ∩ B_{3r_{j}/2}(y_{0})) ⊃ σ_{j}(M_{j−1}) ∩ B_{11r_{j}/8}(y_{0}), and hence (13) and (14) imply

M_{j} ∩ B_{11r_{j}/8}(y_{0})) = G_{j}, (15)

with G_{j} still as in (14).

Now we actually need to establish a result like this over the ball B_{2r_{j}}(y_{0}) rather than merely over B_{11r_{j}/8}(y_{0}); to achieve this, we observe that each y_{ji} is contained in one of the balls B_{r_{j−1}}(y_{j−1 i_{0}}) for some i_{0} ∈ {1, …, Q_{j−1}}, and so B_{r_{j−1}/4}(y_{ji}) ⊂ B_{5r_{j−1}/4}(y_{j−1 i_{0}}). Also, by using the above argument with j − 1 in place of j and with i_{0} in place of i, we deduce that

(15)′  M_{j−1} ∩ B_{11r_{j−1}/8}(y_{j−1 i_{0}})) = G_{j−1},

where G_{j−1} = {x+g_{j−1}(x) : x ∈ Ω_{j−1}} is the graph of a C^{∞} function g_{j−1} defined over a domain Ω_{j−1} contained in the affine space y_{j−1 i_{0}} + L_{j−1 i_{0}} with B_{11r_{j−1}/8}(y_{j−1 i_{0}}) ∩ (y_{j−1 i_{0}} + L_{j−1 i_{0}}) ⊂ Ω_{j−1} and with

r_{j−1}^{−1}|g_{j−1}| + |∇g_{j−1}| ≤ C(n)ϵ, x ∈ B_{11r_{j−1}/8}(y_{j−1 i_{0}}) ∩ (y_{j−1 i_{0}} + L_{j−1 i_{0}}). (14)′

But then by using the coarse estimates (9), (9)′ we deduce that in fact (7) holds with k = j and a fixed constant γ which depending only on n and not on γ. Notice that since S_{∗} ⊂ ∪_{i=1}^{Q_{j}}B_{r_{j}}(y_{ji}) it is clear from (7) and the ϵ-Reifenberg condition in the ball B_{2r_{j}}(y_{ji}), that

S_{∗} ⊂ B_{C(n)ϵr_{j}}(M_{j}), j ≥ 0. (16)

Notice also that (7) tells us that for j ≥ 2

M_{j} ∩ (∪_{i=1}^{Q_{j}}B_{2r_{j}}(y_{ji})) ⊂ (∪_{i=1}^{Q_{j}}B_{C(n)ϵr_{j}}(y_{ji} + L_{ji})) ⊂ B_{C(n)ϵr_{j}}(S),

and hence, since M_{j} \ (∪_{i}B_{2r_{j}}(y_{ji})) = M_{j−1} \ (∪_{i}B_{2r_{j}}(y_{ji})) by mathematical induction it follows that

M_{j} ∩ B_{1+r_{j}/2} ⊂ B_{C(n)ϵ r_{j}}(S) (17)

for each j = 0, 1, …, provided ϵ ≤ ϵ_{0}, where ϵ_{0} = ϵ_{0}(n).

Next we want to show that the sequence τ_{j} = σ_{j} ∘ σ_{j−1} ∘ ⋯ σ_{0}|T_{0} is a sequence of C^{∞} diffeomorphisms of T_{0} onto M_{j} which converge uniformly on T_{0} to a homeomorphism τ of T_{0} onto a closed set M. In fact notice that by (8) we have

|τ_{j}(x) − τ_{j−1}(x)| ≤ C(n)ϵ (\frac{1}{8})^{j} , j ≥ 1, x ∈ T_{0},

and hence by iterating we get

|τ_{j+k}(x) − τ_{j}(x)| ≤ C(n)ϵ (\frac{1}{8})^{j} , j ≥ 0, k ≥ 1, x ∈ T_{0}, (18)

which shows that τ_{j} is Cauchy with respect to the uniform norm on T_{0}, and hence τ_{j} converges uniformly to a continuous map τ : T_{0} → R^{n}. Of course τ is the identity outside B_{2} because each σ_{j} is the identity outside B_{2}. We let M = τ(T_{0}), so that M is a closed subset of R^{n} and in fact is the Hausdorff limit (with respect to the Hausdorff metric d_{H}) of the sequence M_{j} = τ_{j}(T_{0}). Notice in particular that setting j = 0 and taking limit as k → ∞ in the above inequality, we get

|τ(x) − x| ≤ C(n)ϵ, x ∈ T_{0}. (19)

(Thus τ is in the distance sense quite close to the identity if ϵ is small.) Next we want to discuss injectivity of τ_{j}, τ; in fact we’ll show that τ_{j}, τ are injective and that both τ and τ^{−1} are Hölder continuous. To establish this, we first claim

(1 − C(n)ϵ)|x − y| ≤ |σ_{j}(x) − σ_{j}(y)| ≤ (1 + C(n)ϵ)|x − y|, x, y ∈ M_{j−1}, (20)

or equivalently

(20)′  |σ_{j}(x) − σ_{j}(y) − (x − y)| ≤ C(n)ϵ|x − y|, x, y ∈ M_{j−1}.

To prove this, note that if |x − y| ≥ r_{j} with x, y ∈ M_{j−1}, we can write

|σ_{j}(x) − σ_{j}(x) − (x − y)| = |(σ_{j}(x) − x) − (σ_{j}(y) − y)|
≤ |σ_{j}(x) − x| + |σ_{j}(y) − y|
≤ C(n)ϵ r_{j} ≤ C(n)ϵ|x − y|,

where we used (8) in the second inequality. Now if |x − y| < r_{j} we use the definition (4) to write

(σ_{j}(x) − σ_{j}(y)) − (x − y) = ∑_{i=1}^{Q_{j}}(ψ_{ji}(x)p_{j,x}^{⊥}(x − y_{ji})
− ψ_{ji}(y)p_{j,y}^{⊥}(y − y_{ji})), x, y ∈ R^{n},

and note that we can rearrange the sum here to give

(σ_{j}(x) − σ_{j}(y)) − (x − y) = ∑_{i=1}^{Q_{j}} (ψ_{ji}(x)(p_{j,x}^{⊥}(x − y)
+ ψ_{ji}(x)(p_{j,x}^{⊥} − p_{j,y}^{⊥})(y − y_{ji}) + (ψ_{ji}(x) − ψ_{ji}(y))p_{j,y}^{⊥}(y − y_{ji})) .

Now the second group of terms is (by (3)) trivially ≤ C(n)ϵ |x − y| in absolute value for any x, y ∈ R^{n} with |x − y| ≤ r_{j}. Further if x, y ∈ M_{j−1}, then by virtue of (7) (used with y in place of z) we see that the first and third group of terms on the right is ≤ C(n)ϵ |x − y| in absolute value. Thus we again get (20).

Now it is easy to establish the required injectivity and continuity of τ. In fact by iterating the inequality (20) we get

|τ_{j}(x) − τ_{j}(y)| ≤ (1 + Cϵ)^{j}|x − y|, x, y ∈ T_{0}, j ≥ 1, (21) and by (8) we have

|τ_{j}(x) − τ_{j−1}(x)| ≤ Cϵr_{j}, x ∈ T_{0}, j ≥ 1,

and so (Cf. the discussion of uniform convergence of the τ_{j} above)

|τ_{j}(x) − τ(x)| ≤ Cϵr_{j}. (22)

Then by the triangle inequality, for any j ≥ 0 we have

|τ(x) − τ(y)| ≤ |τ(x) − τ_{j}(x)| + |τ_{j}(x) − τ_{j}(y)| + |τ_{j}(y) − τ(y)|
≤ 2C(n)ϵr_{j} + (1 + C(n)ϵ)^{j}|x − y|
≤ r_{j} + (1 + C(n)ϵ)^{j}|x − y| if 2ϵC(n) ≤ 1.

Now let α ∈ (0, 1) be arbitrary and take x, y ∈ T_{0} with 0 < |x−y| < \frac{1}{2}. Choose j such

that r_{j} ≤ |x − y|^{α} and (1 + C(n)ϵ)^{j} ≤ |x − y|^{−(1−α)}; thus we need j ≥ \frac{α}{log 8} log (\frac{1}{|x−y|})

and also j ≤ \frac{(1−α)}{log(1+C(n)ϵ)} log (\frac{1}{|x−y|}). Since log(1 + C(n)ϵ) → 0 as ϵ ↓ 0, we see that such a choice of j ∈ {1, 2, … } exists provided ϵ ≤ ϵ_{0}, where ϵ_{0} = ϵ_{0}(n, α). Then the above inequality gives

|τ(x) − τ(y)| ≤ 2|x − y|^{α}, x, y ∈ T_{0} with |x − y| < \frac{1}{2}.

Thus we can arrange for Hölder continuity with any exponent α < 1. Similarly we have from the first inequality in (20) and (22) that

|x − y| ≤ (1 + Cϵ)^{j}|τ_{j}(x) − τ_{j}(y)|
≤ (1 + Cϵ)^{j} (|τ_{j}(x) − τ(x)| + |τ_{j}(y) − τ(y)| + |τ(x) − τ(y)|)
≤ (1 + C(n)ϵ)^{j} (C(n)ϵr_{j} + |τ(x) − τ(y)|)

and j is again at our disposal. We in fact first choose ϵ such that C(n)ϵ ≤ 1, so that

|x − y| ≤ (1 + C(n)ϵ)^{j} (r_{j} + |τ(x) − τ(y)|) ,

and then choose j such that α ∈ (0, 1)

4^{−j} ≤ \frac{1}{2}|x − y| and (1 + C(n)ϵ)^{j} ≤ |x − y|^{−(α/(1−α)}.

Notice that this requires j ≥ log(2/|x−y|)/ log (\frac{8}{1+C(n)ϵ}) and j ≤ α^{−1}(1−α) log(1/|x− y|)/ log(1+C(n)ϵ), and again certainly such a choice of j exists provided 0 < |x−y| < \frac{1}{2} and provided we take ϵ ≤ ϵ_{0} for suitable ϵ_{0} = ϵ_{0}(n, α). In this case the above inequality gives

\frac{1}{2}|x − y| ≤ |x − y|^{−α/(1−α)}|τ(x) − τ(y)|, |x − y| < \frac{1}{2},

which of course gives

|x − y|^{α} ≤ 2|τ(x) − τ(y)|, |x − y| < \frac{1}{2}.

Thus τ is injective, and the inverse is Hölder continuous with exponent α, for any given α ∈ (0, 1), provided the ϵ-Reifenberg condition holds with ϵ ≤ ϵ_{0}, where ϵ_{0} = ϵ_{0}(n, α).

Now the proof of the Reifenberg inequality is complete, because we have shown that τ maps T_{0} Hölder continuously onto M with Hölder continuous inverse, and by (16) and (17) we have

M ∩ B_{1} = S_{∗},

because (by (19)) M_{j} converges to M with respect to the Hausdorff distance metric.