On geometrical characterizations of R -linear mappings
aa r X i v : . [ m a t h . G M ] A ug On geometrical characterizations of R -linear mappings Saka´e Fuchino
Contents
1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2. Mappings which preserve proportions on each line . . . . . . . . . . . . . . . . . . . . .
3. Geometry of line preserving mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4. A partial answer to Problem 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . In this note, we consider several properties characterizing linear mappings f : X → Y for R -linear spaces X and Y in terms of points and lines in R -linear spaces X and their images in Y by the mappings.Theorem 4.1 and its corollaries generalize the Fundamental Theorem of AffineGeometry, and characterize the R -linear mappings f : X → Y with dim( f ′′ X ) > ∗ Graduate School of System Informatics, Kobe UniversityRokko-dai 1-1, Nada, Kobe 657-8501 Japan [email protected]
Date:
March 9, 2020
Last update:
August 6, 2020 (00:34 JST)
MSC2020 Mathematical Subject Classification:
Keywords: linear mapping, affine mapping, mapping preserving linesThe research was supported by JSPS Kakenhi Grant No.20K03717.
This is an extended version of the paper with the same title.All additional details not to be contained in the submitted version of the paperare either typeset in typewriter font (the font this paragraph is typeset) orput in separate appendices. The numbering of the assertions is kept identicalwith the submitted version.The most up-to-date file of this extended version is downloadable as:https://fuchino.ddo.jp/papers/linear-mappings-x.pdf .In the following, we shall use standard conventions and notation in set theorywhich might be slightly different from everyday mathematics. In particular, wedistinguish sharply between elements x ∈ y and singletons { x } ⊆ y . The imageof a subset U ⊆ X of the domain of a mapping f : X → Y is denoted by f ′′ U with is also often denoted as f [ U ] in the literature. Thus, for a closed interval[ a, b ], we write f ′′ [ a, b ] to denote the image of the interval by the function f insteadof writing f [ a, b ] or even, more correctly, f [[ a, b ]]. A natural number n is the set { , , ..., n − } and thus i ∈ i = 0 or i = 1.We work in Z (the modern version of Zermelo’s set theory without Axiom ofChoice) if not mentioned otherwise. lin-section-1 Let X be a linear space over the scalar field R (or R -linear space). A subset P ⊆ X is a point if P is a singleton. That is, if it is an affine subspace of X ofdimension 0. A subset L is a line if it is an affine subspace of X of dimension0. Thus, L ⊆ X is a line if and only if there are vectors a , d ∈ X such that L = R d + a = { r d + a : r ∈ R } . Similarly E ⊆ X is a plane if it is an affinesubspace of X of dimension 2. Two lines L , L ⊆ X are parallel, if either L = L or there is a plane E ⊆ X with L , L ⊆ E and L ∩ L = ∅ . Note that the Parallel Actually, the author began to write the present text as an additional teaching material to hisonline lecture of an introductory course in linear algebra when he realized that there are severalstatements like that of Problem 2.5 and Theorem 4.1 whose status was not immediately clear andwhose treatment might slightly exceed the capacity of students in the first semester. Most of thedetails of the following text is still kept elementary and detailed in such a way that even somemotivated undergraduate students should be able to follow. f : X → Y for R -linear spaces X , Y is said to be a linear mapping , if we have(2.1) f ( c a ) = cf ( a ) for all a ∈ X and c ∈ R ; and lin-1 (2.2) f ( a + b ) = f ( a ) + f ( b ) for all a , b ∈ X . lin-2 To emphasize that we are talking about linear mappings between R -linear spaces,we shall also say R -linear mappings in contrast to the Q -linear mappings whichare linear mappings between Q -linear spaces. Here, a mapping f : X → Y is a Q -linear mapping, if (2.1) ′ and (2.2) hold where(2.1) ′ f ( c a ) = cf ( a ) for all a ∈ X and c ∈ Q .Note that, for any additive groups X , Y and mapping f : X → Y , if f satisfies(2.2), then f ( X ) = Y holds, and this fact is proved in most of the standard textbooks of linear algebra. [ We have f ( X ) = f ( X + X ) = f ( X )+ f ( X ) . Thus,by subtracting f ( X ) we obtain Y = f ( X ) .] The following Lemma is also elementary and well-known.
P-lin-0
Lemma 2.1
Suppose that X , Y are R -linear spaces, and f : X → Y a linearmapping. Then (2.3) the image f ′′ L of any line L ⊆ X is either a point or a line in Y ; and lin-3 (2.4) If f ′′ L for a line L ⊆ X is a line in Y then f ↾ L is 1-1. lin-3-0 Proof.
Suppose that L = R d + a with d = X . Then, by linearity of f , f ′′ L = { f ( r d + a ) : r ∈ R } = { rf ( d ) + f ( a ) : r ∈ R } = R f ( d ) + f ( a ). Thus, f ′′ L = { f ( a ) } if f ( d ) = Y . Otherwise, f ′′ L is the line R f ( d ) + f ( a ). In the latter case, f ↾ L : L → Y ; r d + a rf ( d ) + f ( a ), r ∈ R , is 1-1. (Lemma 2.1) For two distinct vectors a , b ∈ X , a b denotes the line R ( b − a ) + a = { t ( b − a ) + a : t ∈ R } . a b is the unique line L with a , b ∈ L . a [ b denotes the closed interval between a and b ( a [ b = { t ( b − a ) + a : 0 ≤ t ≤ } ). a ⌢ b denotes the open interval between a and b : a ⌢ b = a [ b \ { a , b } . c ∈ X divides a [ b in ratio r : s for r , s ∈ R with r + s = 0, if c = rr + s ( b − a ) + a .Note that, if rr + s [0 , c dividing a [ b in ratio r : s is on the line a b outside the interval a [ b and that any point on the line a b can be representedas a point dividing a [ b in some ratio.For convenience, we shall also use the notation a b , a [ b , and a ⌢ b for a , b with a = b defining a a = a [ a = { a } and a ⌢ b = ∅ in this case.The proof of Lemma 2.1 actually shows the following:3 -lin-1 Lemma 2.2
Suppose that X , Y are R -linear spaces, and f : X → Y a linearmapping. Then, for any a , b , c ∈ X with a = b and f ( a ) = f ( b ) , and for any r , s ∈ R , if c divides a [ b in ratio r : s then f ( c ) divides f ( a ) [ f ( b ) in ratio r : s . The property of the linear mapping mentioned in Lemma 2.1 does not charac-terize linear mappings even when we add the condition f ( X ) = Y : P-lin-2
Lemma 2.3
For any non zero-dimensional R -linear spaces X , Y , there are non-linear mappings f : X → Y satisfying (2.3) the image f ′′ L of any line L ⊆ X is either a point or a line in Y ; (2.4) If f ′′ L for a line L ⊆ X is a line in Y then f ↾ L is 1-1; and (2.5) f ( X ) = Y . lin-4 Proof.
Suppose that e ∈ X \ { X } and d ∈ Y \ { Y } . Let B be a linear basis of X with e ∈ B . For x ∈ X , let ϕ ( x ) ∈ R be the e -coordinate of x with respect to B . That is, let ϕ ( x ) ∈ R be such that there is a linear combination c of elementsof B \ { e } such that x = ϕ ( x ) e + c .Let ψ : R → R be any non-linear bijective function with(2.6) ψ (0) = 0. lin-5 Let f : X → Y be defined by f ( x ) = ψ ( ϕ ( x )) d . Clearly f is not a linear mapping. f ( X ) = Y by (2.6) and the definition of f .For a line L ⊆ X with L = R b + a for some a , b ∈ X , if ϕ ( b ) = 0 then f ′′ L = { ψ ( ϕ ( a )) d } . Otherwise, by bijectivity of ψ , we have f ′′ L = R d and f ↾ L is 1-1. (Lemma 2.3) In contrast, the property mentioned in Lemma 2.2 does characterize affine map-pings. In particular, if we add the condition (2.5), then we obtain a characterizationof linear mappings.
P-lin-3
Lemma 2.4
For any R -linear spaces X , Y , suppose that f : X → Y is such that f satisfies (2.5) f ( X ) = Y ; (2.7) for any a , b ∈ X with a = b and f ( a ) = f ( b ) , c ∈ X , and r , s ∈ R , if c lin-6 divides a [ b in ratio r : s , then f ( c ) divides f ( a ) [ f ( b ) in r : s .Then f is a linear mapping. roof. Suppose that f : X → Y satisfies (2.5) and (2.7). If f ′′ X = { Y } , then f is a linear mapping. Thus, let us assume that there is b ∈ X such that f ( b ) = Y . b = X by (2.5).For c ∈ R b , if c = c b for some c ∈ R , then c divides X [ b in ratio c : 1 − c ,since c = c b = cc +(1 − c ) ( b − X ) + X . By (2.7), it follows that f ( c b ) divides f ( X ) [ f ( b ) in ratio c : 1 − c . That is, f ( c b ) = cc +(1 − c ) | {z } =1 ( f ( b ) − f ( X ) | {z } = Y ) + f ( X ) | {z } = Y = cf ( b ). Together with (2.5), this implies that f satisfies (2.1).If a = b , then f ( a + b ) = f (2 a ) = |{z} by (2.1) f ( a ) = f ( a ) + f ( a ) = f ( a ) + f ( b ).If a = b , then a + b divides 2 a [ b in ratio 1 : 1. Thus f ( a + b ) divides f (2 a ) [ f (2 b ) = |{z} by (2.1) f ( a ) [ f ( b ) in ratio 1:1 by (2.7). This means that f ( a + b ) = (2 f ( b ) − f ( a )) + 2 f ( a ) = f ( a ) + f ( b ). This shows that f also satisfies(2.2). (Lemma 2.4) Lemma 2.3 and Lemma 2.4 suggest the following question. open-p
Problem 2.5
Suppose that X , Y are R -linear spaces and f : X → Y is such that (2.3) the image f ′′ L of any line L ⊆ X is either a point or a line in Y ; (2.5) f ( X ) = Y ; and (2.8) there are a , a ∈ X such that f ( a ) and f ( a ) are linearly independent. lin-8 Does it follow that f is an R -linear mapping? In the next sections, we shall prove a characterization of linear mappings f : X → Y satisfying (2.8) which is slightly stronger than the conditions in Problem 2.5(see Theorem 4.1).A mapping f : X → Y on linear spaces X , Y is said to be an additive functionif f satisfies (2.2). Note that, for R -linear spaces X , Y , f : X → Y is additive ifand only if it is Q -linear.Whether all additive functions f : R → R are linear mappings is a questionwhose answer depends on the axioms of set-theory. In the Zermelo’s axiom system ZC of set theory with full Axiom of Choice ( AC ), we can use a Hamel basis of R over Q , to construct 2 ℵ many Q -linear (and hence additive) functions from R to R . Since there are only 2 ℵ many R -linear functions from R to R , there are 2 ℵ many additive functions from R to R which are not R -linear. Since a continuousfunction f : R → R is decided by the information on f ↾ Q , the next Lemmafollows: 5 -lin-4 Lemma 2.6
Suppose that f : R → R is an additive function. Then the followingare equivalent : (a) f is R -linear. (b) f is continuous. (c) f is monotonous. The equivalence of (a) ⇔ (b) and (c) above follows from the fact that (1) if anadditive function is discontinuous at a point, then it is everywhere discontinuous;and (2) a monotone function can be discontinuous at most at countably manypoints.More generally, for mappings f : R m → R n , the equivalence of (a) and (b) in theprevious Lemma still holds. Further, by theorems of Steinhaus and Kuratowski,the continuity of f in (b) of Lemma 2.6 in this generalized setting can be replacedeither by being a Baire function or by being a measurable function (Kuczma [3].[2] contains a slightly simplified the proof of these results). Thus we obtain P-lin-4-a-0
Theorem 2.7
For any R -linear spaces X , Y such that Y has a linear base , andfor any aditive f : X → Y , the following are equivalent : (a) f is a linear mapping; (b) f is continuous; (c) f is Lebesgue measurable; (d) f is a Baire function. Solovay [5] constructed, arguing in the theory
ZFC + “there is an inaccessiblecardinal”, a model of ZF together with the Axiom of Dependent Choice ( DC , aweakening of the full AC which covers most of the usages of AC in everyday mathe-matics) which also satisfies the statement that “all subsets of R have Baire propertyand are Lebesgue measurable”. Note that, in this model, all functions f : R m → R n are Baire functions and also measurable, and thus all additive functions from an R -linear space X to an -linear space Y with a linear base are linear.Shelah [6] proved that the statement “all subsets of R are Lebesgue measurable”is equiconsistent with ZF + “there is an inaccessible cardinal”. Thus, the inacces-sible cardinal in Solovay’s model is unavoidable for the statement about Lebesguemeasurability. However, Shelah [6] also shows, that ZF + DC + “all subsets of R have Baire property”, hence also the theory ZF + DC + “all additive functionon R is linear” is equiconsistent with ZF . This is one of the most grave signs ofasymmetry lying between measure and category in spite of the strong similaritybetween them as is seen in Oxtoby [4].Note that Lebesgue measurability and Baire property of all functions R → R are theorems under ZF + the Axiom of Determinacy ( AD ). This is a result by JanMycielski and Stanis law Swierczkowski for Baire property and by Banach, Mazurfor Lebesgue measurability. Thus, the assertion “all additive function from an Of course, under AC , this extra assumption of the existence of a linear base is superfluous. -linear space to an R -linear space with a linear base is a linear mapping” is atheorem under this axiom system.The consistency strength of the Axiom of Determinacy, however, is much higherthan that of the statement that all subsets of R are Lebesgue measurable: Woodin’sfamous theorem states that the Axiom of Determinacy (over ZF ) is equiconsistentwith infinitely many Woodin cardinals which is much stronger than, e.g. classmany measurable cardinals.Although, in the following, we never rely on these deep results in set-theory, theyremind the possibility that a set-theoretic subtlety of this type can lurk anywhere ina discussion like the following, and warns us that careful examination is necessary.The next lemma is an easy application of Lemma 2.6: Lemma 2.8 (Kuczma [3], Theorem 14.4.1) suppose that f : R → R is an additive P-lin-4-a function and also multiplicative, i.e. we have that (2.9) f ( rs ) = f ( r ) f ( s ) for all r , s ∈ R . lin-8-0 Then f is either the constant zero function (i.e. f ( r ) = 0 for all r ∈ R ) or theidentity function (i.e. f ( r ) = r for all r ∈ R ). Proof.
For any x ∈ R ,(2.10) if x ≥
0, then f ( x ) = f (( √ x ) ) = ( f ( √ x )) ≥ lin-8-1 by the multiplicativity (2.9). It follows that, for any x , y ∈ R with x ≤ y ,(2.11) f ( y ) = f ( x + ( y − x )) = f ( x ) + f ( y − x ) ≥ f ( x ) lin-8-2 by additivity of f . Thus f is a monotone function. By Lemma 2.6, and since f isan additive function, there is c ∈ R such that f ( x ) = cx holds for all x ∈ R .By (2.9), it follows that(2.12) c = f (1) = f (1 ·
1) = f (1) f (1) = c . lin-8-3 Since f (1) ≥ c = 0 or c = 1. If c = 0, f is theconstant function f ( x ) = 0 for all x ∈ R . If c = 1, f = id R . (Lemma 2.8) P-lin-4-0 Theorem 2.9
Suppose that X and Y are R -linear spaces and f : X → Y is anadditive function (i.e. it satisfies (2.2) ). If there is a function ϕ : R → R with (2.13) f ( r a ) = ϕ ( r ) f ( a ) for all a ∈ X and r ∈ R , lin-9-0 then f is an R -linear mapping. roof. If f ′′ X = { Y } , then f is a linear mapping. Thus, we may assume that f ′′ X = { Y } .Then we have ϕ (1) = 1. Hence, by Lemma 2.8, it is enough to show that ϕ isadditive and multiplicative.To show that ϕ is additive, suppose that r , s ∈ R . Let a ∈ X be such that f ( a ) = Y . By additivity of f , we have ϕ ( r + s ) f ( a ) = f (( r + s ) a ) = f ( r a + s a ) = f ( r a ) + f ( s a ) = ϕ ( r ) f ( a ) + ϕ ( s ) f ( a ) = ( ϕ ( r ) + ϕ ( s )) f ( a ). It follows that ϕ ( r + s ) = ϕ ( r ) + ϕ ( s ).Multiplicativity of ϕ can be shown similarly: Suppose r , s ∈ R , and let a ∈ X be such that f ( a ) = Y . Then we have ϕ ( rs ) f ( a ) = f ( rs a ) = ϕ ( r ) f ( s a ) = ϕ ( r ) ϕ ( s ) f ( a ). It follows that ϕ ( rs ) = ϕ ( r ) ϕ ( s ). (Theorem 2.9) lin-section-2P-lin-5 Lemma 3.1
For R -linear spaces X , Y let f : X → Y be a mapping such that (2.3) the image f ′′ L of any line L ⊆ X is either a point or a line in Y ; and (3.1) for any line L ⊆ X , if f ′′ L is also a line in Y , then f ↾ L is 1-1. lin-7 Then, ( 1 ) f maps any plane E ⊆ X to either a plane or a line or a point of Y .If f ′′ E is a plane then f ↾ E is 1-1. ( 2 ) Suppose that L , L ⊆ X are lines such that f ′′ L and f ′′ L are also linesin Y . If L and L are parallel to each other then f ′′ L and f ′′ L are also parallelto each other. Proof. (1): Suppose that E ⊆ X is a plane. Let a , b , c ∈ E be such that b − a and c − a are independent. Case 1. f ( a ) = f ( b ) and f ( a ) = f ( c ). By (3.1) and (2.3), we have f ′′ ( a b ) = f ′′ ( a c ) = { f ( a ) } . Suppose p ∈ E \ ( a b ∪ a c ). Then there is a line L ⊆ X going through p and crossing a b and a c at two different points. Since thevalue of f at both of these two points is f ( a ), f ↾ L is not 1-1. By (3.1) and (2.3)it follows that f ( p ) = f ( a ). Thus we have f ′′ E = { f ( a ) } . Case 2. f ( a ) = f ( b ) and f ( a ) = f ( c ). Then f ′′ ( a b ) = { f ( a ) } , f ′′ ( a c ) = f ( a ) f ( c ), and f ′′ ( a c ) is a line in Y . For p ∈ E \ ( a b ∪ a c ), there is a line L ⊆ X going through p and crossing a b and a c at two different points, sayat { p } and { p } respectively. If f ( p ) = f ( p ), then we have f ( p ) = f ( p ) = f ( a ). Otherwise, since f ( p ) and f ( p ) belong to f ( a ) f ( c ) we have f ( p ) ∈ f ( p ) f ( p ) = f ( a ) f ( c ). This shows that f ′′ E = f ( a ) f ( c ).8 ase 3. f ( a ) = f ( b ) and f ( a ) = f ( c ). This case can be treated similarly tothe Case 2. to conclude that f ′′ E = f ( a ) f ( b ). Case 4. f ( a ) = f ( b ) and f ( a ) = f ( c ). Then we have f ′′ ( a c ) = f ( a ) f ( b ), f ′′ ( a c ) = f ( a ) f ( c ), and both f ′′ ( a b ) and f ′′ ( a c ) are lines in Y . If f ( a ) f ( b ) = f ( a ) f ( c ) then we can argue similarly to Case 2. and show f ′′ E = f ( a ) f ( b ).Suppose now f ( a ) f ( b ) = f ( a ) f ( c ). Then we have f ( a ) f ( b ) ∩ f ( a ) f ( c )= { f ( a ) } . Let E ∗ be the plane with f ( a ) f ( b ), f ( a ) f ( c ) ⊆ E ∗ .If p ∈ E \ ( a b ∪ a c ) then we can take a line L ⊆ X going through p andcrossing a b and a c at two different points, say at { p } and { p } respectively.Since f ( p ), f ( p ) ∈ f ( a ) f ( b ) ∪ f ( a ) f ( c ), we have f ( p ) ∈ f ( p ) f ( p ) ⊆ E ∗ .Conversely, suppose that q ∈ E ∗ . Then there is a line L ∗ in Y going through q andcrossing f ( a ) f ( b ) and f ( a ) f ( c ) at two different points say { q } and { q } . Let p ∈ a b , p ∈ a c be the unique vectors such that f ( p ) = q and f ( p ) = q .Since f ′′ ( p p ) = q q , there is p ∈ p p ⊆ E such that f ( p ) = q .To see that f ↾ E is 1-1, suppose that p , p ∈ E are distinct vectors. If f ( p ) = f ( p ) then we can produce the constellation of Case 1. or Case 2. with p and p together with some third point ∈ E . This is a contradiction since wealready know that f ′′ E is a plane in Y .(2): If f ′′ L = f ′′ L then f ′′ L and f ′′ L are parallel.Assume otherwise. Let E be the plane in X with L , L ⊆ E . Since f ′′ L , f ′′ L ⊆ f ′′ E , and since f ′′ L and f ′′ L are two distinct lines in f ′′ E , f ′′ E is a planein Y and f ↾ E is 1-1, by (1). Since L ∩ L = ∅ . We also have f ′′ L ∩ f ′′ L = ∅ .Thus, also in this case, f ′′ L and f ′′ L are parallel to each other. (Lemma 3.1) P-lin-5-0 Lemma 3.2
Suppose that X and Y are R -linear spaces, and f : X → Y a mappingsatisfying (2.5) , (2.3) and (3.1) . If a , b ∈ X are such that a = X , b = X , f ( a ) = Y , and f ( b ) = Y , then we have f ( a + b ) = f ( b ) = f ( a ) + f ( b ) . Proof. a and b are linearly independent: f ↾ R b is 1-1 by (2.3) and (3.1). Thus,if a ∈ R b , we would have f ( a ) = Y by (2.5).Let E ⊆ X be the plane with X , a , b ∈ E . By Lemma 3.1, (1), we have f ′′ E = f ( b ) Y = R f ( b ). Note also that f ′′ R a = { Y } .Suppose, toward a contradiction, that(3.2) f ( a + b ) = f ( b ). lin-9-1 Then f ↾ ( a + b ) b is not a constant function. Thus, we have f ′′ ( a + b ) b = R f ( b ). Let c ∈ ( a + b ) b be such that f ( c ) = Y . Then we have f ′′ c 0 X = { Y } .9or each d ∈ ( a + b ) b , let L ⊆ X be a line going through d which crosses c 0 X and a 0 X at different points, say at points p and p respectively. Thenwe have f ( p ) = f ( p ) = Y . Thus f ′′ L = { Y } and f ( d ) = Y . In particular,we have f ( a + b ) = f ( b ) = Y . This is a contradiction to the assumption (3.2). (Lemma 3.2) lin-section-3 The following theorem gives a positive partial answer to Problem 2.5
P-lin-6
Theorem 4.1
Suppose that X , Y are R -linear spaces and f : X → Y satisfies (2.5) f ( X ) = Y ; and, (2.8) there are a , a ∈ X such that f ( a ) and f ( a ) are linearly independent.Then f is a linear mapping if and only if f satisfies (2.3) the image f ′′ L of any line L ⊆ X by f is either a point or a line in Y ;and, (3.1) for any line L ⊆ X , if f ′′ L is also a line in Y , then f ↾ L is 1-1. Proof.
Linear mappings satisfy (2.3) and (3.1) (Lemma 2.1).Suppose that f : X → Y satisfies (2.5), (2.8), (2.3) and (3.1). By Theorem 2.9,it is enough to show that there is ϕ : R → R with (2.13) for this f , and f isadditive.To prove that there is a mapping ϕ : R → R with (2.13), suppose a ∈ X . Notethat a 0 X = R a . If f ( a ) = Y , then f ′′ ( a 0 X ) = { Y } by (2.3), (2.5) and (3.1).Thus we have(4.1) f ( r a ) = Y = r f ( a ), for any r ∈ R (under f ( a ) = Y ). lin-10 Suppose now that f ( a ) = Y . Then f ′′ R a = f ( a ) Y (= R f ( a )) and f ↾ R a is 1-1 by (2.3), (2.5), (3.1). For each a ∈ X with f ( a ) = Y and r ∈ R , let ϕ ( a , r )be s ∈ R such that f ( r a ) = ϕ ( a , r ) f ( a ). Cl-lin-0
Claim 4.1.1 ϕ ( a , r ) does not depend on a ∈ X with f ( a ) = Y . ⊢ we first prove the following: Subclaim 4.1.1.1
For any a , b ∈ X such that f ( a ) and f ( b ) are independent in Y , we have ϕ ( a , r ) = ϕ ( b , r ) for all r ∈ R . For a ∈ X with f ( a ) = Y , f (0 a ) = Y by (2.5). Thus ϕ ( a ,
0) = 0 for all such a ∈ X .For r = 0, let us consider the four lines L = a 0 X , L = b 0 X , L = a b ,and L = r a r b . We have L ∩ L = { X } , L ∩ L = { a } , L ∩ L = { b } , L ∩ L = { r a } , L ∩ L = { r b } , and, L and L are parallel to each other. By Lemma 3.1,it follows that f ′′ L = f ( a ) Y , f ′′ L = f ( b ) Y , f ′′ L = f ( a ) f ( b ), and f ′′ L = f ( r a ) f ( r b ) = ϕ ( a , r ) f ( a ) ϕ ( b , r ) f ( b ). f ′′ L ∩ f ′′ L = { Y } , f ′′ L ∩ f ′′ L = { f ( a ) } , f ′′ L ∩ f ′′ L = { f ( b ) } , f ′′ L ∩ f ′′ L = { f ( r a ) } = { ϕ ( a , r ) f ( a ) } , f ′′ L ∩ f ′′ L = { f ( r b ) } = { ϕ ( b , r ) f ( b ) } , and, f ′′ L and f ′′ L are parallel to eachother. The last condition is only possible when ϕ ( a , r ) = ϕ ( b , r ). ⊣ (Subclaim 4.1.1.1) Let a , a ∈ X be as in (2.8). By Subclaim 4.1.1.1, we have ϕ ( a , r ) = ϕ ( a , r )for any r ∈ R . Suppose that a ∈ X is such that f ( a ) = Y . Then there is i ∈
2, such that f ( a ) and f ( a i ) are linearly independent. By Subclaim 4.1.1.1,we have ϕ ( a , r ) = ϕ ( a i , r ). Thus, for any a ∈ X with f ( a ) = Y and r ∈ R , ϕ ( a , r ) = ϕ ( a , r ). ⊣ (Claim 4.1.1) Let ϕ : R → R be defined by ϕ ( r ) = ϕ ( a , r ). By Claim 4.1.1 and the argumentabove it, this ϕ satisfies (2.13).To prove the additivity of f , suppose a , b ∈ X . Case 1. f ( a ) and f ( b ) are independent in Y . In this case, a and b areindependent and, letting E be the plane in X with a , b , X ∈ E , we have that f ↾ E is a 1-1 mapping and f ′′ E is the plain in Y with f ( a ), f ( b ), Y (= f ( X )) ∈ f ′′ E by Lemma 3.1, (1).Let L = a 0 X , L = b 0 X , L = a ( a + b ), L = b ( a + b ). Then we have L ∩ L = { X } , L ∩ L = { a } , L ∩ L = { b } , and L ∩ L = { a + b } . We alsohave that L and L are parallel to each other, and, L and L are parallel to eachother.By (2.3) and (2.5), and since f ↾ E is 1-1, it follows that f ′′ L = f ( a ) Y , f ′′ L = f ( b ) Y , f ′′ L = f ( a ) f ( a + b ), f ′′ L = f ( b ) f ( a + b ). We alsohave f ′′ L ∩ f ′′ L = { Y } , f ′′ L ∩ ′′ L = { f ( a ) } , f ′′ L ∩ f ′′ L = { f ( b ) } , and f ′′ L ∩ f ′′ L = { f ( a + b ) } . f ′′ L and f ′′ L are parallel to each other, and, f ′′ L and f ′′ L are parallel to each other by Lemma 3.1, (2). This implies that f ( a + b ) = f ( a ) + f ( b ). 11 ase 2. a and b are not linearly independent. If one of a , b equals to X thenthe additivity(4.2) f ( a + b ) = f ( a ) + f ( b ) lin-11 trivially holds by (2.5). Thus we may assume without loss of generality that a = X , b = X , and b ∈ R a . If f ( a ) = Y or f ( b ) = Y , then f ′′ R a = { Y } by (2.3), (2.4)and (2.5) and the equation (4.2) again holds. Thus we may assume that f ( a ) = Y , f ( b ) = Y and f ( b ) ∈ R f ( a ).Let a , a ∈ X be as in (2.8). Then there is i ∈ f ( a i ) and f ( a ) areindependent. Let E be the plane with a i , a , X ∈ E . By Lemma 3.1, (1), f ↾ E is1-1 and f ′′ E is a plane with f ( a i ), f ( a ), f ( b ), Y ∈ f ′′ E .Let L = a 0 X , L = ( a i + a ) a i , L = a i X , L = ( a i + a ) X , L =( a i + a ) a , L = ( a i + a + b ) b , and L = ( a i + a + b ) ( a + b ).We have L ∩ L ∩ L = { X } , L ∩ L = { a } , L ∩ L = { b } , L ∩ L = { a + b } , L ∩ L = { a i } , L ∩ L ∩ L = { a i + a } , and L ∩ L ∩ L = { a i + a + b } .We also have that L and L are parallel to each other, L , L and L are parallelto each other, as well as L and L are parallel to each other. Since f transfers thisconstellation keeping all the parallelisms, it follows, similarly to the Case 1., we canconclude in turn that f ( a i + a ) = f ( a i ) + f ( a ), f ( a i + a + b ) = f ( a i ) + f ( a ) + f ( b ),and finally f ( a + b ) = f ( a ) + f ( b ). 12 ase 3. a and b are independent in X but f ( a ) and f ( b ) are not independentin Y . If f ( a ) = f ( b ) = Y then f ( a + b ) = Y = f ( a ) + f ( b ) by the the Case1 in the proof of Lemma 3.1, (1). If one of f ( a ) and f ( b ) is Y , then we have f ( a + b ) = f ( a ) + f ( b ) by Lemma 3.2. Thus we may assume that f ( a ) = Y and f ( b ) = Y . Let E ⊆ X be the plane with X , a , b ∈ E . By the assumption of thepresent case, we have f ′′ E = R f ( a ) = R f ( b ). Thus, there is i ∈ f ( a i )is independent from f ( a ) (and also from f ( b ) ).In this case, some of the lines connecting X a , b , a i , a i + a , a i + a + b , a + b aresent to the same lines by f . However, parallelism of some lines survive to concludein turn that f ( a i + a ) = f ( a i ) + f ( a ), f ( a i + a + b ) = f ( a i ) + f ( a ) + f ( b ), andfinally f ( a + b ) = f ( a ) + f ( b ) just as in Case 2. (Theorem 4.1) g : X → Y for R -linear spaces which can be represented as a f ( a ) + b for a linear mapping f : X → Y and b ∈ Y ) with the following conditioncorresponding to (2.8): P-lin-7
Corollary 4.2
Suppose that X , Y are R -linear spaces and g : X → Y satisfies (2.8) ′ there are a ∗ a ∗ , a ∗ ∈ X such that g ( a ∗ ) − g ( a ∗ ) and g ( a ∗ ) − g ( a ∗ ) arelinearly independent.Then g is an affine mapping if and only if g satisfies (2.3) the image g ′′ L of any line L ⊆ X by g is either a point or a line in Y ;and (3.1) for any line L ⊆ X , if g ′′ L is also a line in Y , then g ↾ L is 1-1. Proof. If g : X → Y is an affine mapping then g clearly satisfies (2.3) and (3.1).Conversely, suppose that g : X → Y satisfies (2.8) ′ , (2.3) and (3.1). Let a ∗ , a ∗ , a ∗ ∈ X be as in (2.8) ′ . Let f : X → Y be defined by(4.3) f ( a ) = g ( a + a ∗ ) − g ( a ∗ ) for a ∈ X . lin-11-0 Clearly f also satisfies (2.3) and (3 . f ( X ) = g ( a ∗ ) − g ( a ∗ ) = Y . f ( a ∗ i − a ∗ ) = g ( a ∗ i − a ∗ + a ∗ ) − g ( a ∗ ) = g ( a ∗ i ) − g ( a ∗ ) for i ∈
2. Thus, letting a i = a ∗ i − a ∗ for i ∈
2, we see that f satisfies (2.8). By Theorem 4.1, it follows that f is a linear mapping. Since we have(4.4) g ( a ) = g (( a − a ∗ ) + a ∗ ) = f ( a − a ∗ ) + g ( a ∗ ) = f ( a ) + f ( − a ∗ ) + g ( a ∗ ) = f ( a ) + g ( X )for all a ∈ X , g is an affine mapping. (Corollary 4.2) The following (4.5) apparently holds for any affine mapping f : X → Y . Con-versely, it is easy to see that (2.3) and (4.5) imply (2.4). Thus we obtain: P-lin-8
Corollary 4.3
Suppose that X , Y are R -linear spaces and g : X → Y satisfies (2.8) ′ there are a ∗ a ∗ , a ∗ ∈ X such that g ( a ∗ ) − g ( a ∗ ) and g ( a ∗ ) − g ( a ∗ ) arelinearly independent.Then g is an affine mapping if and only if g satisfies (2.3) the image g ′′ L of any line L ⊆ X by g is either a point or a line in Y ;and, for any a , b , c ∈ X with c ∈ [ ⌢ b , g ( a ) , either g ( a ) = g ( b ) = g ( c ) , or lin-12 g ( c ) ∈ g ( a ) ⌢ g ( b ) holds. The following characterization of linear mappings consisting of a mixture ofalgebraic and geometric properties is also easy to prove.
P-lin-9
Proposition 4.4
Suppose that X and Y are R -linear spaces and f : X → Y .Then f is linear if and only if (4.6) f is an additive function; and (4.5) ′ for any a , c ∈ X , if c ∈ a ⌢ X , then, either f ( a ) = f ( c ) = Y or c ∈ f ( a ) ⌢ Y . Proof.
For a ∈ X , let ϕ a : R → R be such that ( ℵ f ( r a ) = ϕ a ( r ) f ( a ) for all r ∈ R . lin-13 We assume, without loss of generality, that ϕ a ( r ) = r for all r ∈ R ,if f ( a ) = Y . Claim 4.4.1 ϕ a is additive and monotonically increasing. ⊢ If f ( a ) = Y , the assertion is trivial. So assume that f ( a ) = Y .For r , s ∈ R , we have ϕ a ( r + s ) f ( a ) = f (( r + s ) a ) = f ( r a + s a ) = f ( r a ) + f ( s a ) = ϕ a ( r ) f ( a )+ ϕ a ( s ) f ( a ) = ( ϕ a ( r )+ ϕ a ( s )) f ( a ) . This implies that ϕ a ( r + s ) = ϕ a ( r ) + ϕ a ( s ) .To show that ϕ a is monotonically increasing, suppose that r , s ∈ R with r < s . For simplicity, consider the case < r < s . All othercases can be treated similarly. Since a ∈ r a ⌢ X , we have f ( a ) ∈ f ( r a ) ⌢ Y by (4.5) ′ . By definition ( ℵ , it follows that < ϕ a ( r ) . Since r a ∈ s a ⌢ X , it follows that ϕ a ( r ) < ϕ a ( s ) . ⊣ (Claim 4.4.1) By Lemma 2.6, it follows that ϕ a is R -linear since ϕ a (1) = 1 , ϕ a mustbe the identity function. (Proposition 4.4) References ref [1] Shiri Artstein-Avidan, and Boaz A. Slomka, The fundamental theorems ofaffine and projective geometry revisited, Communications in ContemporaryMathematics Vol. 19, No. 5, (2017).152] Saka´e Fuchino, On continuity of additive functions (in Japanese), Memoirs ofCollege of Engineering, Chubu University, Vol.37, (2001), 55–64. https://ci.nii.ac.jp/naid/120006519383/enhttps://ci.nii.ac.jp/naid/120006519383/en