RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM LUCAS BORBOLETA Abstract.
theorem.
This work revisits the proof given in 1932 by Mazur and Ulam for their
1. Introduction In 1932, Mazur and Ulam proved that any isometry
f (0) = 0,
is necessarily a linear transformation [1].
f
of a normed space
E,
with
This article revisits that theorem
and its proof. First, the conditions of the theorem are precisely rephrased. Second, the auxiliary de nitions, lemmas, and proof plan of [1] are reproduced . Third, the proof and its steps are analyzed and commented in details. 2. Definition and Theorem The vector spaces are considered on the real eld
R.
Their norms are general, so they
can be non-euclidean.
De nition 1 (Isometry).
f : E → F , from a vector normed space E onto F , is an isometry if ∀ (x, y) ∈ E 2 , kf (x) − f (y)k = kx − yk.
An application
another vector normed space
Theorem 2. An isometry f : E → F , from a vector normed space E onto another vector normed space F , with f (0) = 0, is always a linear application.
3. Auxiliary definitions, lemmas, and proof plan Let us rephrase the proof of [1], focusing on accuracy, with a few adaptation of notations, but keeping its essential steps and concepts, like metric center and symmetric center of a bounded subset.
De nition 3 noted
(Diameter)
Diam (A),
.
For any subset
A
of a vector normed space
Diam (A) , Sup kx − yk : (x, y) ∈ A By convention,
De nition 4
Diam (Ø) , 0.
(Metric Center)
.
A subset
A
its diameter,
M etc0 (A),
A
Diam (A) < ∞.
of a vector normed space
G,
is a set de ned as:
1 M etc0 (A) , z ∈ A : ∀x ∈ A, kx − zk ≤ Diam (A) 2 ∗ center of order n ∈ N , for any bounded subset A ⊂ G, noted M etcn (A),
de ned recursively as:
M etcn (A) , M etc0 (M etcn−1 (A)) Date
A:
2
is said bounded if
For any bounded subset
its metric center of order zero, noted
A metric
G,
is de ned as the supremun of distance between any two points of
: 2012-10-14 (typo xing 2012-12-23). Norms, Euclidean Spaces, Inner Product Spaces, Linear Isometries.
Key words and phrases.
1
is
RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM
A ⊂ G,
The metric center, for any bounded subset
noted
M etc (A),
2
is de ned as the
intersection of the metric centers of all orders:
\
M etc (A) ,
M etcn (A)
n∈N
Lemma 5. If Diam (A) is nite then the sequence of diameters from metric centers of order n converges to zero:
Diam (M etcn (A)) −→ 0 n→+∞
Corollary 6. The metric center of any bounded subset A ⊂ G of a normed vector space is either empty or a singleton.
Lemma 7. If f : E → F is an isometry, and if {x0 } is the metric center of a bounded subset A ⊂ E , then {f (x0 )} is the metric center of the (bounded) subset f (A) ⊂ F : M etc (A) = {x0 } ⇒ M etc (f (A)) = {f (x0 )}
De nition 8 center, noted
(Symmetric Center)
Symc (A),
.
For any subset
A
of vector space
G,
its symmetric
is a set de ned as:
Symc (A) , {z ∈ A : ∀x ∈ A, (2z − x) ∈ A}
Lemma 9. For any bounded subset A ⊂ G of a normed vector space, if its symmetric center is not empty, then it is also its metric center. ∀A ⊂ G, Diam (A) < ∞ and Symc (A) 6= Ø ⇒ M etc (A) = Symc (A)
Corollary 10. For any bounded subset A ⊂ G of a normed vector space, its symmetric center is either empty or a singleton.
De nition 11 middle, noted
(Middle)
.
M id (a, b),
For any two distinct elements
M id (a, b) , 12
.
The element
and
b
of vector space
G,
its
is a set de ned as:
Claim
a
1 x ∈ G : kx − ak = kx − bk = ka − bk 2
a+b is in the symmetric center of the subset 2
M id (a, b):
a+b ∈ Symc (M id (a, b)) 2
Fact 13. From the lemma 9, since the symmetric center Symc (M id (a, b)) is not empty, it is also the metric center of M id (a, b), and by the way reduced to a singleton: Symc (M id (a, b)) = M etc (M id (a, b)) = f : E → F, a+b
From this point, let us consider an isometry
Fact 14. From the lemma 7, the singleton f
2
a+b 2
with
f (0) = 0.
is also metric center of f (M id (a, b)):
a+b M etc (f (M id (a, b))) = f 2
RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM
Claim
15
.
The singleton
n
o
f (a)+f (b) 2
is the symmetric center of the subset
Symc (f (M id (a, b))) =
f (a) + f (b) 2
3
f (M id (a, b)):
Fact 16. oFrom the lemma 9, M etc (f (M id (a, b))) = Symc (f (M id (a, b))), so f n
a+b 2
=
, which implies:
f (a)+f (b) 2
f
a+b 2
=
f (a) + f (b) 2
Fact 17. Considering new variables x and y, such that a = 2x and b = 2y, one gets . Considering the case y = 0, and remembering the theorem hypothesis f (0) = 0, one ⇔ f (2x) = 2f (x). gets f (x) = f (2x) 2 So one concludes f (x + y) = f (x) + f (y). f (2x)+f (2y) 2
f (x + y) =
Claim
18
.
The transformation
f
, being additive and continuous (as an isometry), is
linear. 4. Analysis and comments of the proof 4.1.
About de nition 1.
Being an isometry, the application
f,
with
f (0) = 0,
ver-
i es ∀x ∈ E, kf (x)k = kxk. Conversely, if f would be just linear and such that ∀x ∈ E, kf (x)k = kxk, then it would imply f (0) = 0 and ∀ (x, y) ∈ E 2 , kf (x − y)k = kf (x) − f (y)k = kx − yk. So de ning an isometry by the sole requirement ∀x ∈ E, kf (x)k = kxk only works (equivalently to 1) for a linear application. So the de nition 1 cannot be simpli ed. 4.2.
About de nition 4.
The de nition of [1] is ambiguous regarding metric center of
M etc0 (A) ⊂ A or just M etc0 (A) ⊂ G ? However, for n ≥ 1, [1] clearly requires M etcn (A) ⊂ requires M etc0 (A) ⊂ A.
order zero. Should one understand
the recursive de nition of metric center of order
A.
So by extension, one also
4.3. About lemma 5. Here is a proof of this lemma. Let us consider M etc0 (A) = z ∈ A : ∀x ∈ A, kx − zk ≤ 21 Diam (A) . • If M etc0 (A) has less than two elements then Diam (M etc0 (A)) = 0. • On the opposite, if M etc0 (A) has at least two distinct elements z1 and z2 , then they 1 1 satisfy the property kz1 − z2 k ≤ Diam (A), so Diam (M etc0 (A)) ≤ Diam (A). 2 2 1 In both cases, Diam (M etc0 (A)) ≤ Diam (A). By recurrence, one gets for any n ∈ N: 2 1 Diam (M etcn (A)) ≤ 2n+1 Diam (A). And one concludes Diam (M etcn (A)) −→ 0.
n→+∞
4.4.
About corollary 6.
Here is a proof of this corollary. Let us make the hypothesis
z2 are members of M etc (A). This means two distinct elements z1 and z2 are members of M etcn (A) for any n ∈ N. This implies kz1 − z2 k ≤ 1 Diam (A), for any n ∈ N. However, z1 and z2 are distinct, so kz1 − z2 k = δ > 0. 2n+1 1 Combining these properties yields to ∃δ > 0, ∀n ∈ N, δ ≤ n+1 Diam (A). But this is 2 incompatible with the lemma 5: Diam (M etcn (A)) −→ 0. This proofs that M etc (A) that two distinct elements
z1
and
n→+∞
cannot have two distinct elements
z1
and
z2 .
RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM
4.5.
About lemma 7. •
4
Here is a proof.
kf (x) − f (y)k = kx − yk implies Diam (f (A)) = Diam (A). f (A). • Second, for any z ∈ M etc0 (A), and for any x ∈ A, kx − zk ≤ 21 Diam (A), so 1 Diam (f (A)), and then f (z) ∈ M etc0 (f (A)). equivalently kf (x) − f (z)k ≤ 2 This means M etc0 (f (A)) = f (M etc0 (A)). • Third, by de nition, M etcn (A) = M etc0 (M etcn−1 (A)), so by recurrence one obtains M etcn (f (A)) = f (M etcn (A)). Then by taking the intersection over the T T family n ∈ N, one gets M etc (f (A)) = M etcn (f (A)) = f (M etcn (A)) = First, let us remark that By the way, if
A
is bounded then be also
n∈N
n∈N
f (M etc (A)). Indeed, since f is an isometry, it is injective: f (x1 ) = f (x2 ) ⇔ kf (x1 ) − f (x2 )k = 0, kf (x1 ) − f (x2 )k = kx1 − x2 k, kx1 − x2 k = 0 ⇔ x1 = x2 . So if for any n ∈ N, y ∈ f (M etcn (A)), it implies ∃x ∈ E, with y = f (x), such that ∀n ∈ N, x ∈ M etcn (A), so x ∈ M etc (A). 4.6.
About lemma 9. •
Here is a proof.
z ∈ Symc (A) and x ∈ A. This implies z ∈ A and (2z − x) ∈ A. (2z − x) ∈ A then k(2z − x) − xk ≤ Diam (A). This inequality 1 can be rewritten kz − xk ≤ Diam (A). This implies z ∈ M etc0 (A). 2 • If one checks that Symc (A) is a singleton then necessarily z ∈ M etc (A). One can understand that[1] makes the hypothesis that Symc (A) is a singleton. Here, one obtains the same result by requiring Diam (A) < ∞ in the lemma. Let us proof that if two distinct elements z1 and z2 are members of Symc (A) then (0) (0) , (z1 , z2 ), ∀n ∈ Diam (A) = ∞. Indeed, let us construct a sequence: z1 , z2 (n) (n) (n−1) (n−1) (n−1) N∗ , z1 , z2 , 2z1 − z2 , 2z2 − z1 . By de nition of Symc (A), if z1 and (n−1) (n) (n) (n) (n) z2 are in the symmetric center, so be also z1 and z2 . So all the pairs z1 , z2 are (n) (n) = (z1 + n (z1 − z2 ) , z2 + n (z2 − z1 )). in Symc (A) . By recurrence, one checks that z1 , z2
(n)
(n) (n) (n) This implies z1 − z2 = (2n + 1) · kz1 − z2 k and z1 − z2 −→ ∞. Let us consider
Since
x∈A
and
n→∞
4.7.
About corollary 10.
This corollary appears as a combination of lemmas 6 and 9.
However, the proof of lemma 9 already discovered it as a by product: For any bounded subset 4.8.
A, Symc (A)
is either empty or singleton.
About de nition 11.
Let us remark that
is a bounded subset. Indeed,
x and y . The distance kx − yk can kx − yk = k(x − a) + (a − y)k ≤ kx − ak + ky − ak ≤ 2 ka−bk . 2
let us consider two of its elements follows:
M id (a, b)
be bounded as
About claim 12.
Here is a proof. The symmetric center of subset M id (a, b) is Symc (M id (a, b)) = {z ∈ M id (a,b) : ∀x ∈ M id (a, b) , (2z − x) ∈ M id (a, b)}. a+b Let us consider z ∈ M id (a, b) and y , 2 − z = a + b − z . One checks y − a = b − z 2 1 and y − b = a − z . Since z ∈ M id (a, b), it implies kz − ak = kz − bk = ka − bk. So one 2 1 a+b checks ky − ak = ky − bk = ka − bk . One concludes z ∈ M id (a, b) ⇒ 2 − z ∈ 2 2 M id (a, b), and so a+b ∈ Symc (M id (a, b)) . 2 4.9.
de ned as
4.10.
About claim 15.
Here is a proof.
One makes the auxiliary hypothesis that
f (M id (a, b)) = M id (f (a) , f (b)), then being in transposed conditions of the claim 12, f (a)+f (b) one concludes that ∈ Symc (M id (f (a) , f (b))) = Symc (f (M id (a, b))). So let 2 us proof the auxiliary hypothesis.
RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM
5
Let us consider the condition x ∈ M id (a, b). It is equivalent to kx − ak = kx − bk = ka − bk. But f being an isometry, it is also equivalent to kf (x) − f (a)k = kf (x) − f (b)k = kf (a) − f (b)k. Which is in turn equivalent to f (x) ∈ M id (f (a) , f (b)). This means f (M id (a, b)) = M id (f (a) , f (b)).
1 2 1 2
4.11.
About fact 17.
Let us remember that fact 17 is derived from considering two
a and b, or equivalently, two distinct elements x and y . Without breaking x = 6 y , the fact 17 establishes, as an intermediate result, that f (2x) = 2f (x). So nally, the properties f (x + y) = f (x) + f (y) , is proved both for x 6= y and x = y . Let us recall that, by hypothesis of theorem 2, f (0) = 0. distinct elements
the hypothesis
4.12. is
About linearity claim 18.
The original article [1] focused on additivity of
∀ (x, y) ∈ E 2 , f (x + y) = f (x)+f (y).
f , that
The linearity property is mentioned in a footnote
of[1]: this transformation, being additive and continuous (as isometry), is linear. Here are detailed arguments for checking the property
∀x ∈ E, ∀λ ∈ R, f (λx) = λf (x).
It
proceeds by successive enlargement of scalar sets. (1) (2)
∀x ∈ E, f (−x) = −f (x). Proof: f (x − x) = 0 = f (x) + f (−x), so f (−x) = −f (x). ∀x ∈ E, ∀n ∈ N, f (nx) = nf (x). Proof: f (0 · x) = 0, f (1 · x) = f (1 · x), then the additivity of f , f ((n + 1) · x) = f (n · x) + f (x), and a recurrence over n gets it.
(3) (4) (5) (6)
∀x ∈ E, ∀n ∈ Z, f (nx) =nf (x). Proof: by combining the previous results. 1 1 1 ∗ ∀x ∈ E, ∀n ∈ N , f n x = n f (x). Proof: f n n x = f (x) = nf n1 x , so f n1 x = n1 f (x). ∀x ∈ E, ∀r ∈ Q, f (rx) = rf (x). Proof: by combining the previous results. ∀x ∈ E, ∀λ ∈ R, f (λx) = λf (x). Proof: since Q is dense in R, it is enough to argue than the application g : λ ∈ R 7−→ f (λx) ∈ F , de ned for a given x ∈ E , 2 is continuous. ∀ (x, y) ∈ R , kg (λ) − g (µ)k = kf (λx) − f (µx)k = kλx − µxk = |λ − µ| · kxk, so it proofs g (λ) −→ g (µ). λ→µ
4.13.
About the role of the theorem hypothesis f (0) = 0.
If one removes the
f (0) = 0 then one can check that g (x) , f (x) − f (0) is linear. Indeed, rst g (x) − g (y) = f (x) − f (y) implies that the application g is an isometry. Second, since g (0) = 0, the application of the theorem 2 yields to the linearity of g . So removing the hypothesis f (0) = 0 yields to the conclusion that f could be expressed as f (x) = f (0) + g (x), with g being linear. Such application f is said a ne. So the discussed hypothesis
hypothesis is not essential in order to obtain an interesting conclusion. 4.14.
About the role of the triangular inequality.
norm is used once: for checking that
M id (a, b)
The triangular inequality of the
is a bounded subset. Is it essential? Or
does it mean that the theorem could be extended to a domain that is larger than the vector normed spaces? 4.15.
About metric center.
subset of
A.
A
Lemma 9 indicates that the metric center of a bounded
does not depend of the kind of norm
Is it still true when
k·k
if such center is also symmetric center
Symc (A) = Ø? 5. License
This work by Lucas Borboleta (http://lucas.borboleta.blog.free.fr) is licensed under a
Creative Commons Attribution-ShareAlike 3.0 Unported License (http://creativecommons.org/licenses/ sa/3.0/).
RETURN ON THE 1932 PROOF OF THE MAZUR-ULAM THEOREM
6
References
[1] S. Mazur and Stanisªaw M. Ulam. Sur les transformations isometriques d'espaces vectoriels normés. Comptes Rendus de l'Académie des Sciences, 194:946948, 1932. E-mail address :
[email protected]
