Expository Notes on Introduction to Harmonic Analysis

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1 Expository Notes on Introduction to Harmonic Analysis Matthew Begué December 8, 202 he following contains results with explanations from Yitzhak Katznelson s An Introduction to Harmonic Analysis [Katz]. I present some of the results with Katznelson s proof often with expanded details. In some cases, I even provide alternate proofs to some of the results. Fourier Analysis on. Preliminaries he group is defined as the quotient /Z. here is a clear identification between functions on and -periodic functions on. his periodicity property gives us translation invariance of the integral. hat is, for every t 0 and f defined on we have ft t 0 ) dt ft) dt. We denote the L norm on as follows: for f L ) we let f L ) ft) dt. It is well-known that L ) is a Banach space. Let f L ). We define the nth Fourier coefficient by ˆfn) ft)e int dt. he following are some elementary properties of Fourier coefficients: heorem.. For f, g L then. f + g)n) ˆfn) + ĝn). 2. For any α C, αf)n) α ˆfn). 3. Let f denote the complex conjugate of f. hen ˆ fn) ˆfn). 4. For x let τ x f)t) ft x). hen τ x fn) ˆfn)e inx.

2 5..) ˆfn) f L. 6. Assume f, f L ) and ˆf0) 0. hen for n 0, ˆfn) in ˆf n). Proof. Parts and 2 follow immediately from linearity and homogeneity of the integral. Parts 3 and 4 follow from the definition of the Fourier coefficients of f. Part 5 follows since ˆfn) ft)e int dt ft)e int dt f L ). For Part 6 we take the definition of ˆfn) and integrate by parts to obtain ˆfn) 0 ft)e int dt ft) e int in 0 0 f t) e int in dt in ˆf n). Notice that the periodicity of f L ) was explicitly used to make boundary terms vanish. heorem.2. [he iemann-lebesgue Lemma] Let f L ), then lim n ˆfn) 0. Lebesgue-iemann heorem Proof. his is the proof given in [Katz]. First, recall that trigonometric polynomials on are dense in L ). hat is, for ɛ > 0 there exists a trigonometric polynomial P such that f P L ) < ɛ. If n > deg P then ˆfn) ˆfn) P t)e int dt f P )n) he integral above is equal to 0 since n > deg P and e ikt dt 0 for any k 0. his gives ˆfn) f P )n) f P L ) < ɛ by.). Lebesgue-iemann heorem Proof 2. Let A denote the indicator function of the Lebesgue measurable set A. Suppose A [ c, c] is a closed interval symmetric about the origin. hen the Fourier coefficients of [ c,c] are: ˆ [ c,c] [ c,c] e int dt c in e int c c c e int dt e inc e inc sin nc) sinnc) πn 2i πn πn. hen by heorem., we can write the Fourier coefficients of the indicator on any interval as: [a,b] n) τ b+a 2 [ b a 2, b a 2 ] ) hen for any α we see that lim α [a,b] )n) n 2 n) e lim n b+a in 2 α πn 0. ) sinn b a πn 2 ).

3 hen for any simple function i.e. finite linear combination of indicator functions), ϕt) α i Ii t) then ˆϕn) 0 as n. i Simple functions are dense in L ); that is, given f L and ɛ > 0 there exists a simple ϕt) such that f ϕ L ) < ɛ. hen by the triangle inequality and heorem. we have ˆfn) ˆϕn) f ϕn) f ϕ L ) < ɛ. his implies that lim n ˆfn) < ɛ which proves the theorem. Lebesgue-iemann heorem Proof 3. Yet another way to prove the Lebesge- iemann heorem is to study functions in CC ), that is, functions on with a continuous first derivative and that vanish at zero. Let g CC ). hen we can integrate by parts to obtain ĝn) gt)e int dt [ 0 + ] g t) eint in dt. here are no boundary terms since we chose g to vanish at 0. hen we can estimate ĝn) [ ] g t) eint in dt max t [0,) g t) 0 as n. n Note that Cc ) is also dense in L ). herefore, for any f L and ɛ > 0 there exists a gt) Cc ) such that f g L ) < ɛ. hen by the triangle inequality and heorem. we have ˆfn) ĝn) f gn) f g L ) < ɛ. his implies that lim n ˆfn) < ɛ which proves the theorem. Definition.3. For f, g L ) we define the convolution f g as f gt) ft x)gx) dx. he following theorem contains standard properties about the convolution which we provide without proof at this time: heorem.4. For f, g L ), then the following properties hold: a) ft τ)gτ) is integrable as a function of τ. b) f g L ) f L ) g L ). c) he convolution operation in L ) is commutative i.e. f g g f). d) f gn) ˆfn)ĝn) for all integers n. 3

4 .2 Summability Kernels Definition.5. A summability kernel is a sequence {k n } of continuous functions on hence -periodic) satisfying:. k nt) dt 2. k nt) dt C 3. For all 0 < δ < π, lim n δ δ k n t) dt 0. hat is, a summability kernel is a sequence of functions that integrate to one, are uniformly bounded in the L norm, and integrate to 0 off of a neighborhood about 0. he following is of great importance and makes summability kernels extremely useful to study: heorem.6. For f L ) and {k n } any summability kernel, then.2. Féjer Kernel lim f k n f n L ) 0. One of the best known summability kernels is the Féjer kernel defined by.2) K n t) n j n j ) e ijt. n + Before we show that {K n t)} is a summability kernel, the following Lemma is useful. [ Lemma.7. K n t) sin n+ t) ] 2 2 n+ sint/2). Proof. ecall the following trigonometric identity: sin 2 t/2) /2 cost)) /2 e it +e it 4 where the last equality comes from Euler s formula cost) /2e it + e it )). hen sin 2 t/2)k n t) /2 e it + e it ) n j ) e ijt 4 n + j n /2 e in+)t + e in+)t ) n + 4 ) n + n + sin2 t 2 which proves the desired equality. Notice that one result of Lemma.7 is that K n t) is a positive summability kernel i.e. K n t) 0 for all n and for all t ). heorem.8. he Féjer kernel {K n } is a summability kernel. 4

5 Proof. We need to verify the three conditions given in Definition.5. Firstly, K n t) dt n j n n j n j n + j n + ) ) e ijt dt e ijt dt his integral is zero for all values of j except for j 0. hus K n t) dt 0 ) dt. n + he second property follows immediately from the first since Lemma.7 tells us that {K n t)} is a positive summability kernel i.e. K n t) K n t) for all t and for all n). Moreover, this tells us that K n L ) for all n. he third condition also follows from Lemma.7 since δ δ [ K n t) dt sin n+ 2 t) ] 2 δ δ n + sint/2) dt δ [ ] 2 dt 0 as n. n + sinδ/2) δ Figure shows the graphs of K n t) for a few choices of n. he figures clearly exhibit the positivity and delta-like behavior as n grows. We will need the Fourier coefficients of the Féjer kernel for proofs in the subsequent sections. Proposition.9. ˆ K n m) { ) m n+ if n m 0 otherwise Proof. By definition of Fourier coefficients, we have K n m) n j n n j n m n + m n + ) ) e ij m)t dt e ij m)t dt. he integral eij m)t dt equals 0 if j m and equals only if j m. hus provided n m. K n m) m ) n + 5

6 Figure : Plots of the Féjer kernel K n t) on t [ π, π] for n, 5, 0, Dirichlet Kernel Another important function is the Dirichlet kernel defined by n.3) D n t) e ijt sinn + /2)t). sint/2) j n Figure 2 shows the graphs of D n t) for a few choices of n. he figures clearly exhibit the positivity and delta-like behavior as n grows. emark.0. It is important to note that the Dirichlet kernel is not a summability kernel. Proof. he Dirichlet kernel does satisfy condition in Definition.5; that is, n e ijt dt j n since each of the 2n + integrals equal zero except when j 0. However D n t) does not satisfy condition number 2 or Homogeneous Banach Spaces Definition.. A homogeneous Banach space on is a linear subspace B of L ) that is Banach under the norm B L ) with the following two properties: 6

7 Figure 2: Plots of the Dirichlet kernel D n t) on t [ π, π] for n, 0, 50, 00. i) If f B and a, then τ a ft) ft a) B and τ a f B f. ii) For all f B, a, a 0 then lim a a0 τ a f τ a0 f B 0. An example of a homogeneous Banach space is C) with the norm f L ). Another is L p ) for p < with the norm f L p ) ft) p dt) /p. We can generalize heorem.6 to homogeneous Banach spaces on : heorem.2. For B a homogeneous Banach space on, f B, and {k n } a summability kernel. hen lim f k n f n B 0. Corollary.3. For any homogeneous Banach space B on, the set of trigonometric polynomials are dense in B. Proof. For any f B, we have lim n K n f f in the B-norm and K n f is a trigonometric polynomial for any n. As discussed, since C), L )) is a homogeneous Banach space, we get a classical result: Corollary.4 Weierstrass Approximation heorem). Every continuous periodic function can be approximated uniformly by trigonometric polynomials. 7

8 Exercise.5 Katznelson.2.3). Let B be a homogeneous Banach space on. Show that if g L ) and f B then g f B and.4) g f B g L ) f B Solution. Suppose first, that g is continuous on. hen g f B gτ)ft τ) dτ gτ)ft τ) B B dτ gτ) ft τ) B dτ gτ) ft) B dτ g L ) f B. In the case where g is not continuous, we use heorem.2 and its corollaries to approximate g by K n g..3 Order of Magnitude of Fourier Coefficients hus far we ve established two immediate facts about the magnitude of ˆfn). Given f L ) we have ˆfn) f L ) for all n and lim n ˆfn) 0 by the iemann-lebesgue heorem. Some natural questions arise. he first question is: Can we get an estimate of the rate of vanishing of ˆfn)? he answer to this is no. In the following theorem, we show that there exist functions whose Fourier coefficients decay to zero arbitrarily slow. heorem.6. Let {a n } n N be an even sequence of nonnegative numbers tending to 0. Assume that {a n } is a convex sequence i.e. a n + a n+ 2a n 0 for all n). hen there exists a nonnegative f L ) such that ˆfn) a n. Proof. he convexity condition gives us that a n a n a n a n+, that is, {a n a n+ } is a sequence of positive numbers monotonically decreasing to 0. We claim, next, that lim n na n a n+ ) 0. o see this, observe first that since lim n a n 0, given any ɛ > 0 there exists some N > 0 such that for all n > N we have a n < ɛ. Since the sequence {a n } is positive and decreasing, then a n a n+k < ɛ/2 for any n > N and k N. We can write this quantity as the telescoping sum a n a n+k k a n+i a n+i ka n+k a n+k ) i since a n+k a n+k ) a n+i a n+i ) for every i,..., k. Now let n > N be fixed and perform a change of variables; let m n + k. hen we have earranging gives m + n + ))a m a m+ ) < ɛ/2. ma m a m+ ) < ɛ/2 + n )a m a m+ ). Since {a m a m+ } 0 there exists an M > 0 such that for all m > M we have n )a m a m+ ) < ɛ/2. hus for picking n > N and then m > M we have ma m a m+ ) < ɛ. Hence.5) lim n na n a n+ ) 0 8

9 as desired. Now consider the following series: na n + a n+ 2a n ) n a 0 + a 2 2a + 2a + 2a 3 4a 2 + 3a 2 + 3a 4 6a 3 + +N )a N 2 + a N 2a N ) + Na N + a N+ 2a N ) a 0 + N )a N + Na N+ 2a N ) a 0 a N + Na N+ a N ) As we let N, the second and third terms vanish by.5). hus lim N a n+ 2a n ) a 0. Now consider the function where K n t) ft) na n + a n+ 2a n )K n t) n n j n + )eijt is the Féjer kernel. j n We claim that f L ). o see this, consider the sequence of functions f N t) na n + a n+ 2a n )K n t). Each f N L ) since n N f N L ) N na n + a n+ 2a n ) K n L ) na n + a n+ 2a n ) n since K n L ) for all n by.8. Furthermore, each f N converges pointwise to some function f. o see this we use the form of K n t) given by Lemma.7 to get: lim f Nt) lim N N n lim N n n na n + a n+ 2a n ) n sin/2t) 2 sin ) 2 nt 2 na n + a n+ 2a n ) sin/2t) 2 a 0 sin/2t) 2 which is finite a.e. on. In addition, by Lemma.7 we see that K n t) is a positive kernel. hus,f N f N+ for all N. hen we can apply the Lebesgue Monotone Convergence heorem Beppo Levi) to conclude that lim N fn f. hus, f L ) na n + a n+ 2a n ) a 0 <. n Since f L ), we can now compute its Fourier coefficients using Propo- N n na n + 9

10 sition.9): ˆfj) na n + a n+ 2a n ) K n j) n n j + n j + na n + a n+ 2a n ) j n ) n j )a n + a n+ 2a n ). Every term in this series cancels except for the first, a j. hus, given any even convex sequence {a n } n N that tends to zero arbitrarily slowly), we can construct a function f L ) such that ˆfj) a j for every integer j. We will show in Proposition.32) that a sequence {b n } l ) is the Fourier coefficients of a function in L ). Sequences in l ) necesarilly tend to zero and can do so arbitrarily slowly. his provides an alternate proof showing that we cannot estimate the rate of vanishing of Fourier coefficients. However the proof given is stronger since the set of convex sequences vanishing are proper subset of l Example.7. Show a convex sequence of numbers that is not in l. Consider the real function fx) log x +2). Clearly fx) is positive and tends to infinity as x. his function is also convex since f 2 x) 2+x) 2 log2+x) x) 2 log2+x) 0) but is not in L ), i.e. 2 log x +2) dx diverges. Now consider the sequence of real numbers {a n } n N defined by a n fn). he sequence clearly tends to 0 as n and the convexity follows from the convexity of f. But by the integral convergence test for series, {a n } / l. hus far, we have demonstrated that convex sequences that tend to zero are the Fourier coefficients of some function f L ) We also mentioned that the same is true for equences in l ). But is it true that any sequence {a n } which tends to zero is the Fourier coeffieicents of some f L )? he answer to this question is no. his can be proved with the contrapositive of the following theorem: heorem.8. Assume f L ) and ˆf n ) ˆf n ) 0. hen n ˆfn) <. n 0 Proof. Define the function F t) t fτ) dτ. By construction, F t) is absolutely) continuous. hen by heorem. ˆF n) ˆfn) in 0 for any n 0. Since F is continuous, then by Féjer s heorem?? at t 0, lim K N F 0) N j N lim N j ) ˆF j)e ij0 N + ˆF 0) + 2 j j ) N + ij ˆfj) F 0) 0

11 his gives which proves the theorem. j 0 ˆfj) if 0) j ˆF 0)). is, he answer to our question is the contrapositive of the above theorem. hat Corollary.9. If a n > 0 and n 0 Fourier series. a nn. hen a n sinnt) is not a In other words, arbitrary sequences that decay to zero are not necessarily the Fourier coeffieicents of an L function. A third natural question to ask is: How does smoothness affect the decay of ˆfn)? heorem.20. If f L ) is absolutely continuous, then ˆfn) o/n). Proof. Again, by heorem. we have ˆfn) ˆf in n). he iemann-lebesgue lemma heorem.2) says that ˆf n) 0 as n. his tells us that ˆfn) o/n). Suppose instead f C k ) and f k) L ) then we can conclude that ˆfn) o/n k ) by following the proof above with repeated integration by parts. heorem.2. If f BV ), then ˆfn) Varf) n. Proof. We have ˆfn) ft)e int dt and integrating by parts gives ˆfn) e int f t) dt in f t) dt Varf) n n. n

12 .4 Fourier Series of Square Summable Functions his section corresponds to Section.5 of [Katz]. Let H be a complex Hilbert space, i.e. a complex inner product space that is complete with respect to the norm induced by the inner product. he norm of f H is defined as f f, f ) /2. We say that f, g H are orthogonal if the inner product f, g 0. he function f H is orthogonal to the set E H if f is orthogonal to each element of E. he set E H is orthogonal if any two elements of E are orthogonal. he set E H is orthogonormal if it is orthogonal and the norm of each vector is. heorem.22. Let {ϕ n } n be an orthonormal system in H and let {a n } n l 2, i.e. n a n 2 <. hen n a nϕ n converges in H. Proof. First notice that for any finite subset of {ϕ n } we have 2 a n ϕ n n N a n ϕ n, a j ϕ j, by definition of norm on H n n j a n ϕ n, a j ϕ j, by linearity of the inner product n j j a n a j ϕ n, ϕ j, a n a n n also by linearity of the inner product a n 2 by the orthogonality of{ϕ n } N n. n Because Hilbert spaces are complete, it suffices to show that the partial sums S N N n a nϕ n form a Cauchy sequence in H. For any N, M > 0 we have S N S M 2 nm+ a n 2. Since {a n } l 2, for any ɛ > 0 we can select N ɛ > 0 so that nn ɛ a n 2 < ɛ which gives us that the above equation is also less than epsilon for N, M > N ɛ. hus S N is Cauchy in H and thus converges in H by completeness. Lemma.23 Bessel s Inequality). Let {ϕ α } be an orthonormal system in Hilbert space H. For f H define a α f, ϕ α. hen.6) a α 2 f 2. α Proof. First, suppose that {ϕ n } N n is a finite orthonormal system. hen by the 2

13 definition of the H-norm and linearity of the inner product: N 0 f 2 a n ϕ n n f a n ϕ n, f a j ϕ j n j j f, f a j ϕ j + a n ϕ n, f a j ϕ j j n f, f a j f, ϕ j + a n ϕ n, f a n a j ϕ n, ϕ j j n n f, f a j a j + a n a n a n a n f 2 a n 2 n n j n j which proves.6) for the case where our orthonormal system is finite. Suppose now that {ϕ α } is infinite countable or uncountable) and that.6) does not hold. hen there must exist some finite subcollection of {ϕ α } for which a α 2 > f 2, which is a clear contradiction. α Definition.24. An orthonormal system {ϕ α } in H is said to be complete provided that the only vector in H orthogonal to it is the zero vector. Lemma.25. Let {ϕ n } be an orthonormal system in H. hen the following are equivalent: a) {ϕ n } is complete. b) For every f H we have f 2 n f, ϕ n 2. c) For every f H we have f n f, ϕ n ϕ n. Proof. a c): Let {ϕ n } be an orthonormal complete system and let f H. By Bessel s Inequality, n f, ϕ n 2 f 2 <. hen by heorem.22, n f, ϕ n ϕ n converges in H to some function g H. Notice now that g, ϕ n f, ϕ j ϕ j, ϕ n f, ϕ j ϕ j, ϕ n f, ϕ n. j j hus, by linearity, f g H is orthogonal to ϕ n. Since {ϕ n } is complete, this means that f g 0. hat is, f n f, ϕ n ϕ n. 3

14 c b): Assume f n f, ϕ n ϕ n. hen f 2 f, f f, ϕ n ϕ n, f, ϕ j ϕ j n j f, ϕ n f, ϕ j ϕ n, ϕ j n j f, ϕ n f, ϕ n f, ϕ n 2. n n b a): If f is orthogonal to all of {ϕ n }, i.e. if f, ϕ n 0 for all n, then by assumption b), f 2 f, ϕ n 2 0. n Since is a norm on H, then f 0 if and only if f 0. hus, {ϕ n } is complete. Exercise.26 Katznelson.5.). Let {ϕ n } be an orthogonal system in a Hilbert space H. Let f H. Show that f a j ϕ j achieves a minimum at j the point a j f, ϕ j for j, 2,..., N and only there. Solution. : We can write f a j ϕ j f f, ϕ j ϕ j + c j ϕ j j where c j f, ϕ j a j. hen by the triangle inequality, f a j ϕ j f f, ϕ j ϕ j + c j ϕ j. j j j hen, since {ϕ} is an orthogonal system, the last norm will equal zero if and only if c j 0 for all j, 2,..., N..4. he Hilbert space L 2 ) It is known that L p ) are complete for all p by the iesz-fischer heorem). However, only L 2 ) is a Hilbert space because it is equipped with the inner product f, g ft)gt) dt. Proposition.27. For H L 2 ), the exponentials {e int } n form a complete orthonormal system. j j Proof. Orthonormality follows easily since e int, e imt e in m)t dt δ n,m. If f is continuous, then we can approximate f in L 2 by trigonometric polynomials. hat is, for any small ɛ > 0 there exists a trigonometric polynomial 4

15 P of degree N such that f P L 2 ) < ɛ. hen by Exercise.26 we also get f N ˆfn)e n N int L < ɛ. his verifies condition b) of Lemma ) Suppose f L 2 ) is not continuous. We can approximate f in the L 2 norm by a continuous function. hat is, for ɛ > 0 there exists a g C) such that f g L 2 ) < ɛ/3. For sake of notation let S Nf) ˆfn)e int ) hen n N f S N f) L 2 ) f g L 2 ) + g S Ng) L 2 ) + S N g) S N f) L 2 ). Look at the three terms on the left hand side. By our choice of g, the first term is less than ɛ/3. By the previous argument, the second term is less than ɛ/3 for N sufficiently large. As for the third term, S N g) S N f) L 2 ) S N g f) L 2 ) g f)n)e int 2 dt n N n N g f)n) 2 f g 2 L 2 ) < ɛ2 /9 ) /2 where the last inequality follows from Bessel s inequality. herefore we ve shown N that f ˆfn)e int ɛ/3 + ɛ/3 + ɛ 2 /9 < ɛ which verifies condition L2) n N b) in Lemma.25. heorem.28. Let f L 2 ). hen a) ˆfn) 2 ft) 2 dt. n b) f lim N N n N ˆfn)e int in the L 2 ) norm. c) Given any complex sequence {a n } n l 2 C), there exists a unique f L 2 ) such that a n ˆfn). d) Let f, g L 2 ). hen ft)gt) dt ˆfn)ĝn). n Proof. a): We have just shown that {e int } is a complete orthonormal system in L 2 ). hen the statement a) of heorem.28 is precisely statement b) of Lemma.27 and statement b) of heorem.28 is precisely statement c) of Lemma.27. 5

16 c): ake {a n } l 2. Define the function f N t) for any N > M > 0, f N f M L 2 ) M N + n N /2 a n 2 e int dt) 2 M+ a n e int.. hen M N + ) a n 2. M+ Since {a n } l 2 then for any ɛ > 0 there exists some N ɛ such that for all N > M > N ɛ, f N f M L2 ) < ɛ. hus, {f N} is Cauchy in L 2 ). Since L 2 ) is a Banach space, {f N } converges in L 2 to a function f and we necessarily have ˆfn) a m e imt e int dt a n. m emark.29. heorem.28 says that L 2 ) is isometric to l 2. We know that any function in f L 2 ) has Fourier coefficients and by part a), { ˆfn)} n l 2. By the uniqueness theorem, the Fourier coefficients are unique to f. Similarly, by part c) any l 2 sequence is precisely the Fourier coeffieicents of some f L 2 ). Finally, part a) tells us that preserved. herefore, L 2 is isometric to l 2. ˆfn) l 2 f L2 ), so norm is Exercise.30 Katznelson.5.5). Let f be absolutely continuous on and assume f L 2 ). Prove that ˆfn) f L ) + π f 3 L2 ). n Z Solution. By the Differentiation heorem, Part 6 of heorem., for n 0 we have ˆf n) in ˆfn). So then n ˆfn) 2 ˆf n) 2 f 2 L 2 ) by n Z n Z heorem.28. We next break up the sum and use the estimate ˆf0) f L ) to get: ˆfn) ˆf0) + ˆfn) + ˆf n) ) n Z n f L ) + ˆfn) + ˆf n) ). We next apply the Cauchy-Schwarz inequality to the above sum. We use n 6

17 the form of Cauchy-Schwarz that says a n n 2 nan 2 to get: ˆfn) f L ) + n 2 n ˆfn) + ˆf n) ) ) 2 n Z n n f L ) + n 2 n ˆfn) 2 + n ˆf n) n n f L ) + n 2 2 f 2 L 2 ) n f L ) + 2 n 2 f L 2 ) Finally, since n n 2 π 2 /6 we the above expression simplifies to: ˆfn) f L ) + π f 3 L2 ). n Z n 2.5 Absolutely Convergent Fourier Series Definition.3. Let A) denote the space of continuous functions on with absolutely convergent Fourier series. hat is, ˆfn)e int ˆfn) <. We induce the norm on A) by f A) n n ˆfn). Proposition.32. he space A) is isomorphic to l. n Proof. Consider the mapping f { ˆfn)} n Z. his mapping is linear by linearity of Fourier coefficients) and one-to-one by the Uniqueness heorem). By a similar argument in heorem.28, if n a n < then the series n a ne int converges on to, say, g and ĝn) a n. his shows that the above mapping is an isomorphism of A) onto l. Since l is a Banach space, then the above proposition tells us that A) is as well. he following Lemma asserts that A) is also an algebra. Lemma.33. Assume that f, g A). hen fg A) and fg A) f A) g A). Proof. We have ft) ˆfn)e n Z int and gt) n Z ĝn)eint and both converge absolutely. herefore, fgt) ˆfn)ĝk)e in+k)t n Z k Z 7

18 also converges absolutely. Collect all terms in which n + k m to obtain fgt) ˆfn)ĝm n)e imt. m Z n Z his gives fgl) n ˆfn)ĝl n). herefore, fgl) ˆfn) ĝl n) ˆfk) ĝl) <. l Z n Z n Z l Z l Z Definition.34. We denote by C 0,α ) the space of all Hölder continuous functions. hat is f C 0,α ft+h) ft) ) if sup x,h,h 0 h <. Equivalently, α we say f C 0,α ) if fx) fy) C x y α for some finite constant C. We call the smallest such constant C the Hölder coefficient of f, denoted f C 0,α ) ft+h) ft) sup x,h,h 0 h. If f C 0,α ), we define the Hölder norm of f as α f C 0,α ) f L ) + f C 0,α ). he following theorem is due to Bernstein: heorem.35 Bernstein). Assume f C 0,α ) for some α > /2. hen f A) and f A) c α f C 0,α ) Proof. If f C 0,α ) then f is certainly continuous and hence in L 2 ). hen by heorem.28 we can write f f, e int e int ˆfn)e int. hen using n Z n Z the notation and results from heorem., we have ft h) ft) n Z τ h fn)e int n Z ˆfn)e int n Ze ihn ) ˆfn)e int. Let h hen 3 2. hen if n is such that 2 m n 2 m+ then inh [ i m 3, 4πi 3 ]. e inh + cos herefore, ˆfn) 2 e inh ) ˆfn) 2 2 m n 2 m+ n Z )) 2 )) ) 2 /2 + sin τ h f f 2 L 2 ) by Lemma.25 τ h f f L ) h 2α f 2 C 0,α ) 3 2 m ) 2α f 2 C 0,α ). by Hölder continuity 8

19 Figure 3: he real and imaginary parts and the magnitude on of the Hardy- in log n e Littlewood series f /2 t) e int pictured is the 2000th partial sum). n n he sum on the left-hand side has at most 22 m+ 2 m ) 2 m+ positive terms. herefore, by Cauchy-Schwarz, /2 ˆfn) f 2 2 m n 2 m+ 2 m n 2 m+ 2 m n 2 ) m+ α 2 m+) m f C 0,α) 2 m/2 α) ) α 2 f C 0,α) 3 We cam sum the above inequality over m 0,, 2,... and the sum remains finite since α > /2. Finally, we observe that by definition f C 0,α ) f C 0,α ) and that ˆf0) f L ) f L ) f C 0,α ) and the theorem is proved. Corollary.36. If f C 0,α ) for some α > /2 and ˆf0) 0, then f A) and we obtain the stricter estimate f A) c α f C 0,α ) Proof. Follow the entire proof of heorem.35 omitting the last paragraph. his proof of Bernstein s theorem relies on the fact that α > /2. he result fails to hold when α /2. One example of a function in C 0,/2 )\A) is the in log n e Hardy-Littlewood series f /2 t) e int. It takes three pages and 4 n n lemmas in [Zyg] to prove that f /2 t) is Hölder continuous with coefficient /2. /2 9

20 We are capable of calculating the Fourier coeffieicents though. By applying the definition directly, we see that So then n Z f /2 n) e in log n /n n in n. ˆfn) + 2 n in + 2 n. n herefore, f /2 t) / A) despite being in C 0,/2 ). he Hardy-Littlewood series is quite an exotic looking function. We show plots of its real and imaginary parts in Figure 3. 2 Fourier Coefficients of Linear Functionals his corresponds to Section.7 of [Katz]. Let B be a homogeneous Banach space on. For simplicity, assume that e int B for all n. We denote the dual space of B by B. We differ from [Katz] in that we denote the evaluation of a fucntional µ B at f B by µ[f] instead of f, µ to avoid confusion with the inner product notation introduced in Section.4. Definition 2.. he Fourier coefficients for n Z of a linear functional µ B are defined as 2.) ˆµn) µ[e int ]. We define the Fourier series of µ as the trigonometric series S[µ]t) n ˆµn)eint. An immediate consequence of 2.) is ˆµn) µ B e int B. Example 2.2. For < p <, let B L p ). It is known that B is isomorphic to L q ) for p + q. For g Lq ) the linear functional g : L p ) defined by g f) ft)gt) dt. hus g n) g [e int ] e int gt) dt gt)e int dt ĝn). hus, the nth Fourier coefficient of the functional g coincides with the nth Fourier coefficient of the function g L q ). It is clear that for any g L q ), the functional g : L p ) is a continuous linear functional g is bounded, hence continuous by Hölder s inequality), thus g L p )). Moreover, for all < p <, this is precisely all of the linear functionals as a result of the Hahn-Banach heorem. Specifically, it is due to the fact that the Banach space L p is reflexive for < p <. For p, these results are not true. he reader is turned to books on functional analysis [Lax, -F]) for more on this. n heorem 2.3 Parseval s formula:). For f B, µ B we have 2.2) µ[f] lim N n N n ) ˆfn)ˆµn). N + 20

21 Proof. First notice that for any trigonometric polynomial, P t) M ˆP n M n)e int then we have [ M ] M M µ[p ] µ ˆP n)e int ˆP n)µ[e int ] ˆP n)ˆµn). n M n M n M hen by heorem.2 we have f lim N K N f in the B norm. herefore µ[f] lim N µ[k N f] since µ in the dual space of B is continuous. his gives µ[f] lim N µ lim [ N n N N n N n ) ] ˆfn)e int N + n N + ) ˆfn)ˆµn). For µ B we define the partial sums in the natural way as we did for functions in B. hat is, we define and let σ n µ, t) S n µ, t) n j n n j n ˆµj)e ijt j ) ˆµj)e ijt. n + Both S n µ, t) and σ n µ, t) are fucntions with domain and codomain C. Let us return to Example 2.2. If B L p for < p < then the dual space B is identified with L q. he functionals g L p ) are defined by g f) ft)gt) dt for g L q. As an abuse in notation, it is sometimes written g B when we actually mean g. We make precisely the same abuse in notation to define S n µ), σ n µ) as elements in B. We define, as elements in B, n S n µ) ˆµj)e ijt and σ n µ) n j n j n j ) ˆµj)e ijt. n + As functionals in B, the evaluations of S n µ) and σ n µ) with function f B are defined by: 2.3) S n µ)[f] ft)s n µ, t) dt n ˆfj)ˆµj) j n n j n ft)ˆµj)e ijt dt 2

22 and 2.4) σ n µ)[f] n j n ft)σ n µ, t) dt n j n ft) j n + j n + ) ˆfj)ˆµj). ) ˆµn)e ijt dt Consider the mapping S n : B defined by S n µ) S n µ). Linearity follows immediately from the definition. Also by definition, we have S n µ) n j n ˆµj)e ijt 2n + ) max n j n e ijt B µ B. herefore S n is a bounded, hence continuous, functional on B i.e. S n : B ) B ) ). heorem 2.4. Let B be a homogeneous Banach space on. Assume that e int B for all n. Let {a n } n be a sequence of complex numbers. hen the following are equivalent: i) here exists µ B with µ C such that ˆµn) a n for all n. ii) For all trigonometric polynomials P M where the subscript M degp )), we have M ˆP M n)a n C P M B. n M Proof. i) ii)): Assuming µ C such that ˆµn) a n, for any trigonometric polynomial, P M, we have M M ˆP M n)a n ˆP M n)ˆµn) by hypothesis) n M n M µ[p M ] by Parseval, hereom 2.3 ) µ B P M B since µ B ) and the desired result is obtained. i) ii)): Conversely, if we assume the linear functional 2.5) P M M n M M n M ˆP M n)a n ˆP M n)a n C P M B then is bounded in the B norm over all trigonometric polynomials. he space of trigonometric polynomials is a linear subspace of B. herefore by the Hahn- Banach heorem see Functional Analysis or [Lax, -F]) then the bounded 22

23 linear functional on the subspace of polynomials extends to a linear functional, µ on all of B with the same norm, C. his gives us the norm estimate in part i); now to show that ˆµn) a n for all n. Since e int is a polynomial, then by 2.5) we have ˆµn) µ[e int ] n j n ê ijt n)a j n j n δ n a j a n. An immediate result of this theorem is: Corollary 2.5. A trigonometric series S n a n e int is the Fourier series of some µ B with µ B C iff σ N S) B C for all N where σ N S) denotes the element of B the Fourier series of which is j ) a j e ijt ). N + Proof. ): Suppose S.4), for any f B σ N S)[f] n ˆµn)e int and µ B n N n ) N + j N C. hen by 2.4) and ˆfn)ˆµn) µ[k N f] µ B K N f B C K N L ) f B C f B. ): Now suppose σ N S) B C. hen for any trigonometric polynomial P M of degree M), by 2.4) lim σ M NS)[P M ] lim P N N M n) n ) a n N + n M M P M n)a n C P M B n M We are now in the situation where we can apply heorem 2.4 directly, thus guaranteeing the existence of a µ with µ B C such that ˆµn) a n which proves the theorem. Example 2.6. In the case B C) the dual space B is identified with M), the space of Borel measures on. We shall refer to the Fourier coefficients and series) of measures as Fourier-Stieltjes coefficients and series, respectively). he mapping f ft) dt is an isometric embedding of L ) to M). his map is surely injective since if ft) dt, gt) dt are equal as measures in M) then we necessarily must have ft) gt) a.e. on. Furthermore, norm is preserved since gt) dt gt) dt g L ). However, an example of a measure not obtained this way is the Dirac measure δ defined by f, δ f0). hen by definition we have ˆδn) for all n. 23

24 n ecall that a measure µ is positive if µe) 0 for every measurable set E. If µ is absolutely continuous, i.e. µ gt) dt for g L ), then µ is positive iff gt) 0 a.e. We will need the following Lemma to prove Herglotz s theorem: Lemma 2.7. S a n e int is the Fourier-Stiltjes series of a positive measure µ iff for all n, σ n S, t) n j n j ) a j e ijt 0 on. n + Proof. ): Suppose S S[µ] for a positive measure µ M). Let f be a nonnegative continuous function on. hen σ n S)[f] ft)σ n S, t) dt n j n n j n ˆfj) ft) j n + j n + ) ˆµj). ) ˆµj)e ijt dt We claim that σ n S)[f] µ[k n f]. o see this, we use Parseval s formula 2.2): µ[k n f] lim N j N lim n N j n j N + j N + ) K n fj)ˆµn) ) j n + ) ˆfj)ˆµn). where the limits in the sum changed since K n fj) 0 for all j > n. his n gives µ[k n f] j ) ˆfj)ˆµn) and the claim is proved. n + hus we have j n σ n S)[f] ft)σ n S, t) dt µ[k n f] 0 Since f was an arbitrary nonnegative continuous function on, this guarantees that σ n S, t) 0 on. ): Assume now that σ n S, t) 0. ecall from Corollary 2.5 that σ n S) is the element in B M) with Fourier Series σ n S)[f] j N ft)σ n S, t) dt f which gives us that the norm of the functional σ n S) is σ n S) M) σ n S, t) dt n j n j N + ) a j e ijt. Since σ n S, t) dt j ) a j e ijt dt a 0. n + 24

25 his allows us to apply Corollary 2.5 to conclude that there exists some µ such that S S[µ]. o check that µ is a positive measure, observe that by Parseval, for any continuous nonnegative function f, f dµ µ[f] lim N n N n N + lim N σ N µ)[f] lim N ) ˆfn)ˆµn) ft)σ n µ, t) dt 0. Since f was an arbitrary nonnegative continuous function, this give µ is a positive measure. Notice that in the proof above we did not use the fact that σ n S, t) 0 for all n. It would suffice to have σ n S, t) 0 for infinitely many n and the result would still hold. Definition 2.8. A sequence of numbers {a n } n is positive definite if for any sequence {z n } having only a finite number of nonzero terms we have a n m z n z m 0. n,m heorem 2.9 Herglotz). A sequence {a n } n is positive definite iff there exists a positive measure µ such that a n ˆµn) for all n. Proof. ) : Assume µ a positive measure with a n ˆµn) for all n. hen a n m z n z m e in m)t z n z m dµ z n 2 0. n,m n,m n ): Assume {a n } is a positive definite sequence. Let S Fix some N > 0 and let hen z n a n m z n z m n,m { e int n N 0 n > N j C j,n a j e ijt n a n e int. where C j,n denotes the number of way to write j n m for n, m N. It is easy to show that C j,n max{0, 2N + j }. hen σ 2N S, t) 2 j 2N 2N + 2N + j ) Ŝj)e ijt 2N + 2 j 2N 2N + j )a j e ijt j C j,n a j e ijt n,m a n m z n z m 0. 25

26 hen by Lemma 2.7, there exists a positive measure µ such that S S[µ] which gives the theorem. heorem 2.0. Let B be a homogeneous Banach space on and µ M). hen there exists a unique linear operator on B with the following two properties: i) µ M). ii) [f]n) ˆµn) ˆfn) for all f B. Proof. We begin first by showing uniqueness of assuming that such a exists. For any polynomial P M ˆP n)e int then by condition ii) we M must have [P M ] M n M n M ˆµn) ˆP M n)e int. herefore, is completely determined n M for polynomials on B. By condition i), is a bounded hence continuous) linear operator. herefore must be determined on all of B by the density of polynomials. Now to show existence: It suffices to show that if we define [P M ] M ˆµn) ˆP M n)e int on the space of polynomials of B then we get the estimate B µ M). his estimate is exactly condition i) and then by construction will satisfy condition ii) on the space of polynomials, and then by the previous argument, for all of B. hen, by the Hahn-Banach heorem [Lax, -F], can be extended to all of B with the same norm. Case : Suppose µ is absolutely continuous with respect to the Lebesgue measure; that is, µ gt) dt for some g C). hen [P M ] M n M ˆµn) ˆP M n)e int hen by.4), we have the estimate M n M g P M n)e int g P M t) [P M ] B g P M B g L ) P M B µ M) P M B. his gives the desired result. Case 2: µ M) is arbitrary. Notice that for each n, σ n µ) is of the form n g nt) dt where g n t) j ) ˆµj)e ijt. Furthermore, n + gn t) dt j n σ n µ) M) µ M). hen by the argument in Case, we have [K n P M ] B g n P M B µ M) P M B. his gives us the norm estimate i) on the space of trigonometric polynomials. hen since P M is a trigonometric polynomial and is now continous, we have [P M ] lim n [K n P M ] µ M) P M B and the theorem is proved. 26

27 Corollary 2.. Let f B and µ M ). hen {ˆµn) ˆfn)} is the sequence of Fourier coefficients of some function in B. In the proof of heorem 2.0 we saw when µ is absolutely continuous) that [f] g P. hen adapting this notation, we define the convolution of µ and f as µ f [f] where is exactly the operator established in heorem 2.0. Definition 2.2. For µ M) we define µ # M) by µ # E) µ E) for every Borel set E, or equivalently by ft) dµ # f t) dµ for every f C). It follows from this definition that 2.6) µ# n) e int dµ # e int dµ # e int dµ ˆµn). Corollary 2.3. Let B be a homogeneous Banach space on and B be its dual space. Let µ M) and v B. hen n ˆµn)ˆvn)eint is the Fourier series of an element µ v B with the estimate µ v B µ M) v B. Proof. As we saw in the proof of heorem 2.0, there exists an operator which sends v B to the element in B with Fourier series ˆvn)ˆµn)e int n n ˆvn) µ # n)e int. As discussed, we call this element µ # v B. With this knowledge, the the corollary follows immediately from heorem 2.0 and the fact that µ # M) µ M). In particular, we have that for any µ, v M), then ˆµn)ˆvn)e int is the Fourier-Stieltjes series of the measure µ v B. n hus far, we ve defined the convolution of two measures in terms of its Fourier-Stieltjes series. It is worth noting that the convolution can be done directly as well. Given µ, v M), f C), the double integral If) ft + τ) dµt) dvτ) is well defined. By linearity of the integral, I is linear and continuous since If) f L ) µ M) v M) <. Since M) is the dual of C), then by the iesz epresentation heorem, there exists some λ M) such that If) ft) dλ. If we let ft) eint we see that ˆλn) Ie int ) e int+τ) dµt) dvτ) e inτ ˆµn) ˆµn)ˆvn). hat is to say, λ µ v. his gives f dµ v) ft + τ) dµt) dvτ). If we let f n be a sequence of continuous functions which converge to E for E measurable. hen by the Lebesgue Dominated Convergence heorem WLOG each f n ) then this definition becomes equivalent to µ v)e) µe τ) dvτ). hese can be taken to be alternate definitions of the convolution µ v. 27

28 Definition 2.4. A measure µ M) is said to be discrete if µ a j δ τj where a j C. In this case µ M) a j. A measure µ M) is said to be continuous if every singleton has measure t+η zero, i.e. µ{t}) 0 for every t. An equivalent definition is lim η 0 t η dµ 0 for all t. Every measure µ M) can be decomposed into a sum of a continuous and discrete measure. From the alternate definition of convolution of measures, if µ is continuous and v is bounded, then µ v is a continuous measure since µ v{t}) µt τ) dvτ) 0. Also δ t δ t )E) δ t E τ) dδ t τ) δ t E t ) δ t+t E). hus if µ j a j δ tj and v k b k δ t k then µ v j,k a j b k δ tj+t k. If µ µ c+µ d the continuous and discrete parts, respectively), then by symmetry arguments, µ # c is the continuous part of µ # and µ # d is the discrete part of µ#. herefore by linearity we have µ µ # µ c + µ d ) µ # c + µ # d ) µ c µ # c + µ c µ # d + µ d µ # c ) + µ d µ # d where the first three terms are continuous and the last is discrete. If µ d a j δ tj then µ # d j a jδ tj since µ # d E) µ d E) j a jδ tj E) j a jδ tj E). Putting this all together, we have µ d µ # d {0}) j a j a j δ 0 j a j 2 δ 0. he same result holds for any bounded measure µ M) since the measure of the continuous part of a singleton vanishes. his gives the following Lemma: Lemma 2.5. Let µ M). hen µ µ # ){0}) t µ{t}) 2. In particular, µ is continuous iff µ µ # ){0}) 0. Proof. he necessity follows from the definitions and the discussion above. Conversely, if µ µ # ){0}) 0, then the discrete part, µ d a j δ tj, must have coefficients a j 0 for all j. hat is to say, µ is continuous. his allows us to recover the discrete part of a measure from its Fourier- Stieltjes series. heorem 2.6. Let µ M), t. hen µ{t}) lim N 2N + n N ˆµn)e inτ. 28

29 Proof. Consider the functions ϕ N t) 2N+ D Nt τ) 2N+ n N e inτ e int. We have ϕ N for all N. Furthermore, off a δ-neighborhood of τ we have ϕ N t) sin N + /2)t τ)) 2N + sin/2t τ)) 2N + sinδ/2) which tends to zero uniformly. For sake of notation, let ν µ µ{τ})δ τ. Notice that ν{τ}) 0; hence ν τ+δ is continuous at τ and thus lim δ 0 dν 0. his gives τ δ lim ν[ϕ N ] lim ϕ N dν 0. N N But ν[ϕ N ] 2N + n N e inτ e int dν e inτ ˆνn) 2N + n N ) ˆµn)e inτ µ{τ}). 2N + n N Sending N gives us the desired result. Corollary 2.7 Wiener). Let µ M). hen τ µ{τ}) 2 lim N 2N + In particular µ is continuous iff 2N+ n N j N ˆµn) 2 0. ˆµn) 2. Proof. In Lemma 2.5 we saw that τ heorem 2.6 and 2.6): µ{τ}) 2 µ µ # ){0}). hen by τ µ{τ}) 2 µ µ # ){0}) lim N 2N + lim N 2N + n N n N µ µ # n)e in0 ˆµn)ˆµn)n) lim N 2N + j N ˆµn) 2. 29

30 3 Fourier ransforms on 3. Fourier ransforms for L ) he space of Lebesgue integrable functions on the real line is denoted by L ). For f L ) we define the norm f L ) fx) dx. For sake of notation, we will sometimes write f L ) f L. Definition 3.. he Fourier ransform, ˆf, of f is defined by 3.) ˆfξ) fx)e iξx dx for all real-valued ξ. Note: we deviate from the definition of ˆf in [Katz] and use the more conventional definition above). Notice that fx) is defined for all x and ˆfξ) is defined for all ξ. herefore it is helpful to distinguish these two domains. We say x and ξ ˆ. Most of the properties of Fourier coefficients from heorem. are valid for Fourier transforms: heorem 3.2. For f, g L ) then. f + g)ξ) ˆfξ) + ĝξ). 2. For any α C, αf)ξ) α ˆfξ). 3. Let f denote the complex conjugate of f. hen ˆ fξ) ˆfξ). 4. For y let τ y f)t) ft y). hen τ y fξ) ˆfξ)e iξy. 5. ˆfξ) f L ). 6. For λ \ {0}, denote f λ x) λfλx). hen ˆf λ ξ) ˆf ξ λ ). Proof. For parts -5, the proofs are exactly as in heorem.. Part 6 comes from a change of variables y λx: ˆf λ ξ) λfλx)e iξx dx fy)e iξ/λ)y dy ˆf ξ λ ). heorem 3.3. Suppose F L ) AC); that is, F x) x fy) dy for some f L ). hen ˆF ξ) ˆfξ). iξ Proof. We integrate by parts to obtain: ˆF ξ) F x)e iξx dx iξ fx)e iξx dx iξ ˆfξ) where the boundary terms vanish since F L ) AC); that is, lim x F x) 0. 30

31 heorem 3.4. Let f L ). hen ˆf is uniformly continous in ˆ. Proof. ˆfξ η) ˆfξ) fx)e iξ η)x e iξx ) dx fx) e iηx dx. Notice that the integral on the right-hand-side is independent of ξ and is bounded by 2 fx) everywhere in. hen by Lebesgue dominated convergence, lim η 0 ˆfξ η) ˆfξ) 0 independent of ξ. heorem 3.5. Let f L ) and xfx) L ). hen ˆf is differentiable in ˆ and d dξ ˆfξ) ixf)ξ). Proof. Consider the Newton quotient: ˆfξ + h) ˆfξ) h fx)e iξx h eihx ) dx. Notice that e ihx h cos hx) h x for all real h. herefore, the integrand on the right-hand side of the Newton quotient is bounded in norm by xfx) L ). herefore, by Lebesgue Dominated Convergence, we have: lim h ˆfξ + h) ˆfξ) h lim fx)e iξx h h eihx ) dx. ) )) cos hx) sin hx) lim h fx)e iξx + i dx. h h fx)e iξx ix) ixf)ξ). heorem 3.6 iemann-lebesgue Lemma). If f L ), then lim ˆfξ) 0. ξ Proof. Let g Cc ). hen by heorem 3.5 and 3.2 then for all ξ ˆ, ξĝξ) < g L ) <. his gives lim ξ ĝξ) 0 and the result hold for g Cc ). Now suppose f L ) is arbitrary. Fix ɛ > 0. Since Cc ) is dense in L, then there exists some g Cc ) such that f g L ) < ɛ. hen by heorem 3.2, ˆfξ) ĝξ) < f g L < ɛ. Since lim ξ ĝξ) 0 this gives lim sup ˆfξ) < ɛ. Since this hold for all ɛ > 0, we have lim ˆfξ) 0. ξ ξ Just as on the torus, we define the convolution on the real line: Definition 3.7. For f, g L ) we define the convolution f g as f gx) fx y)gy) dy. 3

32 he properties of the convolution from the torus also carry over to the real line: heorem 3.8. For f, g L ), then the following properties hold: a) fx y)gy) is integrable as a function of y. b) f g L ) f L ) g L ). c) he convolution operation in L ) is commutative i.e. f g g f). d) f gξ) ˆfξ)ĝξ) for all ξ ˆ. Definition 3.9. We denote by Aˆ) the space of all functions ϕ on ˆ which are the Fourier transforms of functions in L ). We define the norm ˆf Aˆ) f L ) 3.. Summability kernels Definition 3.0. A summability kernel on the real line is a family of continuous functions {k λ } defined on with continuous or discrete parameter λ) satisfying the following conditions:. k λx) dx. 2. k λ L O) as λ. 3. For all δ > 0, lim k λ x) dx 0. λ x >δ he following proposition shows one way to construct summability kernels: Proposition 3.. For any f L ) with fx) dx then k λx) λfλx) is a summability kernel. Proof. By a change of variables u λx, condition is satisfied since k λ x) dx fu) du. Condition 2 is satisfied by the same change of variables since for any λ, k λ L fu) du f L. Finally condition 3 is also satisfied by the same change of variables and the fact that f L ): lim λfλx) dx lim fu) du 0. λ x >δ λ u >λδ Definition 3.2. he Féjer kernel on is defined by K λ x) λkλx) where Kx) ) 2 sinx/2) x/2 32

33 By the previous proposition, to show that the Féjer kernel defined in Definition 3.2 is indeed a summability kernel, it suffices to show that Kx) dx. his follows from a change of variables, trigonometric identity, and the following lemma: Lemma 3.3. cosx) x 2 Proof. We evaluate the integral by means of complex contour integration. Consider the complex function fz) eiz z. For real numbers, 0 < r < we look 2 to examine the integral of fz) along the contour C where fz) dz + fz) dz C C + [, r] π C r + where C is the half circle centered at the origin with radius traversed in the counter-clockwise direction and C r is the half circle centered at the origin with radius r traversed in the clockwise direction. he function fz) has a singularity at z 0 which is not contained in the contour C, therefore we must have fz) dz 0 since f is analytic away from C 0. We first estimate the integral of fz) along the outer loop, C. We parametrize the curve by z e iθ, dz ie iθ dθ where θ [0, π]. hus fz) dz C π 0 π 0 π 0 [r,] expie iθ ) 2 e i2θ ie iθ dθ expicosθ) + i sinθ))) dθ exp sinθ)) dθ π which tends to 0 as. hus, lim C fz) dz 0. o estimate C r fz) dz, we consider the power series of f: fz) eiz z 2 z 2 iz) n n n! i z iz 6 z2 24. Each term will vanish as r 0 except for the first term. herefore, we have i 0 i lim fz) dz lim dz lim r 0 + C r r 0 + C r z r 0 + π re iθ rieiθ dθ π Finally, if we restrict to the real axis and consider z x + 0i then fz) cosx) x 2 So sending r 0, gives cosx) x dx π as desired. 2 We now show some additional examples of summability kernels: 33

34 Example 3.4. he De la Vallée Poussin kernel is given by: 3.2) V λ x) 2K 2λ x) K λ x) with the property, ξ λ ˆV λ ξ) 2 ξ λ, λ ξ λ 0, 2λ ξ. Example 3.5. Poisson s kernel is given by P λ x) λp λx), where 3.3) P x) π + x 2 ) and ˆP ξ) e ξ. Example 3.6. Gauss kernel is given by G λ x) λgλx) where 3.4) Gx) π /2 e x2 and Ĝξ) e ξ2 /4. Figure 4 shows plots of these four real summability kernels. By following the proof methods on we get an analogous result of summability kernels on the real line: heorem 3.7. Let f L ) and {k λ } a summability kernel on, then lim f k λ f L λ ) 0. hen, since the Féjer kernel is a summability kernel, we immediatelly get the following result: Corollary 3.8. If f L ) then λ/ 3.5) f lim ξ ) ˆfξ)e iξx dξ λ λ/ λ in the L ) norm. Proof. First notice that if gx) e iαx, then g fx) e iαx z) fz) dz e iαx ˆfα). his gives that the right hand side of 3.5) is indeed K λ f. hen since the Féjer kernel is a summability kernel, the corollary follows directly from the previous theorem. heorem 3.9 Uniqueness heorem). If f L ) and ˆfξ) 0 for all ξ ˆ, then f 0 a.e. 34

35 Figure 4: Plots of the Féjer kernel, De la Vallée Poussin kernel, Poisson kernel, and Gauss kernel respectively) Proof. his theorem follows immediately from the previous corollary. If ˆfξ) 0 for all ξ then the right hand side of 3.5), then f 0 in the L ) norm. hat is, f 0 a.e. Suppose that ˆf L ˆ). hen ) ˆfξ)e iξx ˆfξ). his al- ξ λ lows us to apply Lebesgue Dominated Convergence to 3.5) which converges uniformly in x ) to give us an inversion formula: 3.6) fx) ˆfξ)e iξx dξ. ˆ By applying 3.6) to the Féjer kernel, we get λ K λ x) ξ ) e iξx dξ ˆK λ ξ)e iξx dξ λ λ { which gives ˆK ξ λ ξ) λ, ξ λ 0, ξ > λ. hen since K λ fξ) ˆK λ ˆfξ) we { ) ˆfξ), ξ λ have K ξ λ fξ) λ 0, ξ > λ 3.7 gives us the following result:. Combining this with heorem 35

36 heorem he space of functions with compactly supported Fourier transforms i.e. ˆf [ Λ, Λ] for some finite Λ ) is a dense subspace of L ). 3.2 L 2 ) heory On the torus,, there was no problem taking the Fourier coefficients of functions in L 2 ) because L p ) L ) for p > since is finite. However, on the real line, L p ) L ) and we need a new way to define the Fourier transform on these spaces. We focus first on functions in L 2 ) and then turn our attention to functions on L p ) for < p < 2. Lemma 3.2. Let f be a continuous function with compact support on. hen fx) 2 dx ˆfξ) 2 dξ. ˆ Proof. Suppose first that suppf) π, π). hen by the L 2 theory on established in heorem.28, f 2 L 2 ) fx) 2 dx ˆfn) 2. eplacing n Z fx) with e iαx fx) gives e iαx fx) 2 dx fx) 2 dx n Z ˆfn + α) 2. Integrating both sides with respect to α from 0 to gives the desired result fx) 2 dx ˆ ˆfξ) 2 dξ. Suppose now that suppf) π, π). Consider gx) λfλx). For λ sufficiently large then suppg) π, π). Furthermore, by elementary changes of variables, we can discover g 2 L 2 ) λ fλx) 2 dx fx) 2 dx f 2 L 2 ) and ĝ 2 L 2 ˆ) ˆ λfλx)e ixξ dx 2 dξ ˆ ξ ˆf λ λ ) 2 dξ ˆfξ) 2 dξ ˆ By the argument above, we know g 2 L 2 ) ĝ 2 L 2 ˆ) which gives the desired result of f 2 L 2 ) g 2 L 2 ) ĝ 2 L 2 ˆ) ˆf. 2 L 2 ˆ) ˆf 2 L 2 ˆ). heorem 3.22 Plancherel). here exists a unique surjective operator F : L 2 ) L 2 ˆ) satisfying the following two properties: Ff L 2 ˆ) f L 2 ) Ff ˆf for f L L 2 ). 36

37 Proof. By the previous lemma, the first condition is satisfied for continuous functions with compact support. Such functions are dense in L L 2 ). hus Ff L2 ˆ) f L 2 ) holds for f L L 2 ). Furthermore, this tells us that F is a continuous operator on a dense subspace of L 2 ). hen by the Hahn-Banach heorem, F extends to all of L 2 ) with the same norm. his gives the uniqueness and both properties. o show that F is onto, we know from heorem 3.20 that every C 2 compactly supported function on ˆ is the Fourier transform of a bounded function in L ). his gives that the image of F is dense in L 2 ˆ). hen by continuity of F, it coincides with all of L 2 ˆ) We will give an alternate proof of the Plancherel heorem in L 2, but first we make a remark with some functional analysis results that will be required. emark It is well known that L 2 ˆ) is reflexive. hat is, every element in the dual space L 2 ˆ) acting on f L 2 ˆ) can be written as f, g ˆ fξ)gξ) dξ for some g L2 ˆ). Suppose V is a closed proper) subspace of L 2 ˆ). As a corollary of the Hahn-Banach heorem, for y L 2 ˆ) \ V, there exists a L 2 ˆ) such that y) and x) 0 for all x V. Alternate proof. Claim : f L2 ) ˆf for f C c ). L 2 ˆ) We first must show that for such an f C c ), we have ˆf L 2 ). Define gx) f fx) fx t)f t) dt where fx) f x). By the properties of convolutions, heorem 3.8, g is continuous and integrable. By construction of g, we have g0) f t)f t) dt ft) 2 dt f 2 L 2 ) and for all ξ ˆ, ĝξ) ˆfξ) ˆ fξ) ˆfξ) By Corollary 3.8, f t)e itξ dt ˆfξ) ˆfξ) ˆfξ) 2. λ/ g0) lim ξ ) λ λ/ λ ˆfξ) 2 dξ a.e. { ) Since [ λ, λ ]ξ) ξ λ ˆfξ) } 2 is a monotone increasing sequence of λ integrable functions that converge pointwise a.e., then by the Beppo-Levy Monotone Convergence) heorem, f 2 L 2 ) g0) ˆf 2 and the claim is L 2 ˆ) proved. We know that C c ) is dense in L L 2 ) which is dense in L 2 ) we take this for granted in this document but the details are given in full on page 52 of [Ben]). 37