Lipschitz Equivalent Separable Banach Spaces Need Not Be Linearly Isomorphic

Proof. Let \ell_1(2^{-i};A) denote the weighted sum space with norm \sum_i2^{-i}\|s_i\|. The map s\mapsto(2^{-i}s_i)_i is an isometry from this space onto \ell_1(A), so it is complete. The map

identifies Z_Q isometrically with a subspace of \ell_1(2^{-i};A)\oplus_\infty c_0(H). This subspace is closed. In fact, suppose (s^{(k)},(Qs_i^{(k)})_i) converges there to (s,y). For each fixed i,

Continuity of Q gives Qs_i^{(k)}\to Qs_i, whereas convergence in c_0(H) gives Qs_i^{(k)}\to y_i. Thus y_i=Qs_i for all i, as required. Completeness follows.

If A,H are separable, truncating a sequence in Z_Q gives convergence in both parts of (2.5): the weighted sum has a vanishing tail, and the Q-image is norm-null. Choose a fixed countable dense subset of A. The finitely supported sequences with entries in that subset form a countable dense subset of Z_Q, because each of their finitely many entries may be approximated in A and Q is continuous. Finally, the two claimed operator bounds are immediate from (2.5).□

Proof. Every finite direct sum of E is weakly sequentially complete: the coordinates of a weakly Cauchy sequence are weakly Cauchy in E, hence have weak limits, and the finite list of limits is the weak limit in the sum. The property also passes to a closed linear subspace F of a weakly sequentially complete space G. Indeed, a weakly Cauchy sequence in F is weakly Cauchy in G; its weak limit lies in F, since a closed linear subspace is weakly closed by Hahn–Banach separation. Every functional on F extends to G by Hahn–Banach, so the same limit is weak in F.

Ordinary c_0 is not weakly sequentially complete. If (e_k) is its usual unit basis, the partial sums e_1+\cdots+e_n are weakly Cauchy: for every (a_k)\in c_0^*=\ell_1, their pairings are the Cauchy partial sums \sum_{k\le n}a_k. A weak limit would have every coordinate equal to one, which is not an element of c_0.

Suppose now that A:c_0\to\ell_1(E) is a linear embedding with \|Ax\|\ge b\|x\|, where b>0. Let P_N be projection onto the first N coordinates, with P_0=0. No P_NA is bounded below, because otherwise its closed range would be a copy of c_0 in the weakly sequentially complete finite sum E^N.

Starting with N_0=0, choose unit vectors x_j\in c_0 and integers N_j>N_{j-1} recursively so that

The first inequality is possible because P_{N_{j-1}}A is not bounded below; after choosing x_j, its \ell_1(E) tail gives the second. Put z_j=(P_{N_j}-P_{N_{j-1}})Ax_j. Then

The z_j have disjoint coordinate supports. For any finite scalar family (\alpha_j),

Consequently,

The finite-sum map therefore extends by completeness to a bounded linear embedding of \ell_1 into c_0, with closed range. This is impossible. If F is any closed subspace of c_0, the restriction map c_0^*=\ell_1\to F^* is onto by Hahn–Banach, so F^* is separable. A subspace isomorphic to \ell_1 would instead have dual isomorphic to \ell_\infty, which is nonseparable: the uncountable set of all 0-1 sequences is pairwise separated by distance one. This contradiction proves the lemma.□

Proof. Suppose that S:c_0(\ell_2)\to Z_q is a linear embedding with \|Sx\|_{Z_q}\ge b_0\|x\|, b_0>0. We will select a scalar copy of c_0 on which the c_0(U) part of S has norm at most b_0/2. The weighted sum part of the graph norm must then retain the lower bound b_0, giving a copy of c_0 in \ell_1(E), contrary to Lemma 2.4.

Write I_n:\ell_2\to c_0(\ell_2) for injection into the nth coordinate and put

We first prove that each T_n fails to be bounded below.

Fix n and an orthonormal sequence (e_k) in \ell_2. It is weakly null, since the scalar coefficients of a fixed Hilbert vector against an orthonormal sequence are square summable. For each fixed i, boundedness of p_iSI_n therefore gives p_iSI_ne_k\to0 weakly in E. Complete continuity of q implies

Set y_k=T_ne_k, so \|y_k\|\le\|T_n\|. Choose k_j increasing and integers 0=m_0<m_1<m_2<\cdots as follows. Having chosen m_{j-1}, choose k_j so large that

The maximum over no coordinates is zero. Then choose m_j>m_{j-1} so large that \sup_{i>m_j}\|(y_{k_j})_i\|<2^{-j}, using y_{k_j}\in c_0(U). Let z_j be the restriction of y_{k_j} to coordinates m_{j-1}<i\le m_j. Thus

Because the z_j have disjoint coordinate supports, for all N

The corresponding input sum has Hilbert norm \sqrt N. No positive lower bound for T_n is compatible with these two estimates.

Choose a unit vector t_n\in\ell_2 such that \|T_nt_n\|\le b_0\,2^{-n-1}. The map

is a linear isometry. For finitely supported a,

Truncation in c_0 and continuity extend this estimate to every a\in c_0. On the other hand \|SRa\|_{Z_q}\ge b_0\|a\|_\infty. The second part of the maximum in (2.5) is at most (b_0/2)\|a\|_\infty, so the first part must satisfy

For a=0 this is immediate; for a\ne0 it follows because the second part is strictly below the required lower bound for the maximum. Hence

is a bounded linear embedding of c_0 into \ell_1(E). Boundedness comes from \sum_i2^{-i}\|(SRa)_i\|\le\|SRa\|_{Z_q}, and the displayed lower bound makes the range closed. This contradicts Lemma 2.4.□

Proof. The space E is separable because M is separable. The two operators Q_1 and q are bounded: in the sum norm,

Thus Lemma 2.3 applies to X and Y.

Define the zero-preserving map

The defining identity q\delta(u,v)=h(u,v)-u gives

For z\in E, the inverse of h is defined at qz; set x=h^{-1}(qz) and

Since

we have Q_1C(z)=x, and substitution in (3.3) gives B(C(z))=z. Conversely, if z=B(e,v), then (3.4) and injectivity of h give h^{-1}(qz)=Q_1(e,v)=(qe,v). Formula (3.5) therefore returns (e,v). Thus C=B^{-1} and

Only the inverse of h was used. In particular, this argument does not require q to be onto.

We next keep track of the two norms. For w=(e,v) and w'=(e',v'), the isometry of \delta and (3.2) yield

For z,z'\in E, put x=h^{-1}(qz), x'=h^{-1}(qz'). The lower bound for h implies

Using the sum norm in (3.5), rather than a Hilbert sum norm on E\oplus V, gives

Neither estimate divides by L_q; they remain valid as written when L_q=0. In the present setting V=\ell_2\ne0, that case cannot in fact occur under the hypotheses, since q=0 would give h(u,v)=u, contradicting injectivity.

Define \Psi coordinatewise by

Because B(0)=0, the bound for B makes \sum_i2^{-i}\|B(s_i)\| finite. Also Q_1s_i\to0 in M; by (3.4) and continuity of h at zero, qB(s_i)=h(Q_1s_i)\to0. Hence \Psi(s)\in Y. Likewise B^{-1}(0)=0 controls the weighted sum for the coordinatewise inverse, and (3.6) shows that if qz_i\to0, then Q_1B^{-1}(z_i)=h^{-1}(qz_i)\to0. Thus the coordinatewise inverse maps Y into X, and the pointwise inverse identities make \Psi a bijection.

For clarity, the estimates apply to every difference of graph sequences. For s,t\in X,

Taking the maximum proves the first inequality in (3.1). For z,z'\in Y, the corresponding two bounds for B^{-1} and Q_1B^{-1}=h^{-1}q are

which prove the second. These are global two-point estimates, not only estimates at the origin.

Finally, define

The sequence belongs to X, and

because \sum_{i\ge1}2^{-i}=1. This is a linear isometry. The Hilbert space M is superreflexive, so Theorem 2.2 makes E=\F(M) weakly sequentially complete. By hypothesis q is completely continuous, and Lemma 2.5 therefore excludes a linear copy of c_0(V) from Y. A linear isomorphism from X onto Y would carry the displayed closed copy to one in Y, which is impossible.□

Proof. For s>0, define

This cutoff is one on B(z,s/2), zero outside B(z,s), takes values in [0,1], and has Lipschitz constant at most 2/s. Whenever f(z)=0 and \Lip(f)\le C, we claim

Let \psi be either cutoff in this claim. If \psi(x)=\psi(y), the difference of the products is at most C\rho(x,y). Otherwise at least one endpoint lies in B(z,s); call it x. Then |f(x)|\le Cs, and

has absolute value at most C\rho(x,y)+Cs(2/s)\rho(x,y)=3C\rho(x,y). Thus (4.2) needs no global bound on f.

Suppose the conclusion fails. Passing to a subsequence, assume

Then C>0. We recursively select increasing indices i_j, numbers 0<s_j<r_{i_j}, and functions

such that

After choosing i_j, approximate \mu_{i_j} by a finite sum of evaluations

Choose s_j\in(0,r_{i_j}) smaller than every positive distance from z to a member of \{0,x_{j1},\ldots,x_{jN_j}\}. If that finite set has no positive such distances, impose only s_j<r_{i_j}. The function p_j=f_{i_j}\phi_{z,s_j} vanishes at every one of these points: outside the ball the cutoff is zero, and at a point equal to z the factor f_{i_j} is zero. Including the basepoint is essential when z\ne0. It gives

By (4.2),

Subtracting from the original pairing proves the first inequality in (4.3); the second is (4.2). Finally choose i_{j+1}>i_j so large that r_{i_{j+1}}<s_j/2. This completes the recursion.

The nonzero set of h_j lies in

These annuli are pairwise disjoint. Every finite signed sum H=\sum_{j\in A}\sigma_jh_j, with \sigma_j\in\{-1,1\}, therefore satisfies \Lip(H)\le6C. Indeed, at a point at most one summand is nonzero. If two points lie in the same nonzero set, the 3C bound for that summand applies. If they lie in different nonzero sets, each of the two contributing functions vanishes at the other point, and their 3C bounds add. The cases where one or both points lie in no nonzero set are included. Subtracting H(0) does not change this Lipschitz bound.

For every \mu\in\F(M) and finite A\subset\N, real signs give

Taking increasing finite sets shows that

is a bounded linear operator of norm at most 6C. Thus T\mu_{i_j}\to0 weakly in \ell_1. By the Schur property of \ell_1, its norms tend to zero. But its jth coordinate has absolute value greater than \epsilon/2 by (4.3), a contradiction.□

Proof of Theorem 4.1. Let E=\F(M) and L=\Lip(D)=\|q\|. If L=0, then D=0 because D(0)=0, and q=0. Assume L>0. If q were not completely continuous, after taking a subsequence there would be \mu_i\to0 weakly in E and \|q\mu_i\|\ge\epsilon_0>0 for all i. Set

The unit vectors b_i are weakly null. For each b\in U,

since a bounded linear map preserves weak convergence. Consequently

The functions d_i tend uniformly to zero on every compact subset of M. They tend pointwise to zero by weak nullity of b_i. Given a compact set and a positive error, take a finite sufficiently fine metric net; pointwise convergence on that net and the common Lipschitz bound control all points of the compact set.

Only finitely many points can retain a large local Lipschitz constant. Choose a countable base \mathcal B for M, which exists by separability. Take the Lipschitz constant on a set of at most one point to be zero. For each O\in\mathcal B, \Lip(d_i|_O)\in[0,L]. Successively take subsequences for the countably many O, then take a diagonal subsequence. Relabeling, we may assume that all limits

exist. This subsequence is fixed before the next parameter is chosen.

Fix \eta>0 and define

Take any finite family of distinct z_1,\ldots,z_m\in Z. Apply (4.1) and shrink the resulting neighborhoods to base neighborhoods O_j containing z_j. Let P_j be projection onto H_j, and put \Pi_{<l}=I-\Pi_{\ge l}. For x,y\in O_j, the tail \Pi_{\ge l}(D(x)-D(y)) lies in H_j. Hence

The first l-1 blocks are finite dimensional. Weak nullity of b_i makes each of its finitely many coordinates there tend to zero, so \|\Pi_{<l}b_i\|\to0. Since a_{O_j}\ge\eta, the preceding inequality gives

The P_j have mutually orthogonal ranges, and \|b_i\|=1. It follows that

The same conclusion holds for every finite family of distinct points of Z, so Z is finite and

Remove small neighborhoods of those finitely many points. Choose r_0>0 so that the balls B(z,r), z\in Z, are pairwise disjoint for every 0<r<r_0. If Z has at most one point, any r_0>0 suffices. With the cutoffs used in Lemma 4.2, put

An empty sum is zero. Equation (4.2) gives \Lip(f_{i,z,r})\le3L. Their nonzero sets lie in disjoint balls, so the two-point argument for finite signed sums in the proof of Lemma 4.2 gives

These constants do not depend on r or on |Z|.

For each fixed z\in Z, we have the ordered limit

If this failed, some a>0 and radii r_j\downarrow0 would satisfy \limsup_i|\langle f_{i,z,r_j},\mu_i\rangle|\ge2a. Choose increasing i_j with |\langle f_{i_j,z,r_j},\mu_{i_j}\rangle|\ge a. These tests vanish at z, have Lipschitz constants at most 3L, and vanish outside B(z,r_j). The weakly null subsequence \mu_{i_j} and Lemma 4.2 give a contradiction. Because Z is finite, the triangle inequality and (4.7) imply

If Z is empty, (4.8) is immediate.

The removed part is now controlled. It remains to show that the remainder has small pairing. This is where completeness of M enters through compact reduction.

Estimate the remainder on one compact set. Fix \tau>0. By Theorem 2.1, choose a compact K\subset M, 0\in K, and \nu_i\in\F(K) with

The set K depends on \tau and the sequence, but not on r. Now fix 0<r<r_0 while taking the sequence limit. Write m=|Z| and

The compact set K\cup Z contains 0, so \epsilon_i\to0, including when Z is empty. Disjointness of the balls gives 0\le\Phi_r\le1, and

For x\in K the right side is a convex combination of values bounded in absolute value by \epsilon_i. Therefore

On each inner ball B(z,r/2), the remainder is the constant d_i(z).

At any point p\notin Z, choose O\in\mathcal B containing p with a_O<\eta. For x,y\in K\cap O, expansion of (4.9) gives

The last factors have absolute value at most 2\epsilon_i, and \Lip(\phi_{z,r})\le2/r. Thus

The radius is fixed here, so the second term tends to zero with i.

The inner balls at points of Z, together with the chosen base neighborhoods at points outside Z, cover K. Choose a finite subcover V_1,\ldots,V_N. The Lebesgue-number property supplies \lambda>0 such that any pair of points of K at distance less than \lambda belongs to some one V_t. For an inner ball put c_{i,t}=0; for a chosen base neighborhood put

For pairs at distance less than \lambda, the Lipschitz quotient of R_{i,r} is at most \max_t c_{i,t}, by constancy on inner balls and (4.11). For pairs at distance at least \lambda, (4.10) bounds it by 2\epsilon_i/\lambda. Therefore

If K has only one point, its Lipschitz constant is zero and the same conclusion holds. There are finitely many V_t, and for each base neighborhood a_{V_t}<\eta. Taking the upper limit gives

Use (2.2) for \nu_i\in\F(K), and use the global bound (4.6) for the approximation error. We obtain

With \eta,\tau,K,r fixed, (4.12) implies

The right side is independent of r. Since d_i=R_{i,r}+g_{i,r}, we may now let r\downarrow0 and use (4.8) to get

The quantifier order is as follows: the diagonal subsequence is fixed first; for each \eta choose Z,r_0; for each \tau choose K; for a fixed r<r_0 take i\to\infty; then take r\downarrow0, \tau\downarrow0, and finally \eta\downarrow0. No uniform convergence in both i and r has been used. The last two limits in (4.13) make its right side zero, contradicting (4.4).□

Proof. All formulas are linear for fixed \xi. Since u_n^\perp\perp\xi_n, the squared norm of (5.2) is

This also proves the output belongs to U.

For y\in U, |\beta_n|\le\|y_n\|, so (\beta_n)\in\ell_2. The first two terms y_m-\beta_m\xi_m in (5.3) are perpendicular to its last term. The prescribed vector therefore belongs to M, since

For this input the old slot coefficient is \beta_{2m}, and the perpendicular component in block m is y_m-\beta_m\xi_m. The even and odd output slots consequently recover every \beta_n, so (5.2) returns y. This proves surjectivity and the inverse formula.□

Proof. For an empty family, take l=1 and the empty family of neighborhoods. For a singleton, take l=1 and O_1=M. In either case there is no disjointness condition to check. Now consider two distinct points. Some p_j has different values at them. Choose k_0\ge j sufficiently large and neighborhoods on which all values of that p_j on the two sides are separated by more than 2/k_0. At any stage k\ge k_0, a coordinate label that occurs on both sides would require the jth factor \psi(kp_j-a_j) to be nonzero at points x,y on opposite sides. Then both p_j(x) and p_j(y) would lie in ((a_j-1)/k,(a_j+1)/k), contradicting |p_j(x)-p_j(y)|>2/k_0\ge2/k. Thus the unions of the supports on the two neighborhoods are disjoint at every common stage k\ge k_0.

Fix 0<\rho<a_{k_0}', below the completed transition into stage k_0. Choose l\ge k_0 so that r_l at each center is less than \rho/2, and shrink both neighborhoods to have radius less than \rho/(2\sqrt3). Since r_n\le r_l for n\ge l and r_l is \sqrt3-Lipschitz, we have r_n(x)<\rho throughout either neighborhood whenever n\ge l. Only stages k\ge k_0 are then active in w_n(x). Distinct stages are orthogonal, and supports at the same stage are disjoint by the first paragraph. This proves the two-point assertion. For finitely many points, intersect the neighborhoods obtained for every pair and take the largest cutoff.□

Proof. Fix a point and a unit direction, and abbreviate the approximant slots w_n^{(N)}(x) and their directional derivatives by w_n,\dot w_n; the slots with n>N have derivative zero. Differentiating the inputs in (5.4) with the slots frozen gives the onto isometry L_{w^{(N)}(x)}. The remaining terms in output block n are

Put R^2=\sum_n a_n(x)^2\|\dot w_n\|^2\le c^2. Since |C_n-\langle w_n,u_n\rangle|\le2a_n, the first term has Hilbert norm at most 2R. Each of the scalar sequences (\dot C_n)_n and (\langle\dot w_n,u_n\rangle)_n has \ell_2 norm at most R, because a_n\ge\|u_n\|. Therefore

A Neumann-series perturbation of the isometry gives

We next establish global bijectivity before using the lower Lipschitz estimate. Put F_N=\bigoplus_{n\le2N}U_n and let P project onto F_N^\perp. We have

For a fixed b\in F_N^\perp, the map h_N thus restricts to

These affine spaces have the same finite dimension. The derivative is an isomorphism between their tangent spaces. Indeed it maps A_0^{-1}(F_N) into F_N; conversely, for z\in F_N, the vector t=Dh_N(x)^{-1}z satisfies PA_0t=PDh_N(x)t=0. Thus (5.19) is a local diffeomorphism. It is proper, since an inverse image of a compact set is closed and norm-bounded by \|h_N(x)\|=\|x\|, hence compact in this finite-dimensional source fiber.

A proper local diffeomorphism between affine Euclidean spaces of the same dimension is bijective. To recall the topological argument, its image is open by local invertibility and closed by properness, so it is the entire target. Each fiber is compact and discrete, hence finite. Local inverse neighborhoods, together with properness to exclude further preimages nearby, make it a covering map. A covering of a simply connected Euclidean space with connected source has one sheet. Applying this to (5.19) proves that h_N is bijective on each fiber and therefore on all of M.

The local inverses supplied by the inverse function theorem agree and form a global continuously differentiable inverse. Integrating the two bounds in (5.18) along line segments, for h_N and for its now-defined inverse, proves (5.16).□

Completion of the proof of Theorem 5.1. For fixed x=(u,v), the first N output blocks of h_N(x) and h(x) agree. Indeed, their own slots agree in those blocks, and the old slot used in an even block 2m\le N has m\le N. For either list \zeta=w(x) or \zeta=w^{(N)}(x), write

Then (L_\zeta x)_n-u_n =\zeta_n(C_n^\zeta-\langle\zeta_n,u_n\rangle). Since |C_n^\zeta|\le a_n(x) and \|u_n\|\le a_n(x), each difference from u_n has norm at most 2a_n(x). For the remaining blocks, the difference between the two maps has norm at most 4a_n(x). Hence

For fixed x,x', take N\to\infty in (5.16). This proves the two-sided bounds (5.1), since 1-4c=24/25 and 1+4c=26/25. In particular, h is injective. It fixes zero by (5.5).

We now prove surjectivity. Fix y=(y_n)\in U and, using Lemma 5.4, put

Freeze the slots of h_N at this particular input:

The directions may vary with N. Nevertheless y=L_{\xi^{(N)}}x_N is an exact equality for each N, so Lemma 5.2 gives

The component y_m-\beta_m^{(N)}\xi_m^{(N)} is perpendicular to \xi_m^{(N)}. Thus its squared norm is \|y_m\|^2-|\beta_m^{(N)}|^2, and the last term in the U_m formula adds orthogonally.

For every integer s\ge1, it follows that

Here |\beta_n^{(N)}|\le\|y_n\|, so all squared-coefficient series converge. The last equality is exact: the even and odd indices 2m,2m-1 for m\ge s partition the integers n\ge2s-1; their sum cancels that tail of \sum_{n\ge s}|\beta_n^{(N)}|^2. The only uncancelled negative terms have s\le n\le2s-2. This finite sum is empty when s=1. Keeping it is what avoids double counting the output’s perpendicular and slot components.

The right side of () tends to zero with s, independently of N. At s=1 the equality gives \|x_N\|=\|y\|. For each s, the heads

lie in a bounded subset of the finite-dimensional space \bigoplus_{m<s}(U_m\oplus\R). By successive subsequence selection in these finite-dimensional spaces and a diagonal choice, there is an increasing sequence N_j for which every fixed head converges.

This diagonal subsequence is Cauchy in the full norm. Given \epsilon>0, choose s so that \sum_{n\ge s}\|y_n\|^2<(\epsilon/4)^2. By (), the tail of each of x_{N_j} and x_{N_k} has norm less than \epsilon/4. For sufficiently large j,k, convergence of the fixed head makes its difference smaller than \epsilon/2. The triangle inequality then gives \|x_{N_j}-x_{N_k}\|<\epsilon. Completeness of M yields x_{N_j}\to x in norm for some x\in M.

Finally, apply the pointwise convergence estimate at this fixed limit x, and use the common upper Lipschitz bound for h_{N_j}:

Thus h(x)=y. Since y was arbitrary, h is onto.

The local orthogonality asserted in the theorem follows from Lemma 5.3. For its neighborhoods O_j and cutoff l, let H_j be the closed span of the union of coordinate supports of w_n(x) with x\in O_j and n\ge l. These are mutually orthogonal coordinate subspaces of U_{\ge l}. Each block D(x)_n is a scalar multiple of w_n(x) by (5.4). Its convergent tail therefore belongs to H_j for every x\in O_j, because H_j is closed. This verifies exactly (4.1) and completes the proof.□

Proof of Theorem 1.1. Take the real Hilbert spaces U, V=\ell_2, M=U\oplus_2V and the map h:M\to U from Theorem 5.1. The blocks U_n are finite dimensional, so U and M are separable; as Hilbert spaces they are complete. Put

The map D is Lipschitz, D(0)=0, and it has the finite-family tail-coordinate orthogonality required by Theorem 4.1. That theorem makes q completely continuous. By (2.3) and (5.1),

Set

All hypotheses of Proposition 3.1 hold with a=24/25, b=26/25. It supplies the coordinatewise bijection \Psi:X\to Y, an isometric copy of c_0(\ell_2) in X, and exclusion of every linearly isomorphic copy from Y. In particular X and Y are not linearly isomorphic.

Since L_q\le51/25 and K_q=\max\{L_q,1\}\le51/25, (3.1) gives

The latter inequality is equivalent to the lower bound \|\Psi(s)-\Psi(t)\|_Y\ge(4/21)\|s-t\|_X. Both bounds hold for all pairs s,t\in X, exactly as in (1.1).□