Fixed Points of Nonexpansive Maps in Reflexive Banach Spaces

Proof. For the empty family, take R=\operatorname{id}_C. Otherwise, Theorem 1.1 applies to every nonempty closed bounded convex subset D of C, since a relatively norm-closed subset of the norm-closed set C is norm closed in X. Thus C has Bruck’s hereditary fixed point property and, by taking D=C, the fixed point property. If f:C\to C is nonexpansive and such a subset D is f-invariant, the same theorem applied to f|_D gives a fixed point in D. Hence C also has Bruck’s conditional fixed point property. The ordinary fixed point and conditional fixed point properties just proved, together with weak compactness of C by reflexivity, satisfy the corresponding hypotheses of Bruck [3, Theorem 1]. That theorem gives the claimed nonempty nonexpansive retract.□

Proof. Start with a fixed-point-free map F:C\to C as in the theorem and a point a\in C. Set C_0=\{a\} and C_{n+1}=\overline{\conv}(C_n\cup F(C_n)). These sets are increasing and separable: continuous images of separable sets are separable, and rational convex combinations of a countable dense set are dense in its convex hull. Hence C'=\overline{\bigcup_n C_n} is separable, bounded, closed, and convex. Norm continuity of F gives F(C')\subseteq C', and Lemma 2.1 makes C' weakly compact.

Every decreasing chain of nonempty norm-closed convex invariant subsets of C' has a nonempty intersection, by weak compactness and the finite intersection property. The intersection is again closed, convex, and invariant. Zorn’s Lemma, with inclusion reversed, therefore supplies a minimal such set K. Put T=F|_K. Its diameter d is positive, since a singleton invariant set would give a fixed point. From now on express distances in units of d, writing \norm{\cdot} for the original norm divided by d. This scalar normalization preserves the topologies, nonexpansiveness, and fixed points; it makes \diam K=1.

Choose a norm-dense sequence (z_i)_{i\geq1} in K. Its pairwise differences have dense linear span in Y, so Y is separable. Choose a dense sequence (u_j) in the unit sphere of Y and norming functionals g_j\in X^* for these vectors. They separate points of Y: if \norm{v}=1 and \norm{v-u_j}<1/2, then g_j(v)>1/2. Thus the map a\mapsto(g_j(a))_{j\geq1} continuously injects weakly compact K into \R^{\N}. The latter is metrizable, for example by \sum_j2^{-j}\min\{1,\abs{s_j-t_j}\}. The injection is a homeomorphism onto its image, since compact subsets of a Hausdorff space are closed. Hence the weak topology on K is metrizable. Finally B_Y is a norm-closed bounded convex subset of X, so is weakly compact by Lemma 2.1. Its inherited weak topology equals its intrinsic weak topology: restriction gives one inclusion of continuous functionals, and the extension argument in Lemma 2.1 gives the other.□

Proof. First K=\overline{\conv}T(K): this hull is nonempty and contained in K, and its image under T is contained in T(K) and hence in the hull. Minimality applies. Consider on K the functions

Both are convex and 1-Lipschitz. For a fixed center, the distance function is convex and continuous, so its supremum is unchanged on passing to a norm-closed convex hull. The hull identity and nonexpansiveness give

where the second inequality also uses \norm{Tu_j-u_j}\to0. Every nonempty sublevel set of either function is norm closed, convex, and invariant. If the function had two different values, a sublevel strictly between them would contradict minimality. Both functions are therefore constant, and s=1 because its supremum is \diam K. Write r=R\leq1.

Take the weak \mathcal U-limit a\in K of (u_j). For every z\in K and \eta>0, all sufficiently large j satisfy \norm{z-u_j}\leq R+\eta. The corresponding closed ball is weakly closed and the cofinite sets belong to \mathcal U, so \norm{z-a}\leq R+\eta. Letting \eta decrease to zero and taking the supremum over z gives 1=s(a)\leq R, hence R=1. This argument applies to every subsequence of (u_j), since each is again an approximate fixed point sequence. As all distances are at most 1, a failure of the asserted limit would give a subsequence bounded above by 1-\eta for some \eta>0, contradicting its limsup of 1.□

Proof. Choose an \varepsilon/2-net a_1,\ldots,a_m in E and set a=m^{-1}\sum_{i=1}^m a_i\in K. Lemma 2.3 implies that, for some \delta>0,

Indeed, otherwise choosing a violating point with defect at most 1/j for each j would contradict that lemma. Choose a norming functional f for y-a. The numbers f(y-a_i) are all at most 1, whereas their sum is m f(y-a)\geq m-\varepsilon/2. Each is therefore at least 1-\varepsilon/2. For z\in E, take a_i with \norm{z-a_i}<\varepsilon/2 and use \norm{f}\leq1 to obtain the claimed bound.□

Proof. For n=1, the defining equation gives R_1(a)=a. For n>1, the map

takes the complete set K into itself and has Lipschitz constant q=1-1/n<1. Starting from any point of K, its iterates have successive displacements bounded by q^r times the first displacement. The geometric series shows that the iterates are Cauchy. Their limit solves (3.1) by norm continuity. Two solutions have distance at most q times their distance and therefore coincide.

For u=R_n(a) and v=R_n(b), comparison of their equations gives

which proves nonexpansiveness. The identity Tu-u=(Tu-a)/n and \diam K=1 give the first estimate in (3.2). For the second, the defining equation and nonexpansiveness of T give

Rearranging proves the claim.□

Proof. For each fixed pair a,b\in K, the condition \norm{h(a)-h(b)}\leq\norm{a-b} is closed in the pointwise weak topology: evaluation at a,b and subtraction are weakly continuous, and norm balls are weakly closed. Intersecting these conditions over all pairs proves the first assertion.

For the second, consider the convex subset

The tuple (h(a_1),\ldots,h(a_q)) belongs to its weak closure, by continuity of evaluation. The weak topology of the finite Banach product X^q is the product of its coordinate weak topologies: each continuous linear functional on X^q is the sum of its coordinate restrictions. By separation, the weak and norm closures of a convex subset of X^q agree. Applying this fact with the maximum coordinate norm gives one element of V satisfying all the desired inequalities.□

Proof. The intersection of a chain of nonempty compact subsemigroups of S is nonempty by compactness and is again a compact subsemigroup. Zorn’s lemma therefore gives a minimal nonempty compact subsemigroup S'\subseteq S. Fix p\in S'. The set S'p is a nonempty compact subset of S' by continuity of right translation. It is a subsemigroup because

Minimality gives S'p=S', so in particular there is an h\in S' with hp=p. The set

is thus nonempty. It is compact by continuity of right translation and is a subsemigroup, since hp=kp=p implies (hk)p=h(kp)=hp=p. Minimality gives E=S', and p\in E proves p^2=p.□

Proof. The sets A_m are nonempty, compact, and decreasing. Every member is nonexpansive by Lemma 3.2.

First, \conv\mathcal H_m is closed under composition. Indeed, if

then k\in\mathcal G and

This identity distributes precomposition over the outer convex combination.

To pass to A_m, fix h,k\in A_m, arguments a_1,\ldots,a_q\in K, and \delta>0. Lemma 3.2 gives k_\delta,h_\delta\in\conv\mathcal H_m with

The second application uses the fixed arguments k(a_i). Nonexpansiveness of h_\delta then gives

The composition h_\delta\circ k_\delta belongs to \conv\mathcal H_m. Since the finite arguments and \delta were arbitrary, these simultaneous norm approximations imply h\circ k\in A_m.

Thus A is a nonempty compact semigroup, by the nested intersection property. For fixed k\in A, the right translation h\mapsto h\circ k is continuous in the pointwise weak topology: its value at an argument a is the evaluation of h at the fixed argument k(a). Lemma 3.3 supplies P\in A with P\circ P=P. Choose any a\in K and set x=P(a); then P(x)=x.

Write

These are nonempty nested weakly compact subsets of K, so D_x=\bigcap_m B_m is nonempty and weakly compact. Since P\in A_m, Lemma 3.2 at the single argument x shows that

Consequently, for each \ell\in X^*,

The maximum exists because B_m is weakly compact. The sets B_m\cap\{z:\ell(z)\geq\ell(x)\} are nonempty nested compact sets; their intersection gives a point of D_x at which \ell\geq\ell(x). Hence

If (3.8) failed, separation from the norm-closed convex hull of D_x would give a functional violating this inequality. This proves the proposition.□

Proof. Choose d_1,\ldots,d_q\in D_x and nonnegative coefficients \lambda_1,\ldots,\lambda_q summing to 1 such that

Fix a metric \rho inducing the weak topology on K. Cycle through the targets d_1,\ldots,d_q: at step i, let r(i) be the member of \{1,\ldots,q\} congruent to i modulo q. Once the threshold m_i has been prescribed, the membership

allows a choice G_i\in\mathcal H_{m_i} with

Any auxiliary choices associated with this map may now be made before the next threshold is prescribed. Stop as soon as the convex hull of the selected images comes within distance \tau of x.

Suppose that the procedure never stops, and let

For each r, the subsequence assigned to target d_r converges weakly to d_r. The set C_0 is weakly closed by separation, so it contains every d_r and their convex combination \sum_r\lambda_rd_r. By the definition of C_0, a finite convex combination of the selected images is within \tau/2 of this last point. It is therefore within \tau of x. All its indices occur in some finite prefix of the selection, contradicting the stopping rule. The procedure thus terminates, and a convex combination at the stopping step supplies (3.9). Deleting its zero coefficients leaves the stated positive list.□

Proof. Start with the root and \tau_0=1/2. Number all candidate selections throughout the construction by \ell=1,2,\ldots, including candidates later discarded. At selection number \ell, assign the budget

before choosing the candidate’s map. The sum of all such budgets is at most 1/256, which is more than sufficient for Equation (4.6) after discarding any candidates.

Suppose the finite levels through L_{k-1} have been constructed and \tau_{k-1} has been fixed. Visit the parents in L_{k-1} from left to right. For a given parent, apply Lemma 3.5 with tolerance \tau_{k-1}. Its successive map choices will be the candidates for this parent’s children. We now specify the threshold to be imposed at every one of these choices.

Let v be the next candidate. Its budget has just been assigned. Using all earlier candidates at the present level, form I_v, \Theta_v, Q_v, and E_v as above. The ancestor maps are fixed, so the number

is finite; take c_{G_r}=0. By Lemma 2.4, there is \delta_v>0 such that any y\in K with \norm{Ty-y}<\delta_v admits a functional f\in X^*, \norm{f}\le1, satisfying f(y-z)\ge1-e_v for every z\in E_v. Choose the integer threshold m so large that

This choice uses only data already available, and is made before selecting G_v.

Require the next map in the adaptive selection to belong to \mathcal H_m. Every such choice has \norm{Ty_v-y_v}\le1/m by Equation (3.5). Equation (3.3) then gives

After selecting G_v, choose the functional f_v given by the compact detector. Both requirements in Equation (4.9) now hold. In particular, they hold for every map choice allowed by this step’s threshold, so they are compatible with the weak-neighborhood choices used in Lemma 3.5.

That lemma terminates after finitely many candidates for the parent under consideration. Only then choose the convex coefficients whose average of the points y_v lies within \tau_{k-1} of x. Discard the candidates with zero coefficient and retain the others, in their original order, as this parent’s children; their positive coefficients are the weights \alpha_v. The retained list is nonempty, and its weights sum to one. Keep the labels of discarded candidates in the finite lists used to form subsequent predictions. Thus no already defined prediction or functional changes when zero coefficients are discarded.

There are only finitely many parents, so this completes a finite, nonempty level L_k. Every parent in L_{k-1} now satisfies Equation (4.8). Define \tau_k by Equation (4.7); its defining minimum is over a nonempty finite set of positive numbers. Proceed to the next level. This constructs the entire tree and proves the stated properties.□

Proof. Fix N. All the maps, weights, functionals, and budgets in \mathcal T_N are already fixed. Put

Choosing a common terminal point. For each v\in\mathcal T_N\setminus\{r\}, use the parameter \theta^v\in\Theta_v from Section 4: it assigns the actual weight \alpha_h to each earlier retained candidate h\in I_v, and zero to each earlier discarded candidate. The point

belongs to E_v by (4.5). By (4.9), f_v(y_v-q_v)\geq 1-e_v. The definition of this weak ultrafilter limit therefore gives

There are only finitely many such conditions. Moreover, (3.3) and the approximate fixed point property of (b_j) give

Intersecting the finitely many ultrafilter members above with a suitable cofinite set, choose one j and write b=b_j so that

The switching experiment. Initially assign the output H_s=y_s to each leaf s\in L_N. Switch these outputs one at a time to H_s=b, proceeding from right to left. At each stage calculate every other output by (4.1). All outputs belong to K. For v\in\mathcal T_N\setminus\{r\}, let U_v and V_v be the root outputs immediately before and immediately after the whole block B_N(v) is switched.

We need two estimates for subtrees whose leaves are all in the same state. If none of the leaves below h\in L_k have switched, then

Here and below an empty sum is zero. At a leaf this is an equality. For an internal node, nonexpansiveness and (4.8) give the induction step

The tolerance schedule (4.7) yields, for k\geq 1 and every v\in L_k,

Thus the error in (5.3) is at most e_v for every unchanged node at depth k, regardless of which v\in L_k is used in the comparison.

If all the leaves below h have switched, put \delta_u=\norm{G_u(b)-b}. At each internal node of that subtree,

The error at each leaf is zero. Iteration gives

The coefficients W_u/W_h are products of weights below h and are at most one.

Detection of an entire block. Fix v\in L_k, where 1\leq k\leq N, and consider the stage immediately before its block switches. Its output is within e_v of y_v by (5.3)–(5.4). Let p_0=r,p_1,\ldots,p_k=v be the path to v, and put A_i=\norm{H_{p_i}-y_v} at this stage. Since every off-path child output and y_v lie in the diameter-one set K, nonexpansiveness gives

The off-path contributions telescope:

Unwinding the recurrence and using (4.9), we obtain

Immediately after the block has switched, precisely the level-k nodes strictly to the left of v have unchanged leaf blocks. By (5.4), their outputs are within e_v of their labels y_h. The outputs at all the other level-k nodes are within M_N\leq e_v of b by (5.5). The ideal level-k assignment consisting of these y_h on the left and b elsewhere has root output Q_v(b;\theta^v) by (4.10). The sensitivity estimate (4.2) and \sum_{h\in L_k}W_h=1 therefore imply

Together, (5.1), (5.6), and (5.7) give

A common witness along one path. For a single leaf s, only its output changes during its switch. By (4.2),

Since W_s>0 and both root outputs belong to K, the vector

belongs to B_Y. The successive root differences over a consecutive block telescope, so

Applying f_v to this identity and using (5.8) yields

Each summand is nonnegative, because \norm{f_v}\leq 1 and \norm{d_s}\leq 1. Summing over the non-root nodes and interchanging the two finite sums gives

If a leaf s has f_v(d_s)\leq 1/2 for any non-root node v on its path, its inner sum in (5.10) is at least 1/2. Consequently, the total weight of all such leaves is at most 5/64. Since the leaf weights sum to one, some leaf has f_v(d_s)>1/2 for every non-root node v on its path. This proves the proposition.□

Proof of Theorem 1.1. Suppose a counterexample exists. The reduction and constructions in Sections 2–4 give the fixed infinite tree to which Proposition 5.1 applies. Call a node v feasible if some d\in B_Y satisfies

The root is feasible, and every ancestor of a feasible node is feasible. Proposition 5.1 gives a feasible node at every depth. These nodes therefore form a finitely branching tree with arbitrarily great depth. Starting from the root, choose a child having feasible descendants at arbitrarily great depths, and repeat. Such a child always exists because each node has finitely many children. This produces a branch r,v_1,v_2,\ldots, with v_k\in L_k, all of whose nodes are feasible.

For each k\geq 1, the set

is nonempty and weakly closed in B_Y, and C_{k+1}\subseteq C_k. The normalized differences used in Proposition 5.1 are controlled in the whole unit ball B_Y. Reflexivity supplies weak compactness of this ball, as established in Section 2. It follows that there is one vector

On the other hand, the functionals on every branch converge pointwise to zero on Y. Indeed, for fixed i,j and k\geq\max\{i,j\}, (4.9) and \diam K=1 give

Subtracting these inequalities yields

where the convergence follows from the summability of the budgets. The linear span of the differences z_i-z_j is dense in Y=\overline{\operatorname{span}}(K-K). Hence the bound \norm{f_{v_k}}\leq 1 extends this convergence to every d'\in Y: for any vector a in that span,

and one first lets k tend to infinity and then lets a approach d'. In particular f_{v_k}(d)\to 0, contradicting (5.11). The assumed counterexample cannot exist.□

Proof of Lemma 2.1. We begin with the Hahn–Banach extension argument underlying the first two assertions. Suppose that a linear functional \ell on a subspace M is dominated by a continuous sublinear function p on X. To extend it to M+\R b, where b\notin M, choose its value at b in the interval

This interval is nonempty because \ell(a)+\ell(c)\leq p(a+c)\leq p(a-b)+p(c+b). Its endpoints are finite, as is seen by setting a=0 or c=0 in this inequality. The displayed bounds give domination on a+b and a-b; positive homogeneity then gives domination on a+tb for every real t. Unions of chains of dominated extensions remain dominated extensions, so Zorn’s Lemma gives an extension to X. It is continuous because domination of both v and -v bounds its absolute value near zero. Applying this construction with p=\norm{\cdot} and \ell(tv)=t\norm{v} gives a norming functional (use the zero functional when v=0). In particular,

For separation, let q lie outside a nonempty norm-closed convex set D. Choose \rho>0 such that the open convex set U=D+\{v:\norm{v}<\rho\} still excludes q, and fix u\in U. The gauge

is finite, nonnegative, positively homogeneous, and subadditive. For the last property, if v\in s(U-u) and w\in t(U-u), convexity gives v+w\in(s+t)(U-u), and one takes infima. Since U-u is open, convex, and contains zero, \{v:p(v)<1\}=U-u: one inclusion follows from convexity, and the other by extending a little past a given point along its ray using openness. A ball about zero contained in U-u gives p(v)\leq c\norm{v} for some c; subadditivity then makes p continuous. On the line through q-u, the functional taking the value p(q-u) at q-u is dominated by p; for negative multiples this follows from -p(q-u)\leq p(u-q). Its extension f therefore satisfies

In particular f\ne0, and taking the supremum over the ball of radius \rho added to any d\in D gives f(d)+\rho\norm{f}\leq f(q). Thus f separates q from D with a positive gap. This proves weak closedness of D. Applied to closed balls it proves weak lower semicontinuity of the norm. Separation also shows that the weak closure of a convex set equals its norm closure. On X^r, equipped with the maximum norm, every continuous linear functional has the form \sum_{i=1}^r f_i(x_i), with f_i\in X^*, by restriction to the coordinate subspaces. This proves the assertion about the finite-product topology and allows the same separation argument there.

Here are the compactness details. A proper filter is a family of subsets closed under finite intersections and supersets, containing the whole space but not the empty set. By Zorn’s Lemma it extends to a maximal proper filter, called an ultrafilter. For every subset an ultrafilter contains either that subset or its complement: if the subset cannot be added while preserving the filter property, an existing filter member misses it. An ultrafilter converges to a point when it contains every neighborhood of that point. In a compact space it converges: the closures of its members have the finite intersection property, and any point in their intersection is a limit by the preceding either-or property. Limits are unique in a Hausdorff space. Conversely, if every ultrafilter converges, extend any family of closed sets with the finite intersection property to an ultrafilter; its limit belongs to all those sets. This is the closed-set criterion for compactness. An ultrafilter on a product of compact spaces has a limit in each coordinate, and the tuple of these limits is its limit, since a basic product neighborhood restricts only finitely many coordinates. This proves product compactness.

Now let D\subseteq X be bounded, norm closed, and convex, with X reflexive. Given an ultrafilter \mathcal V on D, each bounded real evaluation has a limit in a compact interval, and

defines a linear functional with \abs{L(f)}\leq M\norm{f}, where M=\sup_{d\in D}\norm{d}. Linearity follows by intersecting the filter members on which each of two evaluations is close to its limit. Reflexivity supplies a\in X with L(f)=f(a) for every f. Thus \mathcal V converges weakly to a, and weak closedness gives a\in D. The ultrafilter criterion proves weak compactness.□