A note on uniqueness of extension for characteristic functions
aa r X i v : . [ m a t h . C A ] S e p A note on uniqueness of extension for characteristic functions
Saulius Norvidas
Institute of Data Science and Digital Technologies, Vilnius University,Akademijos str. 4, Vilnius LT-04812, Lithuania(e-mail: [email protected])
Abstract
Let f : R → C be the characteristic function of a probability measure. We study the following question: Isit true that for any closed interval I on R , which does not contain the origin, there exists a characteristic function g such that g coincides with f on I but g f on R ? Keywords : Characteristic function; density function; entire function of exponential type; probability measure
Mathematics Subject Classification : 30D15 - 42A38 - 42A82 - 60E10
Let M ( R ) be the set of bounded complex-valued regular Borel measures on the real line R . For µ ∈ M ( R ) , we define its Fourier transform byˆ µ ( t ) = Z R e − itx d µ ( x ) , t ∈ R . If µ ∈ M ( R ) is non-negative and k µ k =
1, then in the language of probability theory, µ and f ( t ) = ˆ µ ( t ) are called a probability measure and its characteristic function, respectively. Anycharacteristic function f satisfies f ( ) = f ( − t ) = f ( t ) , t ∈ R . (1.1)Let L ( R ) ⊂ M ( R ) be the usual Lebesgue space of absolutely continuous measures (with respectto Lebesgue measure dx ). If ϕ ∈ L ( R ) , k ϕ k L ( R ) =
1, and ϕ ≥ R , then ϕ is called aprobability density function, or (probability) density for short. Note that if ϕ is a measurablefunction, then we write ϕ ≥ R , if ϕ ≥ dx -almost everywhere on R .A complex-valued function f defined on ( − A , A ) , 0 < A ≤ ∞ , is said to be positive definite if n ∑ j , k = f ( t j − t k ) c j c k ≥ c , . . . , c n ∈ C and t , . . . , t n ∈ ( − A / , A / ) . The Bochner theorem (see, e.g., [11,p. 71]) states that a continuous function f : R → C is characteristic function if and only if f ispositive definite on R and f ( ) = n this paper, we shall be concerned with a problem of uniqueness for extensions of characteristicfunctions. By a classical results of Krein [8] (see also Akutowicz [1], Kahane [7], Levin [10], andSasv´ari [14]), for any neighbourhood U of the origin and a continuous positive definite function f on U , there exists a continuous extension of f to R so as to retain the positive definite property.In general such an extension is not unique (see, e.g., [11, p. 89]). A question concerning theuniqueness of that extensions was also studied. Conditions for the uniqueness can be found in[8].We consider a certain extension problem in the case if U is a neighbourhood of infinity. Ourstudy is initiated by the question posed by N.G. Ushakov ([16, p. 276]) : Is it true that for anyinterval [ a , b ] ⊂ R , 0 / ∈ [ a , b ] , there exists the characteristic function f such that f e − t / , but f ( t ) = e − t / for all t ∈ [ a , b ] ? A positive answer to this question was given by T. Gneiting in [4].It follows from the following result: Theorem 1.1 [4, p. 360]. Let f : R → C be the characteristic function of a distribution witha continuous and strictly positive density. Then there exists, for each a > , a characteristicfunction g such that f ( t ) = g ( t ) if t = or | t | ≥ a and f ( t ) = g ( t ) otherwise. Repeating the author’s definition [4, p. 361], we say that a characteristic function f has thesubstitution property if, for any a >
0, there exists a characteristic function g such that f ( t ) = g ( t ) for | t | > a but f g . Also in [4, p. 361] we find the following: One might conjecture that anycharacteristic function with an absolutely continuous component has the substitution property .This conjecture fails in general. Indeed, for a > ϕ a ( x ) = ( ( a − | x | ) a , | x | ≤ a , , otherwise,is the density of the triangular probability distribution. Let σ > g is a char-acteristic function such that g = c ϕ a on the neighbourhood of infinity R \ [ − σ , σ ] . If a satisfies a ≤ π / σ , then g = c ϕ a on the whole R (see [13, Example 1, p. 238]). Below we improve thisestimate in our Corollary 1.4. Moreover, in Theorem 1.5 and Proposition 1.8 we show that theexact estimate is a ≤ π / σ .Let us recall the notation that will be used throughout this paper. In view of (1.1), it is enough tostudy the extensions of characteristic functions only from symmetric neighborhoods of infinity.Therefore, in the sequel, U σ = { x ∈ R : | x | > σ } , σ >
0, denotes such a neighborhood. Given a measurable function ϕ : R → R , we denote by S ϕ the essential support of ϕ . By definition, a point x ∈ R belongs to S ϕ if, for any ε >
0, the set ( x − ε , x + ε ) ∩ { t ∈ R : | ϕ ( t ) | > } as positive Lebesgue measure. We call N ϕ = R \ S ϕ the essential zero set of ϕ . The Lebesquemeasure of a measurable E ⊂ R will be denoted by | E | . Let A be a proper subset of R . Forfunctions f and g defined on A , we write f = g on A if f ( x ) = g ( x ) for all x ∈ A . If f ( x ) = g ( x ) for all x ∈ R , then we write f ≡ g . Finally, let Z be the group of integers as usual.We now formulate our results. We start by proving that it is sufficient to study a smaller class ofcharacteristic functions. Proposition 1.2
Suppose that µ and η are probability measures on R . Let µ = µ a + µ s and η = η a + η s be the usual Lebesgue decompositions of µ and η into their absolutely continuousand singular parts, respectively. If there exists a U σ such that b µ = b η on U σ , then µ s = η s . For this reason, we can restrict our extension problem to a smaller class of characteristic functionsof probability densities. The following uniqueness theorem is the main result of this paper.
Theorem 1.3
Let f : R → C be the characteristic function of a probability density ϕ . Supposethat there exist a > and measurable set E ⊂ [ , a ) such that | E | > and (cid:12)(cid:12)(cid:12) ( E + a Z ) ∩ S ϕ (cid:12)(cid:12)(cid:12) = . (1.2) Let g : R → C be a characteristic function, and let f = g on U σ . Ifa ≤ πσ , (1.3) then f ≡ g. Corollary 1.4
Let f , g, and ϕ be the same as that in Theorem 1.3. If f = g on U σ and (cid:12)(cid:12) S ϕ (cid:12)(cid:12) < πσ , (1.4) then f ≡ g. The statements of Theorem 1.3 and Corollary 1.4 are sharp in the sense that the right-hand sidesof (1.3) and (1.4) cannot be replaced by 2 π / σ + ε for any positive ε . Indeed, we can show thefollowing. Theorem 1.5
Suppose that f : R → C is the characteristic function of a continuous probabilitydensity ϕ . Let α , β ∈ R , α < β , be such that ( α , β ) ⊂ S ϕ . (1.5) If β − α > πσ , (1.6) then there is a characteristic function g such that f = g on U σ but f g. t is easy to see that Theorem 1.5 provides the following sufficient conditions under which acharacteristic function has the substitution property. Corollary 1.6
Let f : R → C be the characteristic function of a continuous probability density ϕ . If, for any δ > , there exists A = A ( δ ) ∈ R such that ( A , A + δ ) ⊂ S ϕ , (1.7) then f has the substitution property. Conjecture 1.7
Suppose that f is the characteristic function of a probability measure with anontrivial absolutely continuous component. If ϕ is the density function of this component, thenf has the substitution property if and only if N ϕ does not contain any lattice τ + a Z , τ ∈ R , a > . Of course, if ϕ is continuous, then the sufficiency part of this conjecture follows from Corollary1.6.Finally, we show that for the limit case β − α = π / σ in (1.4) and (1.6), there is no an exactanswer to the question for uniqueness of characteristic extension from U σ . Proposition 1.8
Given σ > and α ∈ R , let ϕ be any probability density withS ϕ = (cid:16) α , α + πσ (cid:17) . (i) There exists a ϕ such that for any characteristic function g : R → C with b ϕ = g on U σ , itfollows that b ϕ ≡ g.(ii) There exist a ϕ and the characteristic function g : R → C such that b ϕ = g on U σ but b ϕ g. We conclude this section by presenting our previous paper [13], where a similar extension prob-lem was studied in the case of continuous density functions. The main result of [13] states that if ϕ is a continuous probability density such that there exist lattices Λ j = τ j + α j Z , τ j ∈ R , α j > α j σ ≤ π , j = , Λ ∩ Λ = /0, and ϕ vanishes on Λ ∪ Λ , then, for any characteristic function g : R → C with g = b ϕ on U σ , we have that g ≡ b ϕ . It is easy to see that for continuous density ϕ this statement is more general than our Theorem 1.3. On the other hand, the formulation of thisstatement (as also as its proof) uses substantially the fact that ϕ is continuos. Let B ( R ) = { b µ : µ ∈ M ( R ) } denote the Fourier-Stieltjes algebra with the usual pointwise multi-plication. The norm in B ( R ) is inherited from M ( R ) in such a way k ˆ µ k B ( R ) : = k µ k M ( R ) . e normalize the inverse Fourier transformˇ ω ( x ) = π Z R e ixt ω ( t ) dt so that the inversion formula d ( ˇ ω ) = ω is true for suitable ω ∈ L ( R ) .As usual, we write S ( R ) for the Schwartz space of test functions on R and S ′ ( R ) for the dual spaceof tempered distributions. Let Ω be a closed subset of R . A function ω ∈ L p ( R ) , 1 ≤ p ≤ ∞ ,is called bandlimited to Ω if b ω vanishes outside Ω . If 2 < p ≤ ∞ , then we understand b ω in adistributional sense of S ′ ( R ) .For σ >
0, we denote by B p σ the Bernstein space of all F ∈ L p ( R ) such that F is bandlimited to [ − σ , σ ] . The space B p σ is equipped with the norm k F k p = (cid:16) Z R | F ( x ) | p dx (cid:17) / p for 1 ≤ p < ∞ and k F k ∞ = ess supp x ∈ R | F ( x ) | . By the Paley-Wiener-Schwartz theorem (see [6, p. 68]), any F ∈ B p σ is infinitely differentiableon R and has an extension onto the complex plane C to an entire function of exponential type atmost σ . Note that 1 ≤ p ≤ r ≤ ∞ implies ([6, p. 49], Lemma 6.6) B σ ⊂ B p σ ⊂ B r σ ⊂ B ∞σ . There are several sampling expansion formulas for functions from B p σ like Whittaker-Kotelnikov-Shannon sampling formulas. For example, if F ∈ B p σ , 1 ≤ p ≤
2, then F can be expanded as thefollowing formula with derivatives (see [5, p. 60]) F ( x ) = ∑ n ∈ Z (cid:18) F (cid:18) πσ n (cid:19) + F ′ (cid:18) πσ n (cid:19)(cid:18) x − πσ n (cid:19)(cid:19) sin (cid:18) σ x − π n (cid:19) σ x − π n ! . (2.1)This series converges absolutely and uniformly on any compact subset of C . For F ∈ B σ , x ∈ R ,and a >
0, the Poisson summation formula reads (see, e.g., [5, p. 63] and [2, p. 509]) ∑ n ∈ Z F ( x + an ) = a ∑ k ∈ Z b F (cid:18) π a k (cid:19) e i π xa k , (2.2)where both sums converge absolutely. Proof of Proposition 1.2
Define ω = b µ − b η . Then ω ∈ B ( R ) and ω is compactly supported.Using Wiener’s local theorem that the local belonging to B ( R ) is equivalent to the local belongingto A ( R ) = { b ϕ : ϕ ∈ L ( R ) } (see [15, p. 258]), we have that µ − η ∈ L ( R ) . Therefore, µ s = η s and Proposition 1.2 is proved. Proof of Theorem 1.3.
By Bochner’s theorem, there is a probability measure µ such that g = b µ .The function f − g is continuous on R and supported on [ − σ , σ ] . Therefore, ( f − g ) ∈ L ( R ) and ϕ − µ = ( f − g ) ∈ C ( R ) , (2.3) here C ( R ) is the usual space of continuous functions on R that vanish at infinity. Hence, (2.3)implies that µ is absolutely continuous with respect to the Lebesgue measure. Let ψ ∈ L ( R ) denote the density of µ . Define ζ = ϕ − ψ . According to the Bochner-Wiener-Schwartz theorem,we conclude that ζ ∈ B σ . Moreover, ζ ≤ ϕ on R (2.4)and Z R ζ ( x ) dx = . (2.5)We claim that Z E ζ ( x ) dx = . (2.6)To verify the claim, first let us define E n = an + E , n ∈ Z . Then (1.2) and (2.4) imply that Z E n ζ ( x ) dx ≤ n ∈ Z . For F = ζ , using the Poisson summation formula (2.2), we get a ∑ n ∈ Z ζ ( x + an ) = ∑ m ∈ Z b ζ (cid:16) π ma (cid:17) e i π ma x . (2.8)By combining the condition (1.3) with the fact that the continuous function b ζ is supported on [ − σ , σ ] , we have that b ζ ( π m / a ) = m ∈ Z \ { } . Moreover, we conclude from (2.5) that b ζ ( ) = R R ζ ( x ) dx =
0. Therefore, (2.8) implies that ∑ n ∈ Z ζ ( x + an ) = x ∈ R . The series on the left-hand side of (2.2) and (2.9) converge in L [ , a ] (see [2, p.509]). Since E ⊂ [ , a ) , it follows from (2.9) that0 = Z E (cid:16) ∑ n ∈ Z ζ ( x + an ) (cid:17) dx = ∑ n ∈ Z (cid:16) Z E ζ ( x + an ) dx (cid:17) = ∑ n ∈ Z (cid:16) Z E n ζ ( x ) dx (cid:17) . Combining this with (2.7), we see that R E n ζ ( x ) dx = n ∈ Z . This proves our claim (2.6),since E = E .Now we claim that ζ ( x ) = x ∈ E . (2.10)To that end, we decompose E into a disjoint union of its parts E + = { x ∈ E : ζ ( x ) > } , E − = { x ∈ E : ζ ( x ) < } , and E = { x ∈ E : ζ ( x ) = } . We need only check that E + = E − = /0. Indeed,using (1.2) and (2.4), we get Z E + ζ ( x ) dx ≤ Z E + ϕ ( x ) dx ≤ Z E ϕ ( x ) dx = . his proves that E + = /0, since ζ is continuous on R and therefore E + is an open subset of R .Now it follows directly from (2.6) that R E − ζ ( x ) dx =
0. We have therefore E − = /0, which provesour claim (2.10).Finally, (2.10) shows that the entire function ζ vanishes on the set E of positive Lebesgue mea-sure. Thus, by the uniqueness theorem for analytic functions, we have that ζ is the zero functionor ϕ ≡ ψ . Theorem 1.3 is proved. Remark 2.1
Suppose that f : R → C is the characteristic function of a density ϕ . The proof ofTheorem 1.3 implies that in order to exist a characteristic function g such that f = g on U σ butf g, it is necessary and sufficient that there is ζ ∈ B σ , ζ , satisfying (2.4) and (2.5).Proof of Corollary 1.4 Choose any a ∈ R such that | S ϕ | < a ≤ πσ . (2.11)Let h a : R → R / a Z denote the natural homomorphism of R onto the quotient group R / a Z . Wecan identify R / a Z with [ , a ) in the usual way. Then let us define E = [ , a ) \ h a ( S ϕ ) . (2.12)We claim that E satisfies the hypotheses of Theorem 1.3. Indeed, E is a measurable subset of [ , a ) , since S ϕ is measurable and h a is an open map. Next, by the definition of h a and (2.11, wehave | h a ( S ϕ ) | ≤ | S ϕ | < a . Hence | E | >
0. Next, suppose that there exists a x ∈ S ϕ such that x ∈ E + a Z . Then h a ( x ) ∈ h a ( S ϕ ) and together h a ( x ) ∈ h a ( E + a Z ) = E . This contradicts the definition of E in (2.12). Therefore,we see that ( E + a Z ) ∩ S ϕ = /0. This implies (1.2), proving our claim. Finally, using Theorem1.3, we complete the proof of Corollary 1.4.Before proving Theorem 1.5, we shall need the following lemma. Lemma 2.2 . Let a , b ∈ ( − ∞ , ∞ ) , a < b. Suppose that F ∈ B σ , F , satisfies Z R F ( x ) dx = and F ( x ) ≤ χ [ a , b ] ( x ) (2.14) for all x ∈ R , where χ [ a , b ] is the indicator function of [ a , b ] . If b − a = π / σ , then there exists a τ ∈ ( , π / σ ] such thatF ( x ) = τ (cid:20) x − a + σ π − σ ( x − a ) (cid:21) sin σ ( x − a ) , (2.15) x ∈ R . roof. Using the change of variables x → x + a , we may assume further without loss of generality,that a =
0. Then b = π / σ . Now we claim that F (cid:18) πσ n (cid:19) = n ∈ Z . Indeed, (2.14) implies clearly that F (cid:18) πσ n (cid:19) ≤ , n ∈ Z \ { , } . (2.17)Moreover, since F is continuous on R , we see that F satisfies (2.17) also at the endpoints 0 and2 π / σ of [ , π / σ ] . Combining this with (2.13) and using the following summation formula for F ∈ B σ (see [3, p. 122]) σ Z R F ( x ) dx = ∑ n ∈ Z F (cid:18) πσ n (cid:19) , we obtain our claim (2.16).By (2.14), we see that F ( x ) ≤ R \ [ , π / σ ] . Then (2.16) implies that F attains a maximum(local) at every point 2 n π / σ , n ∈ Z \ { , } . Hence F ′ (cid:18) πσ n (cid:19) = n ∈ Z \ { , } . Now substituting (2.16) and (2.18) into (2.1), we have F ( x ) = xF ′ ( ) (cid:18) sin σ x σ x (cid:19) + (cid:16) x − πσ (cid:17) F ′ (cid:16) πσ (cid:17)(cid:18) sin σ x σ x − π (cid:19) . (2.19)It is straightforward to check that this function belongs to B σ if and only if F ′ ( ) = − F ′ (cid:16) πσ (cid:17) .Therefore, the function (2.19) reduces to F ( x ) = τ (cid:20) x + σ π − σ x (cid:21) sin σ x τ =
0. Moreover, (2.14) implies that τ >
0. Thenmax x ∈ [ , π / σ ] F ( x ) = F (cid:16) πσ (cid:17) = στπ . Thus, it follows from (2.14) that τ can be an arbitrary number such that 0 < τ ≤ π / ( σ ) . Lemma2.2 is proved. Proof of Theorem 1.5.
Let ε = ( β − α ) − πσ . It follows from (1.6) that ε >
0. Clearly, the length of [ α + ε , β − ε ] is 2 π / σ . Define ρ = min { ϕ ( x ) : x ∈ [ α + ε , β − ε ] } . he quantity ρ is well defined, since now ϕ is continuous. Moreover, (1.5) gives that ρ > F is the function (2.15) with a = α + ε and an arbitrary fixed τ ∈ ( , π / σ ] . Set G = ρ F . Then G ∈ B σ and G
0. In addition, G ( x ) ≤ ϕ ( x ) , x ∈ R , since the function (2.15) satisfies F ( x ) ≤ x ∈ R \ [ a , a + π / σ ] . On the other hand,(2.13) implies that R R G = ρ R R F =
0. Therefore, we see that ϕ − G is a probability density. Set g = \ ϕ − G . Then g is the characteristic function such that f ( x ) = b ϕ ( x ) = \ ( ϕ − G )( x ) = g ( x ) for x ∈ U σ but f g , since G
0. Theorem 1.5 is proved.
Proof of Proposition 1.8.
To simplify the proof, we will show that there exist continuously differentiable functions ϕ ∈ C ′ ( R ) for which the requirements (i) or (ii) of Proposition 1.8 are satisfied. Therefore, throughoutthis proof we assume that ϕ ∈ C ′ ( R ) .By Remark 2.1, we see that each characteristic function g : R → C such that g = b ϕ on U σ hasthe form b ϕ + b ζ , where ζ is any function in B σ satisfying (2.4) and (2.5). Note that now we donot require that g f , i.e., it can be that ζ ≡
0. For any such ζ and ρ = max n ϕ ( x ) : x ∈ h α , α + πσ io , let us define F = ζ / ρ . Then F satisfy all the conditions of Lemma 2.2 except the hypothesisthat F
0. Therefore, we may assume that F is the function (2.15) with a = α and certain τ ∈ ( , π / σ ] or τ = ζ ≡ F ( α ) = F ( α + π / σ ) = F ′ ( α ) = − F ′ (cid:16) α + πσ (cid:17) = τσ . (2.20)(i) Let ϕ ∈ C ′ ( R ) be a density function such that S ϕ = ( α , α + π / σ ) . Suppose that ϕ satisfiesat least one of the following conditions ϕ ′ ( α ) = ϕ ′ (cid:16) α + πσ (cid:17) = . (2.21)Then (2.4) implies that ζ ′ ( α ) ≤ ζ ′ ( α + π / σ ) ≥
0, since ζ ( α ) = F ( α ) = ζ ( α + π / σ )) = F ( α + π / σ ) =
0. Therefore, by the definition of F , we obtain from (2.20) that τ =
0, i.e., ζ ≡
0. Hence, any such ϕ satisfying at least one of the conditions (2.21) also satisfiesthe statement (i) of Proposition 1.8.(ii) Suppose that ϕ ∈ C ′ ( R ) is any density function that satisfies S ϕ = ( α , α + π / σ ) and ϕ ′ ( α ) = ϕ ′ (cid:16) α + πσ (cid:17) = . Now we claim that there exists ζ ∈ B σ , ζ
0, such that ζ satisfies (2.4) and (2.5). Indeed, itis easily see that there exist ε > τ ∈ ( , π / σ ] for which the function (2.15) with a = α atisfies F ( x ) ≤ ϕ ( x ) for x ∈ h α , α + ε i ∪ h α + πσ − ε , α + πσ i . Since S ϕ = ( α , α + π / σ ) and ϕ ∈ C ′ ( R ) , we may suppose (passing to a smaller positive valueof τ in (2.15) if necessary) that the condition F ( x ) ≤ ϕ ( x ) is satisfied for all x ∈ R . Finally, if weset ζ = F and g = b ϕ + b ζ , then these functions satisfy the statement (ii) of our proposition. References [1] Akutowicz, E.J.: On extrapolating a positive definite function from a finite interval. Math.Scand. , 157–169 (1959)[2] Benedetto J.J.: Georg Zimmermann, Sampling multipliers and the Poisson summation for-mula. J. Fourier Anal. Appl. (5), 505–523 (1997)[3] Boas, R.P. Jr.: Summation formulas and band-limited signals. Tˆohoku Math. J. (2), (1972),no. 2, 121–125[4] Gneiting, T.: Curiosities of characteristic functions. Expo. Math. (4), 359–363 (2001)[5] Higgins, J.R.: Five short stories about the cardinal series. Bull. Amer. Math. Soc. (N.S.) (1), 45–89 (1985)[6] Higgins, J.R.: Sampling Theory in Fourier and Signal Analysis. Foundations. ClarendonPress, Oxford (1996)[7] Kahane, J.P.: Sur quelques probl`emes d’unicit et de prolongement relatifs aux fonctionsapprochables par des sommes d’exponentielles. Annales de linstitut Fourier , 39–130 (1954)[8] Krein, M.G.: Sur le probl`eme du prolongement des fonctions hermitiennes positives et con-tinues (French). C. R. (Doklady) Acad. Sci. URSS (N.S.) , 17–22 (1940)[9] Levin, B.Ya: Lectures on Entire Functions. Translations of Mathematical Monographs, 150.American Mathematical Society. Providence, RI (1996)[10] Levin, B.Ya.: On functions determined by their values on a certain interval. (Russian) Dok-lady Akad. Nauk SSSR (N.S.) , 757–760 (1950)[11] Lukacs, E.: Characteristic functions, 2nd edn. Hafner Publishing Co., New York (1970)[12] Norvidas, S.: On extensions of characteristic functions. Lith. Math. J. (2), 236–243(2017)[13] Norvidas, S.: A theorem of uniqueness for characteristic functions. C. R. Math. Acad. Sci.Paris (8), 920–924 (2017)
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