Preliminaries : A function f : R R is additive if it satisfies the - PDF document

1 Preliminaries : A function f : R − → R is additive if it satisfies the Cauchy equation (CE) f ( x + y ) = f ( x )+ f ( y ) for all x, y ∈ R . Cauchy asked under what conditions an addi- tive function must be linear. The following lemma is obvious. Lemma. If f is additive, then f ( qx ) = qf ( x ) for all q ∈ Q . Thus, if, in addition, f is con- tinuous, then f ( x ) = f (1) x . If one is willing to use the axiom of choice, then one can construct a non-linear additive function as follows. By Zorn’s Lemma, there exists a maximal B ⊆ R of numbers which are linearly independent over Q . Thus, for each x , there exists a unique { a b ( x ) : b ∈ B } ⊆ Q such that a b ( x ) = 0 for all but a finite number of b ∈ B and x = � b ∈ B a b ( x ) b . Clearly, for each b ∈ B , x � a b ( x ) is additive. On the other hand, because a b ( x ) ∈ Q for all x , it is obvious that a b is not linear.

2 Theorem. If f is a Lebesgue measurable, ad- ditive function, then f is linear. Proof. Choose R > 0 so that { x : | f ( x ) | ≤ R } has positive Lebesgue measure. Then, Vitalli guarantees that there is a δ > 0 such that [ − δ, δ ] ⊆ { y − x : | f ( x ) | ∨ | f ( y ) | ≤ R } . Hence, | f ( x ) | ≤ 2 R if | x | ≤ δ . Given any x � = 0, choose δ . Then x ′ ≡ x q ∈ Q so that | x | δ ≤ q ≤ 2 | x | q ∈ [ − δ, δ ] and so | f ( x ) | = q | f ( x ′ ) | ≤ 4 R | x | . More δ generally, | f ( y ) − f ( x ) | = | f ( y − x ) | ≤ 4 R | y − x | , δ and so f is continuous. �

3 Another Approach Lemma. If f is a Lebesgue measurable, ad- ditive function that is locally integrable, then f ( x ) = xf (1) . Proof. It suffices to show that f is continu- ous. To this end, let ρ : R − → R be a smooth function with compact support and integral 1. Then � ρ ∗ f ( x ) = f ( t ) ρ ( x − t ) dt = f ( x ) + ρ ∗ f (0) , and so f = ρ ∗ f − ρ ∗ (0) is smooth. � Given a Borel probability measure µ on R , define µ α for α ∈ R to be the distribution of � x ∈ R �− → αx under µ . That is, f dµ α = � f ( αx ) µ ( dx ). Say that ( α, β ) ∈ (0 , 1) is a Pythagorean pair if α 2 + β 2 = 1, and let γ be the standard Gauss measure on R . That is, γ ( ξ ) = e − ξ 2 2 e − x 2 γ ( dx ) = (2 π ) − 1 2 dx and ˆ 2 .

4 Lemma. If ( α, β ) is a Pythagorean pair, then µ = µ α ∗ µ β if and only if µ = γ σ for some x 2 µ ( dx ) = σ 2 . � σ ≥ 0 , in which case γ σ ( ξ ) = e − σ 2 ξ 2 2 , the “if” assertion Proof. Since ˆ is trivial. Thus, assume that µ = µ α ∗ µ β . I begin by proving the last statement un- der the assumption that µ is symmetric. By assumption, ˆ µ ( ξ ) = ˆ µ ( αξ )ˆ µ ( βξ ). Thus, by in- duction on n ≥ 1, n � ( m ) . n � α m β n − m ξ � µ ( ξ ) = ˆ µ ˆ m =0 � Because µ is symmetric, ˆ µ ( ξ ) = cos( ξx ) µ ( dx ), and therefore � � − log µ (1) ˆ n � n � �� � � cos( α n β n − m x ) µ ( dx ) = − log . m m =0

5 Since − log t ≥ 1 − t for t ∈ [0 , 1], this means that � � − log µ (1) ˆ � � n � � n �� � 1 − cos( α m β n − m x ) � ≥ µ ( dx ) , m m =0 and, because n → x 2 � n �� � 1 − cos( α m β n − m x ) � − 2 , m m =0 Fatou’s Lemma guarantees that � x 2 µ ( dx ) ≤ − 2 log � � µ (1) ˆ . To remove the symmetry assumption, set ν = µ ∗ µ − 1 . Then ν is symmetric and ν = x 2 ν ( dx ) < ∞ . Now choose a � ν α ∗ ν β . Thus, median a of µ . That is, ≥ 1 � � � � µ [ a, ∞ ) ∧ µ ( −∞ , a ] 2 .

6 Then � � � � µ { x : | x − a | ≥ t } ≤ 2 ν { x : | x | ≥ t } , and so � � x 2 µ ( dx ) ≤ 2 a 2 + 2 | x − a | 2 µ ( dx ) � ≤ 2 a 2 + 4 x 2 ν ( dx ) < ∞ . Given that µ has a finite second moment σ 2 , note that � � x µ ( x ) = ( α + β ) x µ ( dx ) , � and therefore, since α + β > 1, x µ ( dx ) = 0. Thus, µ ( η ) = 1 − σ 2 η 2 � � ˆ 1 + o (1) as η → 0 , 2

7 and so, since n µ ( α m β n − m ξ )( m ) , n � µ ( ξ ) = ˆ ˆ m =0 one has that = − σ 2 ξ 2 � � log µ ( ξ ) ˆ . � 2 Now suppose that f is a Lebesgue measur- able, additive function, and let µ be the distri- bution of f under γ : � � ϕ ◦ f dγ = ϕ dµ. Take α = 3 5 and β = 4 5 . Then µ ( αξ )ˆ ˆ µ ( βξ ) �� � �� � = exp iξ f ( αx ) + f ( βy ) γ ( dx ) γ ( dy ) �� e iξf ( αx + βy ) γ ( dx ) γ ( dy ) = ˆ = µ ( ξ ) .

8 Hence � � f ( x ) 2 γ ( dx ) = x 2 µ ( dx ) < ∞ , and so f is locally integrable and therefore lin- ear. This approach has several advantages. For example, with hardly any change in the argu- ment, one can use it to show that if f is a Lebesgue measurable function with the prop- erty that for a.e. ( x, y ) ∈ R 2 , f ( x + y ) = f ( x ) + f ( y ) then there is an a ∈ R such that f ( x ) = ax for a.e. x ∈ R . Indeed, using Fubini’s theo- rem and the translation invariance of Lebesgue measure, one can show first that f ( αx + βy ) = αf ( x ) + βf ( y ) for a.e. ( x, y ) ∈ R 2 when α and β are rational. One then proceeds as before to show that f is locally integrable and is there- fore equal to ˜ f ≡ ρ ∗ f − c a.e. for some c ∈ R .

9 Finally, because ˜ f is a continuous and equal a.e. to f , it must be a continuous, additive function and therefore satisfy ˜ f ( x ) = x ˜ f (1). Infinite Demensions Suppose that E and F are a pair of Banach spaces over R and that Φ : E − → F is an additive, Borel measurable function. Then, for each x ∈ E and y ∗ ∈ F ∗ , t � � Φ( tx ) , y ∗ � is an R -valued, Borel measurable, additive function on R and is therefore linear. Hence, Φ is a Borel measurable, linear function on E , and one can ask whether it is continuous. As we will now show, the answer is yes. When E and F are separable, one can use Laurent Schwartz’s Borel graph theorem to prove this. Namely, his theorem says that if E and F are separable Banach spaces and Φ : E − → F is a linear map whose graph is Borel measurable, then Φ is continuous. His proof is a tour de force based on deep properties of Polish spaces, and using those properties it is easy to show

10 that Φ is Borel measurable if and only if its graph is. Indeed, a Borel measurable, one-to- one map from one Polish space to another takes Borel sets to Borel sets. Applying this fact to � � the map x ∈ E �− → x, Φ( x ) ∈ E × F , one sees that the graph G (Φ) of Γ is Borel measurable if Φ is. Conversely, if G (Φ) is Borel measur- able and π E and π F are the natural projection maps of E × F onto E and F , then π E ↾ G (Φ) is a one-to-one, Borel measurable map and, as such, its inverse is Borel measurable. Since � − 1 , this shows that Φ � Φ = π F ◦ π E ↾ G (Φ) is Borel measurable. To prove this result by the technique used earlier, we need to introduce a Gaussian mea- sure on E . For this purpose, take P = γ Z + on Ω ≡ R Z + . Given a sequence { x n : n ≥ 1 } ⊆ E set n for n ∈ Z + and ω ∈ Ω , � S n ( ω ) = ω m x m m =1

11 � � A ≡ ω : lim n →∞ S n ( ω ) exists in E and � lim n →∞ S n ( ω ) if ω ∈ A S ( ω ) = 0 if ω / ∈ A. Since � ∞ ∞ � � 2 � � E P | ω m |� x m � E = � x m � E , π m =1 m =1 P ( A ) = 1 if � ∞ m =1 � x m � E < ∞ . Lemma. If f : E − → R is a Borel measur- able, linear map, then f ( x m ) 2 ≤ E P � ( f ◦ S ) 2 � < ∞ for all m ≥ 1 . Proof. Since, for every ω ∈ Ω, S ( ω ) is an ele- ment of the closed linear span of { x n : n ≥ 1 } , I will, without loss in generality, assume that E

12 is separable and therefore that B E 2 = B E ×B E . In particular, this means that the map ( x, y ) ∈ E 2 �− → x + y 2 ∈ E is B E × B E -measurable. √ Next note that f ◦ S ( ω 1 ) + f ◦ S ( ω 2 ) � S ( ω 1 ) + S ( ω 2 ) � √ √ = f 2 2 for ( ω 1 , ω 2 ) ∈ A 2 , and therefore the distribu- tion µ of f ◦ S under P is γ σ for some σ ≥ 0. = σ 2 < ∞ . ( f ◦ S ) 2 � Hence, E P � To complete the proof, let m ∈ Z + be given, and define ω � S ( m ) ( ω ) relative to the se- quence { (1 − δ m,n ) x n : n ≥ 1 } . Then S ( m ) ( ω ) is P -independent of ω m , and S ( ω ) = ω m x m + S ( m ) ( ω ) for ω ∈ A. Hence ( f ◦ S ) 2 � = f ( x m ) 2 + E P � ( f ◦ S ( m ) ) 2 � ≥ f ( x m ) 2 . E P � �

13 Theorem. If f : E − → R is a Borel measur- able, additive map, then there exists an x ∗ ∈ E ∗ such that f ( x ) = � x, x ∗ � . Proof. We know that f is linear. Suppose it were not continuous. We could then find { x n : 1 n ≥ 1 } such that � x n � E ≤ n 2 and | f ( x n ) | ≥ n . Now define ω � S ( ω ) accordingly, and thereby get the contradiction that m 2 ≤ | f ( x m ) | 2 ≤ E P � ( f ◦ S ) 2 � < ∞ for all m ≥ 1 . � Corollary. Suppose that Φ : E − → F is a ad- ditive map with the property that x � � Φ( x ) , y ∗ � is Borel measurable for each y ∗ ∈ F ∗ . Then Φ is continuous. Proof. By the closed graph theorem, it suffices to show that G (Φ) is closed. By the preceding, we know that x � � Φ( x ) , y ∗ � is continuous for each y ∗ ∈ F ∗ . Now suppose