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“Strange” Limits

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“Strange” Limits c Gabriel Nagy “Strange” Limits Notes from the Functional Analysis Course (Fall 07 - Spring 08) Prerequisites: The reader is assumed to be familiar with the Hahn-Banach Theorem and/or the theory of (ultra)filter convergence. References to some of my notes will be added later. Notations. Given a non-empty set S, we denote by `∞(S) the vector space of all bounded R functions x : S → . On occasion an element x ∈ `∞(S) will also be written as an “S-tuple” R R x = (xs)s∈S. If S is infinite, we denote by Pfin(S) the collection of all finite subsets of S. Assuming S is infinite, for an element x = (x ) ∈ `∞(S), we can define the quantities s s∈S R lim sup xs = inf sup xs , F ∈P (S) S fin s∈SrF lim inf xs = sup inf xs . S s∈S F F ∈Pfin(S) r Definition. Suppose S is infinite. A limit operation on S is a linear map Λ : `∞(S) → , R R satisfying the inequalities: lim inf x ≤ Λ(x) ≤ lim sup x , ∀ x = (x ) ∈ `∞(S). (1) s s s s∈S R S S Comment. Of course, the quantities lim supS xs and lim infS xs can be defined for ar- bitrary “S-tuples,” in which case these limits might equal ±∞. In fact, this allows one to use S-tuples which might take infinite values. In other words, one might consider maps x : S → [−∞, ∞]. We will elaborate on his point of view later in this note. With the above terminology, our problem is to construct limit operations. The key difficulty arises from the linearity requirement. Remark. In order to produce a limit operation Λ : `∞(S) → , besides the linearity of R R Λ, all we need is half of (1), namely Λ(x) ≤ lim sup x , ∀ x = (x ) ∈ `∞(S). (1’) s s s∈S R S (Note that lim infS xs = − lim supS(−xs).) Below we illustrate two methods of finding limit operations. The reader is warned that neither of them is constructive! Both methods are based on the Axiom of Choice/Zorn Lemma. Method 1: Hahn-Banach Approach. Consider the map q : `∞(S) → , defined by R R q(x) = lim sup x , ∀ x = (x ) ∈ `∞(S), (2) s s s∈S R S and the space (S) = x = (x ) ∈ `∞(S) : lim inf x = lim sup x . CvgR s s∈S R s s S S The three key observations are as follows: 1 (i) q is a quasi-seminorm on `∞(S), i.e. R • q(x + y) ≤ q(x) + q(y), ∀ x, y ∈ `∞(S); R • q(tx) = tq(x), ∀ x ∈ `∞(S), t ∈ [0, ∞). R (ii) Cvg (S) is a linear subspace of `∞(S); R R (iii) the restriction Λ0 = q Cvg (S) : CvgR(S) → R is linear. R Since Λ0(x) ≤ q(x), ∀ x ∈ CvgR(S), (in fact we have equality!), by the Hahn-Banach Theorem, Λ0 can be extended to a linear map Λ : `∞(S) → , satisfying R R Λ(x) ≤ q(x), ∀ x ∈ `∞(S), R which is exactly (1’) Method 2: Limits along ultrafilters. Fix U a non-constant ultrafilter on S. For every x = (x ) ∈ `∞(S), let K(x) denote the closure of the set {x : s ∈ S}, so we can s s∈S R s now regard x as a function x : S → K(x). Since (xs)s∈S is bounded, the set K(x) ⊂ R is compact. If we consider then the push-forward −1 x∗U = {V ⊂ K(x): x (V ) ∈ U}, we know that this is an ultrafilter in K(x), so by compactness x∗U converges to some (unique) U point in K(x). This point – uniquely determined by x and U – is denoted by limS xs and is referred to as the limit of (xs)s∈S along U. This way we produce a map lim U : `∞(S) → . (3) S R R The map (3) is not just linear, but also multiplicative. The proof of these two statements is left as an exercise. Finally, the map (3) is in fact a limit operation. By previous discussion, it suffices to show that lim U x ≤ lim sup x , ∀ x = (x ) ∈ `∞(S). (1’) s s s s∈S R S S To prove this, we argue by contradiction. Assume there is some “S-tuple” (xs)s∈S with U limS xs > lim supS xs. Pick then some real number α, such that U lim xs > α > lim sup xs. (4) S s∈S U On the one hand, the set M = (α, ∞) ∩ K(x) is a neighborhood of limS xs in K(x), so since U we assume that x∗U converges to limS xs, it follows that M belongs to x∗U, i.e. the set −1 A = x (M) = {s ∈ S : xs > α} (5) 2 belongs to U. On the other hand, by the definition of lim sup (as an infimum), there exists some finite set F ⊂ S, such that sup xs < α, s∈SrF which obviously entails xs < α, ∀ s ∈ S r F. (6) This clearly forces the set A, defined in (5) to be contained in F , thus A must be finite. But this is impossible, since non-constant ultrafilters do not contain finite sets. Remark. Suppose x = (xs)s∈S is an arbitrary “S-tuple” possibly taking infinite values. This means that xs ∈ [−∞, ∞]. We can still think of [−∞, ∞] as a compact space, so for U an non-constant ultrafilter U on S we can still define limS xs, this time thinking of x∗U as an ultrafilter on [−∞, ∞]. This “outer” limit operation will still satisfy U lim inf xs ≤ lim xs ≤ lim sup xs. (7) S S S for all [−∞, ∞]-valued “S-tuples” x = (xs)s∈S. Exercises. Before continuing, the reader should work all problems below. 1. Prove that the “outer” limit defined above agrees with the one outlined in Method 2 on bounded “S-tuples.” In other words, the compact sets K(x) are not needed: they can all be replaced by [−∞, ∞] U 2. Prove that if x = (xs)s∈S is constant, i.e. xs = k ∈ [−∞, ∞], then limS xs = k, for every non-constant ultrafilter U. 3. Prove that limU : `∞(S) → is linear and multiplicative. S R R U 4. Prove that the usual Calculus rules involving infinite limits are also valid for the limS operations. 5. Prove that in the case when S = N, the definitions of lim supS and lim infS agree with the traditional ones: lim supn→∞ and lim infn→∞. Application: Banach Limits. From this point on we specialize to the case S = N. In this situation we will deal with the space of all bounded real-valued sequences, which is simply denoted by `∞. One operator is of a particular interest for us: the backward shift R S : `∞ → `∞ defined by R R S(x1, x2, x3,... ) = (x2, x3, x4,... ). A limit operation Λ : `∞ → is called a Banach limit, if R R Λ(Sx) = Λ(x), ∀ x ∈ `∞. (8) R With this terminology, Banach’s Limit Theorem simply states the existence of Banach limits. 3 First Proof. Consider the average operator A : `∞ → `∞ defined by R R x + x x + x + x A(x , x , x ,... ) = x , 1 2 , 1 2 3 ,... 1 2 3 1 2 3 Using the quasi-seminorm q defined in (2) we have the inequality q(Ax) ≤ q(x), ∀ x ∈ `∞. (9) R x1+x2+···+xn (This is Cesaro’s Theorem, which states that lim supn→∞ n ≤ lim supn→∞ xn.) If we start now with a limit operation Λ , and we define Λ = Λ ◦ A, then Λ : `∞ → is still 0 0 R R linear, and by (9) it still satisfies the inequality Λ(x) ≤ q(x), ∀ x ∈ `∞, (10) R thus Λ will still be a limit operation. In order to prove (8), start with some sequence x = (x )∞ , so that the sequence y = Sx − x ∈ `∞ is given by y = x − x , and then n n=1 R n n+1 n ∞ the averaged sequence z = Ay is given by zn = (xn+1 − x1)/n. Since (xn)n=1 is bounded, we clearly have limn→∞ zn = 0, so Λ0(z) = 0. Of course, by the construction of Λ this means that Λ(y) = Λ0(Ay) = 0, which then simply reads: Λ(Sx − x) = 0, ∀ x ∈ `∞. (11) R By the linearity of Λ, this yields the desired conclusion (8). Second Proof. Let us forget now everything about the availability of limit operations! Start- ing from scratch, we aim at directly constructing a linear map Λ : `∞ → , satisfying (10) R R and (11). If we consider the linear subspace Y = Range(S − Id) = {Sx − x : x ∈ `∞}, R then all we need to do is check the condition: 0 ≤ q(y), ∀ y ∈ Y. (12) Indeed, if the above inequality holds, then Λ can be any linear extension of the zero map 0 : Y → R satisfying (10), and the existence of such extensions is a consequence of the Hahn-Banach Theorem. (The condition Λ Y = 0 is exactly condition (11).) To check (12), all we need to do (see the notations from the first proof) is prove the inequality lim sup(x − x ) ≥ 0, ∀ x = (x )∞ ∈ `∞. (13) n+1 n n n=1 R n→∞ But this inequality follows trivially from Cesaro’s Theorem arguing by contradiction: if there is some x = (x )∞ ∈ `∞, such that the quantity L = lim sup (x −x ) is negative, then n n=1 R n→∞ n+1 n xn xn lim supn→∞ n ≤ L < 0, which is impossible, since boundedness forces limn→∞ n = 0. 4.
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