Bochner s Theorem on the Fourier Transform on R

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1 Bochner s heorem on the Fourier ransform on Yitao Lei October 203 Introduction ypically, the Fourier transformation sends suitable functions on to functions on. his can be defined on the space L ) + L 2 ), i.e. functions which can be written as the sum of a function in L ) and a function in L 2 ). A celebrated result the Hausdorff Young inequality) states that the Fourier transform takes functions in L p ) to L q ) for p 2, where p + q =. However, this does not extend to the case when p > 2. In addition, the maps F : L p ) L q ) are not surjective when p < 2. herefore it seems natural to try to extend the Fourier transform to objects other than functions. he most complete method of doing this is extending the Fourier transform to the space of tempered distributions, i.e. the space of linear functionals on the Schwartz functions. Instead, we will study the Fourier Stieltjes transform, a slight generalisation of the Fourier transform. We now transform complex finite Borel measures rather than functions, and output a function. Bochner s heorem answers the question of which functions ϕ are the Fourier Stieltjes transform of some positive Borel measure. It states that the function is continuous and positive definite is a necessary and sufficient condition for it to be a Fourier Stieltjes transform. We shall first explore the analogous situation on the torus or the circle here, when the dimension is one). Fourier Stieltjes coefficients will be examined, and are related to Fourier coefficients. here is much similarity between Fourier Stieltjes coefficients and the Fourier Stieltjes transform. However, the theory building up to a Bochner type result on the torus is clearer and simpler than going directly to Bochner s theorem on. 2 Preliminaries Let be the torus [0, 2π] with the points 0 and 2π identified, i.e. the same point. In this paper, we shall be working with two spaces of continuous functions, both equipped with the supremum norm: C), continuous functions on, and C 0 ), continuous functions that vanish at infinity. We will also consider two spaces of finite complex Borel measures on and, namely M) and M). For clarity, µ will denote a measure in M), while ν will denote

2 a measure in M). he norm on both spaces are given by µ M) = µ ) and ν M) = ν ). ecall that if V is a Banach space, then the dual space V is the set of linear functionals ψ : V C which are continuous/bounded. he following theorem is very useful in multiple ways; the proof can be found in [3]. heorem iesz representation theorem). ) Any linear functional ψ C)) can be identified with a unique measure µ M) such that ψf) = ft) dµt). In 2π addition, ψ C) = µ M). 2) Any linear functional φ C 0 )) can be identified with a unique measure ν M) such that φg) = gx) dνx). Furthermore, φ C) = ν M). ecall that the Fourier series on of an integrable function f is given by ˆfn) = ft)e int dt. rigonometric polynomials are functions of the form P t) = N a ne int. It can be easily verified that P t) = N ˆP n)e int. A basic result in Fourier analysis is that the partial sums N ˆfn)e int do not necessarily converge to the function itself. Nevertheless, we have convergence in a related series. A proof of this statement can be found in [2]. Lemma 2 Fejér). For positive integer N, define the Fejér kernel by F N x) = N + sin 2 [N + )x/2] sin 2. [x/2] hen for a continuous function f on, we have the following convolution: F N f)t) = n ) N + ˆfn)e int. Furthermore, F N f)t) converges uniformly to f as N. As the convergents in the previous lemma are trigonometric polynomials, any function can be approximated by trigonometric polynomials. Hence trigonometric polynomials are dense in C). 3 Fourier Stieltjes Coefficients Fourier Stieltjes coefficients are an extension of Fourier coefficients, defined with measures in M): Definition 3. he Fourier Stieltjes coefficients of a measure µ M) is a function on Z given by the expression ˆµn) = e int dµt). 2π 2

3 Note that if we have an integrable function f, we can identify it with a measure dµ = ft) dt dt represents the usual Lebesgue measure). Computing the Fourier Stieltjes coefficients gives ˆµn) = 2π e int dµ = 2π e int ft) dt = ˆfn), so we really do have an extension of Fourier coefficients. For Fourier transforms of functions in L 2 ), an important result is Parseval s formula: fx)gx) dx = ˆfξ)ĝξ) dξ. An analogue holds in the context of Fourier Stieltjes 2π coefficients in the following sense: Proposition 4 Parseval s Formula). If µ M) and f C), then ft)dµt) = lim 2π n ) N + ˆfn)ˆµn). Proof. Suppose f is any continuous function. hen by using lemma 2, we can approximate f uniformly as lim n N+) ˆfn)e int. As µ is a finite measure, we can exchange the limit and integral to yield ft)dµt) = lim = lim = lim [ n N + n N + n N + ) ) ) ˆfn)e int ] dµt) ˆfn) e int dµt) ˆfn)ˆµn). We remark that without the converge. ) n factor, the right hand side may not necessarily N+ Bochner s theorem is about trying to determine which sequences are the Fourier Stieltjes coefficients of a measure. As a first step, we give a necessary and sufficient condition for a sequence to the Fourier Stieltjes coefficients of a measure µ: Proposition 5. Let {a n } n= be a sequence of complex numbers. hen the following are equivalent: a) here exists µ M) with µ C and ˆµn) = a n for all n. b) For all trigonometric polynomials P, ˆP n)an C sup P t). t Proof. a) = b): Given any trigonometric polynomial P, use Parseval s formula to get ˆP n)a n = P t) dµ µ sup P t) t where we used the iesz representation theorem to show that µf) µ f C) = µ sup t fx). 3

4 b) = a): he linear map P ˆP n)an defines a linear functional on the space of trigonometric polynomials. As trigonometric polynomials are dense in C), we can extend our linear functional to C). By the iesz representation theorem our linear functional is of the form f f dµ 2π for a suitable measure µ M) of norm C. Substituting f = e int into both P ˆP n)an and f f dµ immediately gives ˆµn) = a 2π n or ˆµn) = a n. Consider a sequence {a n } n Z which eventually vanishes, i.e. a n = 0 when n > K. hen it can be checked that the measure given by dµ N := K n= K a ne int dt satisfies µ N n) = a n for all n. his allows us to make sense of the following corollary. he proof involves algebraic manipulations and Parseval s formula only, and hence will not be presented here. Corollary 6. A sequence {a n } n Z is the Fourier Stieltjes coefficients of some µ with µ C, if, and only if, µ N M) C for all N. Here, µ N is the measure such that µ N n) = n )a N+ n whenever n N, and zero when n > N. 4 Hergoltz s heorem Hergoltz s theorem is the analogue of Bochner s theorem on the torus, as in it gives necessary and sufficient conditions for a sequence to be the Fourier Stieltjes coefficients of a positive measure. o prove this, we first need the following lemma: Lemma 7. A sequence {a n } n Z is the Fourier Stieltjes series of a positive measure if, and only if, for all N and t, σ N t) := n ) a n e int 0. N + Proof. First suppose ˆµn) = a n for some positive measure µ M). Let f C) be an arbitrary non-negative function. hen ft)σ N t) dt = [ N ft) n ) ] a n e int dt 2π 2π N + = n ) ˆfn)ˆµn) N + = n ) ˆfn) e int dµt) N + = n ) ˆfn)e int dµ N + = F N f)t) dµ from lemma 2). 4

5 Now note that the Fejér kernel is non-negative. As f is non-negative, the convolution F N f) is also non-negative. From the positivity of the measure µ, our quantity is non-negative. As this is true for any non-negative function f, we get that σ N t) 0. Now suppose ) N n a N+ n e int 0. As before, denote µ N as the measure with ) Fourier Stieltjes coefficients n a N+ n. hen we want to compute µ N M). By the iesz representation theorem we just need to compute the norm of the functional ) f f dµ 2π N. However, from earlier discussion, dµ N = n a N+ n dt. herefore µ N M) = sup f C) = ft) dµ N = sup 2π f = ft) σ N t) dt. 2π As σ N t) is positive, µ N is a positive measure and the above quantity is maximised when ft) =. his means that µ N M) = 2π n ) a n e int dt = a 0. N + his must be true for all N. herefore, µ N M) are uniformly bounded by a 0. By corollary 6, we can find a µ M) such that ˆµn) = a n for all n. o show that µ is a positive measure, take an arbitrary non-negative function f. hen f dµ = lim f dµ N 0 as this convergence comes from taking the limit in Parseval s formula. hus µ is positive. We remark in the above proof that one can replace the condition σ N t) 0 for all N with σ N t) 0 for infinitely many N. he proof remains the same, except that we only take a subsequence N j where σ Nj t) 0. Definition 8. A sequence {a n } n Z is positive definite if for all sequences of complex numbers {z n } n Z which have all but a finite number of terms zero, we have a n m z n z m 0. n,m Z he work from the last few pages can now be combined to prove Herglotz s theorem: heorem 9 Herglotz). A sequence {a n } n Z is the Fourier Stieltjes transform of a positive measure µ M) if, and only if, the sequence is positive definite. Proof. If µ is a positive measure with ˆµn) = a n, then a n m z n z m = n,m Z n,m Z e int e imt 2 z n z m dµ = z n e int dµ 0. n Z 5

6 On the { other hand, suppose {a n } is positive definite. Fix values t, N. Define numbers e int, n N z n =. By direct substitution, one can verify that n,m 0, else a n mz n z m = j C j,na j e ijx, where C j,n = max0, 2N + j ). herefore by positive definiteness, σ 2N t) = = 2 j= 2N 2N + j ) a j e ijx = 2N + n,m Z a n m z n z m 0. 2N + 2 j= 2N C j,n a j e ijx We proved that σ N t) 0 for even N. he result then follows from lemma 7. 5 he Fourier Stieltjes ransform We define the main object of interest here, the Fourier Stieltjes transform. Definition 0. he Fourier Stieltjes transform of a measure ν M) is a function on given by the expression ˆνξ) = e iξx dνx). Like Fourier Stieltjes coefficients, there is consistency between the definition here, and the Fourier transform of L functions. If g is an integrable function, identify it with a measure dν = g dx. Hence ˆνξ) = e iξx gx) dx = ĝξ), which shows consistency in the transforms. he iemann Lebesgue lemma on the ordinary Fourier transform states that if g L ), then ĝξ) is uniformly continuous and goes to 0 as ξ. he uniform continuity still holds for the Fourier Stieltjes transform, but ˆνξ) does not necessarily go to 0. A simple example is ν = δ 0, the measure with mass at 0. It can be easily verified that ˆνξ) = for all ξ. We can deduce another version of Parseval s formula in this new context: Proposition Parseval s Formula). If ν M) and both g, ĝ are in L ), then gx)dνx) = ĝξ)ˆν ξ) dξ. 2π Proof. If both g and ĝ are integrable, then the Fourier inversion formula holds: gx) = 2π ĝξ)eiξx dξ. herefore, gx)dνx) = ĝξ)e iξx dνx) dξ = ĝξ)ˆν ξ), 2π 2π where we used the integrability of ĝξ) to justify the usage of Fubini s theorem. 6

7 he following proposition gives a necessary and sufficient statement for a function to be the Fourier Stieltjes transform of a measure, and is the first step to Bochner s theorem on. Proposition 2. If ϕ is a continuous function defined on, then it is the Fourier Stieltjes transform of a measure if, and only if, there exists a constant C such that ĝξ)ϕ ξ) dξ 2π C sup gx) x for every continuous function g L ) such that ĝ has compact support. Proof. Firstly, suppose that ϕ = ˆν. hen the statement follows directly from Parseval s formula by setting C = ν M). On the other hand, suppose our inequality is valid. hen the linear functional ψ which maps g to ĝξ)ϕ ξ) dξ is a bounded, continuous linear functional on the set of 2π continuous functions g such that ĝ is compactly supported. his is a dense subset of C 0 ). herefore we can extend ψ to a bounded functional on C 0 ). By the iesz representation theorem, ψ can be represented by a measure ν M), where ν M) C. herefore ψ maps g to ĝx) dνx). Using Parseval s formula here yields that ĝξ)ϕ ξ) dξ = ĝξ)ˆν ξ) dξ. his must hold for all g, so 2π 2π ϕ = ˆν. 6 Connection to Fourier Stieltjes Coefficients We have a natural covering map p : sending x to x mod 2π. his can be used to pushforward a measure ν M) to a measure µ M). If we extend a continuous function f on to a 2π-periodic function g on which do not necessarily go to zero), then we get that gx) dνx) = ft) dµt). An immediate consequence is that ˆνn) = ˆµn) for all integers n. his allows us to relate the original problem on to the problem on. he following theorem is a key step in doing this: heorem 3. If ϕ is a continuous function defined on, then it is the Fourier Stieltjes transform if, and only if, there exists a constant C > 0 such that for any choice of λ > 0, {ϕλn)} n= are the Fourier Stieltjes coefficients of a measure of norm C on. Proof. First suppose ϕ = ˆν. hen ϕn) = ˆνn) = ˆµn), where µ is the pushforward measure of ν with respect to the covering. Note that µ M) ν M). Denote νx/λ) the measure on which satisfies the following equation for all g: gx)dν x) = gλx) dνx). λ 7

8 We get that νx/λ) M) = ν M) and νx/λ)ξ) = ˆνξλ). Setting ξ = n yields that ϕλn) = µx/λ)n), so {ϕλn)} n Z form the Fourier Stieltjes coefficients of norm at most ν M). o show the converse, we shall use theorem 2. Let g be continuous such that ĝ is continuous and compactly supported. hen by those conditions the integral ĝξ)ϕ ξ) dξ 2π can be approximated by its iemann sums where the width of each rectangle is λ). For arbitrary ɛ, choose sufficiently small λ to obtain ĝξ)ϕ ξ) dξ 2π < λ ĝλn)ϕ λn) 2π + ɛ. Note that λ 2πĝλn) is the Fourier coefficient for the function G λt) = m Z gt + 2πm)/λ) on. If λ is sufficiently small, then by the decay of g, we have sup t G λ t) sup gx) + ɛ. x By assumption, ϕλn) = µ λ n) for some µ λ M), with µ λ M) C. hen Parseval s formula for gives λ ĝλn)ϕ λn) 2π = Ĝ λ n) ˆµ λ n) C sup G λ t). n Combining all inequalities gives λ ĝλn)ϕ λn) 2π < C sup gx) + C + )ɛ. n As ɛ > 0 was arbitrary, the condition for proposition 2 is satisfied. n Now here is a necessary and sufficient condition for a function ϕ to be a Fourier Stieltjes transform of a positive measure. he proof is very similar to theorem 3, so it will not be presented here. Proposition 4. If ϕ is a continuous function defined on, then it is the Fourier Stieltjes transform of a positive measure if, and only if, there exists a constant C > 0 such that for any choice of λ > 0, {ϕλn)} n= are the Fourier Stieltjes coefficients of a positive measure on. n x t 7 Bochner s heorem Definition 5. A function ϕ defined on is positive definite if for all ξ,..., ξ N and z,..., z N C, we have ϕξ j ξ k )z j z k 0. j,k= 8

9 Finally, we have the machinery to prove the main subject of this paper. heorem 6. A function ϕ defined on is the Fourier Stieltjes transform of a positive measure if, and only if, it is continuous and positive definite. Proof. First assume that ϕ = ˆν for a positive measure ν M). ξ,... ξ N and z,... z N C, we get ϕξ j ξ k )z j z k = e iξjx z j e iξkx z k dνx) j,k= = j,k= z j e iξ jx j= 2 dνx) 0. Given numbers For the other direction, if ϕ is positive definite, then by definition, {ϕλn)} is a positive definite sequence for all λ. Herglotz s theorem guarantees a measure µ λ, such that µ λ n) = ϕλn). By proposition 4, there must exist ν such that ˆν = ϕ. eferences [] Katznelson, Y. 2004). An Introduction to Harmonic Analysis 3rd edn), Cambridge Mathematical Library. [2] Grafakos, L. 2003). Classical and Modern Fourier Analysis, Prentice Hall. [3] udin, W. 987). eal and Complex Analysis 3rd edn), McGraw Hill Book Company. 9