A Simple Inductive Proof of Levy-Steinitz Theorem

Home , University of Lviv

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Without loss of generality, each xn is not equal to zero. Also we can assume that the norm of X is generated by a scalar product h·, ·i, so X is a finite-dimensional Hilbert space, whose dual X∗ can be identified ∞ with X. Under such identification, the set Γ coincides with the set {y ∈ X : Pn=1 |hy, xni| < ∞}. Let S = {x ∈ X : kxk =1} denote the unit sphere of the Banach space X. ∞ Claim 1. The sequence (xn)n=1 tends to zero. ∞ N Proof. Assuming that (xn)n=1 does not tend to zero, we can find an ε> 0 such that the set E = {n ∈ : kxnk≥ ε} is infinite. By the compactness of the unit sphere S = {x ∈ X : kxk =1} the sequence {xn/kxnk}n∈E ⊂ S has an accumulating point x∞ ∈ S. Then for the linear functional f : X → R, f : x 7→ hx, x∞i, the series ∞ Pn=1 f(xn) is not potentially convergent as its terms do no tend to zero. But this contradicts our assumption ∞ (that Pn=1 xn is ∗-potentially convergent). divergence direction ∞ A point x of the sphere S is defined to be a of the series Pn=1 xn if for every neighborhood U ⊂ S of x the set xn NU := {n ∈ N : ∈ U} kxnk is infinite and the series kx k is divergent. n∈NU n P ∞ Let D ⊂ S be the (closed) set of all divergence directions of the series Pn=1 xn. If D is empty, then the (compact) sphere S admits a finite cover U by open subsets U ⊂ S for which the series kx k is n∈NU n ∞ ∞ P convergent. Then kxnk ≤ N kxnk < ∞, so the series xn is absolutely convergent, Pn=1 PU∈U Pn∈ U Pn=1 Γ= X and the set Σ is a singleton, equal to Σ =Σ+ {0} =Σ+Γ⊥. So, assume that the (compact) set D is not empty. We claim that the convex hull of D contains zero. In the opposite case we can apply the Hahn-Banach Theorem and find a linear functional f : X → R such that ∞ f(D) ⊂ (0, +∞) hence the series Pn=1 f(xn) fails to converge potentially, which contradicts our assumption. This contradiction shows that zero belongs to the convex hull of the set D. Let D0 be a subset of smallest (finite) cardinality containing zero in its convex hull conv(D0) and let X0 be the linear hull of D0 in X. The minimality of D0 ensures that the set D0 is affinely independent and zero is contained in the interior of the convex hull conv(D ) in X . Let X := {x ∈ X : hx, yi =0} be the orthogonal complement of X in X, 0 0 1 Ty∈X0 0 and let pr0 : X → X0 and pr1 : X → X1 be the orthogonal projections.

Claim 2. For every z ∈ D0 there exists a subset Ωz ⊂ N such that

(1) kpr (xn)k < ∞; Pn∈Ωz 1 (2) for each neighborhood U ⊂ S of z the series N kxnk is divergent. Pn∈Ωz∩ U 1 Proof. For every k ∈ ω consider the neighborhood Ok = {x ∈ S : kx − zk < 2k } of z in S and observe that each point x ∈ Ok has 1 kpr (x)k = kpr (x − z)+pr (z)k = kpr (x − z)+0k≤kx − zk < . 1 1 1 1 2k xn Since z ∈ D0 ⊂ D, the set NO = {n ∈ N : ∈ Ok} is infinite and the series N kxnk is divergent. k kxnk Pn∈ O N k Using Claim 1, we can find a finite subset Fk ⊂ Ok such that

k< X kxnk < k +1. n∈Fk

We claim that the set Ωz = Sk∈ω Fk has the required properties. Indeed, ∞ ∞ xn kpr (xn)k≤ kpr (xn)k = pr ( ) ·kxnk < X 1 X X 1 X X 1 kxnk

n∈Ωz k=0 n∈Fk k=0 n∈Fk ∞ ∞ ∞ 1 1 k +1 < kx k = kx k < < ∞. X X 2k n X 2k X k X 2k k=0 n∈Fk k=0 n∈Fk k=0 So, the first condition is satisfied. To check the second condition, take any neighborhood U ⊂ S of z and find N N k ∈ ω such that Ok ⊂ U. The latter inclusion implies Fm ⊂ On ⊂ U for all m ≥ k and hence

X kxnk≥ sup X kxnk≥ sup m = ∞, m≥k m≥k n∈NU n∈Fm

so the series N kxnk is divergent. Pn∈ U

Claim 2 implies that the set Ω = Ωz has the following properties: Sz∈D0 A SIMPLE INDUCTIVE PROOF OF LEVY-STEINITZ´ THEOREM 3

(Ω1) Pn∈Ω kpr1(xn)k < ∞; (Ω2) for each x ∈ D0 and each neighborhood U ⊂ S of x the series N kxnk is divergent. Pn∈Ω∩ U

Claim 3. There exists a positive constant C such that for every x ∈ X0, ε > 0 and ﬁnite set F ⊂ Ω, there exists a ﬁnite set E ⊂ Ω \ F such that and (1) kx − Pn∈E xnk <ε ′ ′ for any subset . (2) Pn∈E xn ≤ C · max{kxk,ε} E ⊂ E

Proof. Since the interior of the convex hull of the set D0 contains zero, it also contains some closed ball 1 {x ∈ X0 : kxk ≤ δ} of positive radius δ < 4 . Replacing δ by a smaller number, we can assume that Cδ kx − yk > 2δ for any distinct points x, y ∈ D0. We claim that the constant C = δ where

Cδ = sup X tzxz : ∀z ∈ D0 (tz ∈ [0, 1], kxz − zk <δ) z∈D0 satisfies our requirements. Indeed, fix any x ∈ X, ε> 0, and a finite set F ⊂ Ω. Replacing ε by a smaller number, we can assume that 1 ε<δ< 4 . Let 1 c := · max{kxk,ε}. δ x x It follows that k c k ≤ δ and hence the point c belongs to the interior of the convex hull of the affinely x independent set D in X . So, = tzz and hence x = ctzz for some sequence (tz)z D of 0 0 c Pz∈D0 Pz∈D0 ∈ 0 positive real numbers with tz = 1. Pz∈D0 ε For every z ∈ D0 consider the spherical disk Sz = x ∈ S : kx − zk < min{δ, 2c } and its convex cone ˆ 1 ′ ′ ˆ Sz = {t · s : t > 0, s ∈ Sz}. Since δ < 2 kz − z k for any distinct points z,z ∈ D0, the cones Sz, z ∈ D0, are pairwise disjoint. Observe also that for any elements x, y ∈ Sz, we have the lower bound 1 hx, yi = hx, xi + hx, y − xi≥ 1 −kx − yk > 1 − 2δ ≥ . 2

Then for any elements x, y ∈ Sˆz we have the inequality 1 (1) kx + yk ≥ hx + y, x i = kxk + kykh y , x i≥kxk + kyk. kxk kyk kxk 2

xn By the property (Ω2) of the set Ω, for every z ∈ D the set Ωz := {n ∈ Ω: ∈ Sz} = {n ∈ Ω: xn ∈ Sˆz} 0 kxnk is infinite and the series kxnk is divergent. Since the sequence (xn)n is contained in the cone Sˆz, Pn∈Ωz ∈Ωz tends to zero, and the series kxnk is divergent, we can use the inequality (1) and choose a finite subset Pn∈Ωz Ez ⊂ Ωz \ F such that the sum sz = xn ∈ Sˆz has norm in the interval Pn∈Ez ε c · tz − < kszk≤ c · tz 2|D0| where |D0| denotes the cardinality of the finite set D0. Then s s s s ε ε ks − ct zk≤ ks k · z − ct z + ct z − ct z = ks k− ct + ct z − z < + ct z z z ks k z ks k z ks k z z z z ks k 2|D | z 2c z z z z 0 and for the finite set E = Ez ⊂ Ω \ F we get Sz∈D0 x − x = x − ct z ≤ x − ct z = ks − ct zk < X n X n X z X X n z X z z n∈E n∈E z∈D0 z∈D0 n∈Ez z∈D0 ε ε ε < X + tz = (1 + X tz)= ε. 2|D0| 2 2 z∈D0 z∈D0 Now we prove that the set E satisfies the second condition of Claim 3. Take any subset E′ ⊂ E and for every ′ z ∈ D0 let E = E ∩ Ez. Then ′ xn ∈ Sˆz and k ′ xnk≤k xnk≤ ctz ≤ c, which implies that z Pn∈Ez Pn∈Ez Pn∈Ez ′ xn ∈ [0,c] · Sz and thus Pn∈Ez max{ε, kxk} k x k≤ cC = C · = C · max{kxk,ε} X n δ δ δ n∈E′ by definition of the numbers Cδ and C = Cδ/δ. Claim 4. The sets Γ= {y ∈ X : Pn=1 |hy, xni| < ∞} and Γ1 = {y ∈ X1 : Pn=1 |hy, pr1(xn)i| < ∞} coincide. 4 TARAS BANAKH

⊥ Proof. First we show that Γ ⊂ X1 = X0 . In the opposite case we would find points y ∈ Γ and z ∈ D0 such 1 that hy,zi 6= 0. Then U = {x ∈ S : |hy, xi| > 2 |hy,zi|} is an open neighborhood of z in the sphere S and xn by the definition of the set D ∋ z the set NU = {n ∈ N : ∈ U} is infinite and the series N kxnk kxnk Pn∈ U N kxnk diverges. It follows that for every n ∈ U we have |hy, xni| > 2 |hy,zi|, which implies the divergence of the series N |hy, xni| but this contradicts the choice of y ∈ Γ. This contradiction shows that Γ ⊂ X1. Pn∈ U Next, observe that for any y ∈ X1 and n ∈ N we get

hy, xi = hy, x − pr1(x)i + hy, pr1(x)i = hy, pr0(x)i + hy, pr1(x)i =0+ hy, pr1(x)i = hy, pr1(x)i ∞ ∞ and thus Pn=1 |hy, xi| = Pn=1 |hy, pr1(x)i|, which implies the equality Γ1 = X1 ∩ Γ = Γ. ∞ Now we are ready to complete the inductive proof of the theorem. By our assumption, the series n=1 xn ∞ P is ∗-potentially convergent. Then the series Pn=1 pr1(xn) is ∗-potentially convergent in the Hilbert space X1. Since the space X1 has dimension strictly smaller than d, the induction hypothesis and Claim 4 guarantee that the set Σ1 of potential sums of the series Pn∈N pr1(xn) is a non-empty affine subspace of X1 such that ⊥ ⊥ ∞ Σ1 =Σ1 +Γ1 , where Γ1 = {x ∈ X1 : ∀y ∈ Γ1 hy, xi =0} and Γ1 =Γ= {y ∈ X : Pn=1 |hy, xni| according to Claim 4. ∞ It remains to prove that the set Σ of potential sums of the series Pn=1 xn coincides with the non-empty affine subspace X0 ⊕ Σ1 of X. The inclusion Σ ⊂ X0 ⊕ Σ1 is trivial. To prove the reverse inclusion, we need to show that for any points N ∞ x ∈ X0 and y ∈ Σ1 there exists a permutation σ of such that Pn=1 xσ(n) = x + y. It will be more convenient instead of permutation of N, to construct a well-order 4 of order type ω on the N 4 N N set such that Pn∈N xn = x + y, which means that for any ε> 0 there exists k ∈ such that for any m ∈ with k m we get k Pn4m xn − (x + y)k <ε. Let s1 ∈ X1 be the sum of the absolutely convergent series Pn∈Ω pr1(xn). Since y ∈ Σ1, there exists a N N ∞ permutation σ : → such that y = Pn=1 pr1(xσ(n)). The permutation σ determines a well-order on the N set Λ := \ Ω defined by n m iff σ(n) ≤ σ(m). For this well-order on Λ we get Pn∈Λ xn = y − s1. From now on, talking about minimal elements of subsets of Λ we have in mind the well-order . The set Ω ⊂ N is endowed with the standard well-order ≤. By induction, we shall construct decreasing sequences of sets (Λi)i∈ω and (Ωi)i∈ω and an increasing sequence of finite sets (Fi)n∈ω such that the following conditions are satisfied for every n ∈ ω:

(1) Λi+1 =Λi \{min Λi}; (2) Ωi+1 ⊂ Ωi \{min Ωi}; (3) Fi+1 = Fi ∪{min Λi} ∪ (Ωi \ Ωi+1); 1 (4) kx − pr (xn)k < i ; Pn∈Fi+1 0 2 1 (5) for any subset E ⊂ Fi+1 \ Fi we get k Pn∈E xnk≤ C( 2i + kxminΛi + xmin Ωi k). We start the inductive construction, applying Claim 3 and choosing a finite set F0 ⊂ Ω such that kx − pr (xn)k≤kx − xnk < 1. Next, we put Ω =Ω \ F and Λ =Λ= N \ Ω. Pn∈F0 0 Pn∈F0 0 0 0 Assume that for some i ≥ 0 the sets Fi, Ωi and Λi have been constructed. Consider the element ai := x − pr (xmin Λ + xmin Ω + xn) ∈ X0 and observe that the inductive assumption (4) guarantees that 0 i i Pn∈Fi 1 ka k≤kx − pr (x )k + kx + x k < + kx + x k. i X 0 n min Λi min Ωi 2i minΛi min Ωi n∈Fi 1 Applying Claim 3, find a finite subset Ei ⊂ Ωi \ (Fi ∪{min Ωi}) such that kai − xnk < i+1 and for Pn∈Ei 2 every E ⊂ Ei 1 1 X xn < C · max{ 2i+1 , kaik} ≤ C( 2i + kxminΛi + xmin Ωi k). n∈E Let Fi+1 = Fi ∪Ei ∪{min Ωi, min Λi},Ωi+1 =Ωi \(Ei ∪{min Ωi}) and Λi+1 =Λi \{min Λi}. This completes the inductive step. N After completing the inductive construction of the sequence (Fi)i∈ω, on the union Si∈ω Fi = fix any well-order 4 such that n 4 m for any i ∈ ω and numbers n ∈ Fi and m∈ / Fi. Observe that the well- order 4 induces the well-order on the set Λ, which, combined with the absolute convergence of the series 4 Pn∈Ω pr1(xn), ensures that the series Pn∈N pr1(xn) converges to y. On the other hand, the conditions (4), ∞ (5) of the inductive construction and the convergence of the sequence (xn)n=1 to zero imply that the series 4 4 Pn∈N pr0(xn) converges to x. Consequently, the series Pn∈N xn converges to x + y, which yields the desired inclusion x + y ∈ Σ. ⊥ ⊥ ⊥ Finally, observe that Σ + Γ = (X0 +Σ1) + (X0 +Γ1 )= X0 + (Σ1 +Γ1 )= X0 +Σ1 = Σ. A SIMPLE INDUCTIVE PROOF OF LEVY-STEINITZ´ THEOREM 5

3. Generalizing Levy-Steinitz´ Theorem to Abelian topological groups The problem of extending the Levý-Steinitz Theorem to infinite-dimensional Banach space was posed by Stefan Banach in the Scottish Book [9, Problem 106] and to Abelian topological groups by Stanislaw Ulam , ∞ who noticed (without proof) in [10] that for any convergent series n=1 xn in a compact metrizable Abelian ∞ P topological group X the set Σ of potential sums of the series Pn=1 xn coincides with the shift H + Pn=1 xn of some subgroup H ⊂ X. This fact (whose proof was given in Banaszczyk [1, 10.2]) can be considered as an extension of the first part of Levý-Steinitz Theorem (describing the structure of the set Σ of potential sums of a series). On the other hand, the second part of Levý-Steinitz Theorem (characterizing series with non-empty set Σ of partial sums) does not extend to locally compact Abelian topological groups as shown by the following simple example. Example 1. Let T = R/Z be the circle group. The compact topological group X = T × T contains a sequence (xn)n∈ω such that ∞ (1) (xn)n=1 converges to zero; for every continuous homomorphism T the series ∞ is potentially convergent in T; (2) f : X → Pn=1 f(xn) ∞ (3) the series (xn)n=1 is not potentially convergent in X. Proof. Let q : R → R/Z = T be the quotient homomorphism and q2 : R2 → T2 = X be the covering homomorphism defined by q2(x, y) = (q(x), q(y)) for (x, y) ∈ R2. The compactness of the topological group X = T2 implies that the set of all non-zero continuous homomorphisms h : X → T is countable and hence can be 2 2 enumerated as {hn}n∈ω. For every n ∈ ω consider the continuous homomorphism fn = hn ◦ q : R → T. Since 2 the space R is simply-connected, the homomorphism fn can be lifted to a unique continuous homomorphism 2 fˆn : R → R such that q ◦ fˆn = fn. Being continuous and additive, the homomorphism fˆn is linear. Choose 2 any non-zero linear continuous functional f : R → R such that f∈ / {fˆn}n∈ω. Then for every n ∈ ω the sets 2 2 {x ∈ R : f(x) > 0, fn(x) > 0} and {x ∈ R : f(x) > 0, fn(x) < 0} are not empty. This allows us to choose a ∞ R2 sequence (zn)n=1 in the open half-plane {z ∈ : f(z) > 0} such that ∞ (1) the sequence (zn)n=1 converges to zero; ∞ (2) the series Pn=1 f(zn) diverges; ∞ ˆ ∞ ˆ (3) for every k ∈ ω the series Pn=1 max{0, fk(zn)} and Pn=1 min{0, fk(zm)} are divergent. 2 2 In the group T consider the sequence (xn)n∈ω of points xn = q (zn), n ∈ N. The conditions (1) and (2) imply ∞ R2 ∞ that the series Pn=1 zn is not potentially convergent in and hence the series Pn=1 xn is not potentially convergent in T2. On the other hand, the condition (3) combined with the Riemann Rearrangement Theorem ∞ ˆ imply that for every k ∈ ω the series Pn=1 fk(zn) is potentially convergent in the real line and hence the series ∞ ∞ ˆ 2 T Pn=1 hn(xn)= Pn=1 q ◦ fn ◦ q (zn) is potentially convergent in the circle . Therefore, the sequence (xn)n∈ω satisfies the conditions (1)–(3) of Example 1.

4. Acknowledgements The author express his sincere thanks to Vaja Tarieladze for asking the problem and many valuable sugges- tions and to Volodymyr Kadets for his comments on the original Steinitz proof (which is diﬀers from the proof presented in this paper).

References [1] W. Banaszczyk, Additive subgroups of topological vector spaces, Lecture Notes in Mathematics, 1466. Springer-Verlag, Berlin, 1991. [2] W. Gross, Bedingt Konvergente Reihen, Monat. Math. Physik. 28 (1917), 221-237. [3] I. Halperin, Sums of series permitting rearrangements, C.R. Math. Rep. Acad. Sci. Canada, 8:2 (1986), 87–102. [4] M.I. Kadets, V.M. Kadets, Series in Banach spaces. Conditional and unconditional convergence, Birkhauser Verlag, Basel, 1997. [5] P. Levý, Sur les séries semi-convergentes, Nouv. Ann. d. Math. 64 (1905), 506–511. [6] P. Rosenthal, The remarkable Theorem of Levy and Steinitz, Amer. Math. Monthly, 94:4 (1987), 342–351. [7] E. Steinitz, Bedingt Konvergente Reihen und Konvexe Systeme, J. f. Math. 143 (1913), 128–175. [8] V. Tarieladze, Is “weakly good” series in a finite-dimensional Banach space “good”?, Lviv Scottish Book, 24.09.2017; (https://mathoverflow.net/questions/281948/is-weakly-good-series-in-a-finite-dimensional-banach-space-good) [9] S. Ulam, Scottish Book, 1958 (http://kielich.amu.edu.pl/Stefan Banach/pdf/ks-szkocka/ks-szkocka3ang.pdf). [10] S. Ulam, A collection of mathematical problems, Intersci. Publ., NY, 1960.

Jan Kochanowski University in Kielce (Poland) and Ivan Franko National University of Lviv (Ukraine) E-mail address: [email protected]