On a discrete version of the antipodal theorem by

MATHEMATICAE 151 (1996)

On a discrete version of the antipodal theorem

by

Krzysztof O l e s z k i e w i c z (Warszawa)

Abstract. The classical theorem of Borsuk and Ulam [2] says that for any continuous mapping f : S

→ R

there exists a point x ∈ S

such that f (−x) = f (x). In this note a discrete version of the antipodal theorem is proved in which S

is replaced by the set of vertices of a high-dimensional cube equipped with Hamming’s metric. In place of equality we obtain some optimal estimates of inf

kf (x) − f (−x)k which were previously known (as far as the author knows) only for f linear (cf. [1]).

We introduce standard notation: k and n will denote positive integers, C _n ^k = {x ∈ [−1, 1] ⁿ : #{i : |x i | = 1} ≥ n − k} will stand for the k-dimensional skeleton of the cube C = [−1, 1] ⁿ equipped with the stan- dard CW-structure. Let C _n = C _n ⁰ = {−1, 1} ⁿ . We consider Hamming’s metric d on C n defined as d(x, y) = #{i : x i 6= y i }. (X k , k · k) will stand for a normed k-dimensional linear space. By S ^k (resp. B ^k+1 ) we denote the unit Euclidean sphere (resp. ball) with centre at zero in R ^k+1 .

This is the main result of the paper:

Theorem 1. Let f : C n → X k satisfy the following two conditions:

f (−x) = −f (x) and kf (x) − f (y)k ≤ d(x, y) for any x, y ∈ C _n . Then (i) there exists x ∈ C _n such that kf (x)k ≤ ¹ ₂ min(k, n),

(ii) if the norm is Euclidean then there exists x ∈ C _n such that kf (x)k ≤ ¹ ₂ p

min(k, n).

The examples of X _k = l ^k ₁ and X _k = l ^k ₂ with

f (x 1 , . . . , x n ) = ¹ ₂ (x 1 , . . . , x _min(k,n) , 0, 0, . . . , 0) indicate that the constants cannot be improved.

1991 Mathematics Subject Classification: 54H25, 46B09, 05B25.

Supported in part by the Foundation for Polish Science and KBN Grant 2 P301 022 07.

[189]

Corollary 1. Let g : (C n , d) → (X k , k · k) be such that kg(x) − g(y)k ≤ d(x, y)

for any x, y ∈ C _n . Then there exists z ∈ C _n such that kg(z) − g(−z)k ≤ min(k, n).

This antipodal version follows from Theorem 1 when we set f (x) = (g(x) − g(−x))/2. In fact, one can easily see that Theorem 1(i) and Corol- lary 1 are equivalent. Therefore we will be interested only in the “anti- symmetric” case.

Theorem 1(i) is trivial if n ≤ k. If n < k then (i) easily follows from (ii), because for any k-dimensional linear normed space (X _k , k · k) there exists a Euclidean norm | · | such that |v| ≤ kvk ≤ √

k|v| for any v ∈ X _k . Therefore the proof will be devoted to the Euclidean case.

We will need two lemmas.

Lemma 1. If k < n then there exists a continuous mapping h k : S ^k → C _n ^k such that h _k (−x) = −h _k (x) for any x ∈ S ^k .

P r o o f. As the homotopy group π _i depends on the (i + 1)-dimensional skeleton of the CW-complex only, we know that π _i (C _n ^k ) = π _i (C) = 0 for any i < k. We inductively construct continuous mappings h i : S ⁱ → C _n ^k for i = 0, 1, . . . , k such that h _i (−x) = −h _i (x) for any x ∈ S ⁱ . We choose the function h ₀ arbitrarily, just to keep anti-symmetry (S ⁰ = {−1, 1}). Assume h i−1 is well defined, continuous and anti-symmetric. Since π i−1 (C _n ^k ) = 0, there exists a continuous function H _i : B ⁱ → C _n ^k such that H _i (x) = h _i−1 (x) for any x ∈ S ⁱ⁻¹ = ∂B ⁱ . Let G _i : B ⁱ → C _n ^k be defined as G _i (x) = −H _i (−x).

For x ∈ S ⁱ we put h i (x 1 , . . . , x i+1 ) = H i (x 1 , . . . , x i ) if x i+1 ≥ 0 and h _i (x ₁ , . . . , x _i+1 ) = G _i (x ₁ , . . . , x _i ) if x _i+1 ≤ 0. This completes our induction as h k satisfies the desired conditions.

Lemma 2. If k < n and F : C _n ^k → X _k is continuous and such that F (−x) = −F (x) for any x ∈ C _n ^k then there exists z ∈ C _n ^k such that F (z) = 0.

P r o o f. Let h _k be defined as in Lemma 1. The function F ◦ h _k : S ^k → X _k is continuous. Recall that dim X _k = k. Therefore by Borsuk–Ulam’s Theorem

F (h _k (x)) = F (h _k (−x)) = F (−h _k (x)) = −F (h _k (x)) for some x ∈ S ^k . Hence z = h _k (x) satisfies the desired conditions.

The proofs above were given for the sake of completeness and because

of their simplicity, but it should be noticed that they are only special cases

of well known, far more general topological theorems (cf. [3]).

P r o o f o f T h e o r e m 1(ii). First consider the case k < n. Let f : C → X b k be defined by the formula

f (x b 1 , . . . , x n ) = X

y∈C

 Y _n

j=1

1 + y _j x _j 2

 f (y).

One can easily see that b f | C

= f and ∂ ² f /∂x b ² _i = 0 for i = 1, . . . , n (i.e. the function b f is affine with respect to each variable). The function F = b f | _C

n

satisfies the conditions of Lemma 2, hence F (z) = 0 for some z ∈ C _n ^k . This is the crucial point of the proof. Without loss of generality we can assume that |z k+1 |, . . . , |z n | = 1 and define G : [−1, 1] ^k → X k by G(x ₁ , . . . , x _k ) = F (x ₁ , . . . , x _k , z _k+1 , . . . , z _n ). For any i ≤ n and x ∈ C by the triangle inequality we have

k b f (x 1 , . . . , x i−1 , 1, x i+1 , . . . , x n ) − b f (x 1 , . . . , x i−1 , −1, x i+1 , . . . , x n )k

≤ X

y∈C

,y

=1

 Y

j6=i

1 + y j x j

2



kf (y ₁ , . . . , y _i−1 , 1, y _i+1 , . . . , y _n )

−f (y ₁ , . . . , y _i−1 , −1, y _i+1 , . . . , y _n )k ≤ 1, since

X

y∈C

,y

=1

 Y

j6=i

1 + y _j x _j 2



= 1

for any x ∈ C and d((y 1 , . . . , y i−1 , 1, y i+1 , . . . , y n ), (y 1 , . . . , y i−1 , −1, y i+1 , . . . . . . , y _n )) = 1 for any y ∈ C _n .

Therefore for any i ≤ k and x ∈ [−1, 1] ^k we have

kG(x ₁ , . . . , x _i−1 , 1, x _i+1 , . . . , x _k ) − G(x ₁ , . . . , x _i−1 , −1, x _i+1 , . . . , x _k )k ≤ 1.

Consider independent real random variables Y 1 , . . . , Y k such that P (Y i = 1)

= (1+z _i )/2, P (Y _i = −1) = (1−z _i )/2 for i ≤ k. As EY _i = z _i and ∂ ² G/∂x ² _i = 0 for i ≤ k (G is affine with respect to each variable), we have

EkG(Y 1 , . . . , Y k )k ²

= 1 + z _k

2 EkG(Y ₁ , . . . , Y _k−1 , 1)k ² + 1 − z _k

2 EkG(Y ₁ , . . . , Y _k−1 , −1)k ²

= 1 + z _k

2 EkG(Y ₁ , . . . , Y _k−1 , z _k )

+ (G(Y ₁ , . . . , Y _k−1 , 1) − G(Y ₁ , . . . , Y _k−1 , z _k ))k ² + 1 − z _k

2 EkG(Y 1 , . . . , Y k−1 , z k )

+ (G(Y ₁ , . . . , Y _k−1 , −1) − G(Y ₁ , . . . , Y _k−1 , z _k ))k ²

= 1 + z _k

2 E

G(Y ¹ , , . . . , Y _k−1 , z _k )

+ 1 − z _k

2 (G(Y ₁ , . . . , Y _k−1 , 1) − G(Y ₁ , . . . , Y _k−1 , −1))

2

+ 1 − z _k

2 E

G(Y ¹ , . . . , Y _k−1 , z _k )

− 1 + z _k

2 (G(Y ₁ , . . . , Y _k−1 , 1) − G(Y ₁ , . . . , Y _k−1 , −1))

2

= EkG(Y ₁ . . . , Y _k−1 , z _k )k ² + 1 − z ² _k

4 EkG(Y ₁ , . . . , Y _k−1 , 1) − G(Y ₁ , . . . , Y _k−1 , −1)k ²

≤ EkG(Y 1 , . . . , Y k−1 , z k )k ² + 1 4 (by easy induction)

≤ kG(z ₁ , . . . , z _k )k ² + k 4 = k

4 .

Hence there exists y ∈ {−1, 1} ^k such that kf (y ₁ , . . . , y _k , z _k+1 , . . . , z _n )k = kG(y ₁ , . . . , y _k )k ≤ √

k/2, which proves (ii) if k < n. In the case n ≤ k we deal with b f (x ₁ , . . . , x _n ) instead of G(x ₁ , . . . , x _k ) (then Lemmas 1 and 2 are unnecessary since b f (0) = 0 follows just from the anti-symmetry of f ) and the rest of the proof is essentially the same, so that we omit it. This completes the proof.

Corollary 2. If (X _k , k·k) = l ^k _p for p ∈ [1, 2] then under the assumptions of Theorem 1 there exists x ∈ C n such that kf (x)k p ≤ k ^1/p /2.

To see this notice that kvk 2 ≤ kvk p ≤ k ^1/p kvk 2 for any v ∈ R ^k .

Corollary 3. Let P be a convex , centrally symmetric polytope in R ⁿ and let P ₀ be the set of its vertices. Let (X _k , k · k) be a normed linear space of finite dimension k < n. We will say that x ∼ y for x, y ∈ P ₀ if there exists a k-dimensional face of P containing x and y. Then for any function g : P ₀ → X _k there exists z ∈ P ₀ such that

kg(z) − g(−z)k ≤ 2 max

x∼y; x,y∈P

kg(x) − g(y)k.

We sketch a proof which is just a modification of the proof of Theorem 1.

Instead of C _n ^k we consider the k-dimensional skeleton of P . After obvious

changes Lemmas 1 and 2 remain valid. Then we define a continuous function

f by induction: b b f | _P

= g and b f is harmonic (i.e. each of its coordinates is)

inside any i-dimensional cell of P for i = 1, 2, . . . Then Corollary 3 follows

from the maximum property of harmonic functions.

Theorem 2. Let p ∈ [1, 2]. If (X k , k · k) is p-smooth, i.e. there exists K ≥ 1 such that

kx + yk ^p + kx − yk ^p

2 ≤ kxk ^p + Kkyk ^p

for any x, y ∈ X _k , then under the assumptions of Theorem 1 there exists z ∈ C n such that kf (z)k ≤ (K min(k, n)) ^1/p .

P r o o f. Assume that k < n. Precisely as in the proof of Theorem 1 we construct a function G : [−1, 1] ^k → X _k and deduce that G(z ₁ , . . . , z _k ) = 0 for some z ∈ [−1, 1] ^k . Our problem is reduced to the following lemma.

Lemma 3. Let a function G : [−1, 1] ^k → X k satisfy the conditions

G(x) = X

y∈{−1,1}

 Y k j=1

1 + y j x j

2

 G(y)

for any x ∈ {−1, 1} ^k and

kG(x) − G(y)k ≤ #{i : x _i 6= y _i }

for any x, y ∈ [−1, 1] ^k . Let (X _k , k · k) satisfy the conditions of Theorem 2.

Then for any z ∈ [−1, 1] ^k there exists x ∈ {−1, 1} ^k such that kG(x) − G(z)k

≤ (Kk) ^1/p .

P r o o f. According to the properties of the function G shown in the proof of Theorem 1 we have kG(x) − G(z)k ≤ k for each x ∈ {−1, 1} ^k . We will prove that if Lemma 3 is valid with a constant L = L(k) in place of (Kk) ^1/p then it is true also with the constant ¹ ₂ L + ¹ ₂ (Kk) ^1/p . Thus Lemma 3 can be proved by a limit argument.

Without loss of generality we can assume that z 1 , . . . , z k ≥ 0. Applying Lemma 3 to the cube [0, 1] ^k we see that there exists y ∈ {0, 1} ^k such that kG(y) − G(z)k ≤ ¹ ₂ L. Now we only need to show that there exists x ∈ {−1, 1} ⁿ such that kG(x) − G(y)k ^p ≤ (Kk)/(2 ^p ). Without loss of generality we can assume that y ₁ = . . . = y _i = 0, y _i+1 = . . . = y _k = 1. Let r ₁ , . . . , r _i be independent symmetric random Bernoulli variables, i.e. P (r j = ±1) = ¹ ₂ for j ≤ i. From p-smoothness of the norm we easily deduce that

EkG(r ₁ , . . . , r _i , 1, . . . ,1) − G(0, 0, . . . , 1, . . . , 1)k ^p

= E

X i j=1

(G(r ₁ , . . . , r _j−1 , r _j , 0, 0, . . . , 1, . . . , 1)

− G(r ₁ , . . . , r _j−1 , 0, 0, . . . , 1, . . . , 1)) ^p

≤ K X i j=1

EkG(r ₁ , . . . , r _j−1 , r _j , 0, 0, . . . , 1, . . . , 1)

− G(r ₁ , . . . , r _j−1 , 0, 0, . . . , 1, . . . , 1)k ^p

= 1 2 ^p K

X i j=1

EkG(r ₁ , . . . , r _j−1 , 1, 0, 0, . . . , 1, . . . , 1)

− G(r ₁ , . . . , r _j−1 , −1, 0, 0, . . . , 1, . . . , 1)k ^p

≤ Ki 2 ^p ≤ Kk

2 ^p .

We used the same properties of the function G which were verified during the proof of Theorem 1. This completes the proof of Lemma 3. The proof of Theorem 2 in the case n ≤ k follows by the argument ending the proof of Theorem 1.

R e m a r k. Let us consider X _k = R ^k with l _p and l _q norms for p, q ∈ [1, 2].

One can easily deduce from Corollary 2 and elementary inequalities between norms that for f : C n → X k satisfying the conditions

f (−x) = f (x) and kf (x) − f (y)k _p ≤ d(x, y)

for any x, y ∈ C _n , there exists z ∈ C _n such that kf (z)k _q ≤ k ^1/q /2. The constant k ^1/q /2 is optimal with respect to k. It is of interest to know what happens when p = 2 and q = ∞. The well known and still unproved Koml´os conjecture states that in this case the above estimate is universal (does not depend on k or n) if f is linear (as we consider C n embedded in the natural way in R ⁿ ).

Acknowledgements. I would like to thank Zbigniew Marciniak for a helpful topological consultation. Prof. Stanisław Spie˙z turned my attention to classical proofs of Lemmas 1 and 2 using the notion of Smith index.

Remarks of the referee improved the organization of the paper.

References

[1] I. B ´a r ´a n y and V. S. G r i n b e r g, On some combinatorial questions in finite-dimensio- nal spaces, Linear Algebra Appl. 41 (1981), 1–9.

[2] E. H. S p a n i e r, Algebraic Topology, McGraw-Hill, New York, 1966, Theorem 5.8.9.

[3] C.-T. Y a n g, On a theorem of Borsuk–Ulam, Ann. of Math. 60 (1954), 262–282, Theorem 1, (2.7), (3.1).

Department of Mathematics Warsaw University

Banacha 2

02-097 Warszawa, Poland E-mail: koles@mimuw.edu.pl

Received 3 April 1996;

in revised form 24 June 1996