TOPICS IN FOURIER ANALYSIS-II. Contents

TOPICS IN FOURIER ANALYSIS-II M.T. NAIR Contents. Trigonometric series nd Fourier series 2 2. Riemnn Lebesgue Lemm 4 3. Dirichlet kernel 6 4. Dirichlet-Dini criterion for convergence 8 5. Ce`sro summblity of Fourier series 2 6. Divergence of Fourier series 8 7. Uniqueness 20 8. Convolution 23 9. L 2 -Theory 26 References 28 NOTES PREPARED FOR PART OF AN ELECTIVE COURSE FOURIER ANALYSIS FOR M.SC. STUDENTS OF IIT MADRAS, JULY-NOVEMBER, 204.

2 M.T. NAIR. Trigonometric series nd Fourier series Definition.. A series of the form (.) c 0 + ( n cos nx + b n sin nx) n= is clled trigonometric series, where c 0, n, b n re rel numbers. If (.) converges on [, π] to n integrble function f nd if it cn be integrted term by term, then nd c 0 = f(x) cos nxdx, n = π f() = f(π), f(x) cos nxdx, b n = π f(x) sin nxdx. If the (.) converges (pointwise) on [, π] to function f, then f cn be extended s -periodic function by defining If the series f(x + 2nπ) = f(x), n Z. ( n + b n ) converges, then (.) converges uniformly on [, π] n= nd it cn be integrted term by term. We know tht if f L [, π], then the function f : [, π] C defined by stisfies The series (.) cn be written s { f(x), x [, π), f(x) = f(), x = π f() = f(π) nd f = f.e. n= c n e inx. Definition.2. Let f L [, π]. The Fourier series of f is the series (.2) where (.3) n = π 0 2 + ( n cos nx + b n sin nx), n= f(x) cos nxdx, b n = π f(x) sin nxdx.

The series (.4) n= TOPICS IN FOURIER ANALYSIS 3 c n e inx with c n = f(x)e inx dx is lso clled the Fourier series of f. The coefficients c n re clled the Fourier coefficient nd re usully denoted by ˆf(n), i.e., The sum ˆf(n) = S N (f, x) := f(x)e inx dx, n Z. n= N ˆf(n)e inx is clled the N-th prtil sum of the Fourier series (.4)). Nottion: In the bove nd in the following, the integrl re w.r.t. the Lebesgue mesure. The fct tht (.2) is the Fourier series of f is usully written s Equivlently, f(x) 0 2 + ( n cos nx + b n sin nx). n= f(x) n= ˆf(n)e inx. Since cos nx, sin x, e inx re -periodic functions, we cn tlk bout Fourier series of -periodic functions. If (.2) (resp. (.4)) converges t point x [, π], then it converges t x + 2kπ for every k Z. The Fourier series (.4)) converges t x [, π] if nd only if S N (f, x) f(x) s N. If f L [, π], then ˆf(n) 0 s n. If ˆf(n) converges, then ˆf(n)e inx converges uniformly. n= n= Suppose Fourier series of f L [, π] converges uniformly, sy to g. continuous nd π g(x)e imx = ˆf(n) e i(n m)x dx = ˆf(m), n= Then g is

4 M.T. NAIR i.e., ĝ(m) = ˆf(m) for ll m Z. A nturl question would be whether f = g.e. We shll nswer this ffirmtively. We know tht if the Fourier series of f L [, π] converges, then ˆf(n) 0 s n. Cn we ssert this for every f L [, π]? The nswer is in the ffirmtive s proved in the next section. 2. Riemnn Lebesgue Lemm Theorem 2.. (Riemnn Lebesgue lemm) Let f L [, b]. Then b f(t) cos(λt)dt 0 nd b f(t) sin(λt)dt 0 s λ. Corollry 2.2. (Riemnn Lebesgue lemm) Let f L [, b]. Then b f(t) cos(nt)dt 0 nd b For the proof of the Theorem 2., we shll mke use of f(t) sin(nt)dt 0 s n. LEMMA 2.3. The spn of ll step functions on [, b] is dense in L [, b]. Proof of Theorem 2.. First we observe tht if for every ε > 0, there exists function g L [, b] such tht f g < ε nd the the result is true for g, then the result is true for f lso. Indeed, b b b f(t) cos(λt)dt [f(t) g(t)] cos(λt)dt + g(t) cos(λt)dt b ε + g(t) cos(λt)dt. Let λ 0 > 0 be such tht b g(t) cos(λt)dt < ε for ll λ λ0. Then we hve b f(t) cos(λt)dt < 2ε λ λ 0 Step functions re finite liner combintions of chrcteristic functions. Also, recll tht L [, b] is the vector spce of ll Lebesgue mesurble complex vlued functions f such tht f := b f(x) dx <. Here, dx stnds for the Lebesgue mesure.

TOPICS IN FOURIER ANALYSIS 5 so tht b f(t) cos(λt)dt 0 s λ. Similrly, b f(t) sin(λt)dt 0 s λ. Hence, it is enough to prove the result for step functions. Since every step function is finite liner combintion of chrcteristic functions on intervls, it is enough to prove for f of the form f = χ [c,d], [c, d] [, b]. Note tht b d χ [c,d] cos(λt)dt = cos(λt)dt Similrly, b = c sin(λd) sin(λc) λ 2 0 λ s λ. χ [c,d] sin(λt)dt 0 s λ. Remrk 2.4. If f is Riemnn integrble on [, b], then there exists sequence of (f n ) of step functions such tht f f n 0. Thus, conclusion in Theorem 2. holds if f is Riemnn integrble. Proof of Lemm 2.3. If f L [, b] with f 0, then there exists n incresing sequence of non-negtive simple mesurble functions ϕ n, n N such tht ϕ n f pointwise. Hence, by DCT, b f ϕ n 0. From this, for ny complex vlued f L [, b], there exists sequence (ϕ n ) of simple complex mesurble functions We observe (see [3]): b f ϕ n 0. () Every simple rel vlued mesurble function is finite liner combintion of chrcteristic function of mesurble sets. (2) For every mesurble set E (, b) nd ε > 0, there exists n open set G E such tht m(g \ E) < ε. Hence, b χ G χ E = b χ (G\E) m(g \ E) < ε. (3) If G (, b) is n open set, then G = k= I n, where {I n } is countble disjoint fmily of open intervls in (, b); χ G = lim n ψ n, ψ n = n χ Ik, k=

6 M.T. NAIR Since 0 ψ n χ G, by DCT, χ G ψ n 0. (4) By ()-(3), if ϕ is simple mesurble function nd ε > 0, there exists step function ψ such tht Thus, the lemm is proved. b ϕ ψ < ε. Note tht where S N (f, x) := n= N 3. Dirichlet kernel e inx D N (t) := f(t)e int dt = f(t)d N (x t)dt, n= N Redefining f t the end-points if necessry, nd extending it s -periodic function, we cn lso write (verify!), S N (f, x) = e int. f(x t)d N (t)dt. Nottion: We denote by T the unit circle T := {e it : t < π}. Note tht if f : T C nd if we define f : R C by f(t) = f(e it ), then f() = f(π) nd f(t + 2nπ) = f(t) for ll n Z. Tht is, f is -periodic function. In the due course, we shll identify -periodic functions with functions on T. We shll denote L (T ) for the spce of ll -periodic (complex vlued) functions on R (with equlity replced equl.e.) which re integrble on [, π] with norm f f := f(x) dx. Anlogously, for p <, L p (T ) denotes the spce of ll -periodic (complex vlued) functions f on R such tht f p is integrble on [, π] with norm ( π ) /p f f p := f(x) p dx

TOPICS IN FOURIER ANALYSIS 7 The spce L 2 (T ) is lso Hilbert spce with inner product (f, g) f, g := f(x)g(x) dx. Definition 3.. The function D N ( ) is clled the Dirichlet kernel. We observe tht, D N ( t) = D N (t) for ll t [, π] nd π D N (t)dt =. D N (t) = n= N e int = + Remrk 3.2. We shll see tht Theorem 3.3. [e int + e int ] = + 2 n= cos nt. n= D N (t) dt s N. 2N +, t = 0, D N (t) = sin(n + 2 sin( t ), 2 t 0. Proof. Clerly, D N (0) = 2N +. So, let t 0. Note tht (e it )D N (t) = [e i(n+)t e int ] = e i(n+)t e int. n= N But, (e it )D N (t) = e it/2 (e it/2 e it/2 )D N (t) = 2ie it/2 sin(t/2)d N (t). Thus, i.e., 2i sin(t/2)d N (t) = e it/2 [e i(n+/2)t e i(n+/2)t ] = 2i sin(n + /2)t. D N (t) = sin(n + 2 )t sin( t 2 ), t 2kπ.

8 M.T. NAIR 4. Dirichlet-Dini criterion for convergence We investigte the convergence: Since S N (f, x) f(x). D N (t)dt = nd S N (f, x) = π f(x t)d N(t)dt, we hve f(x) S N (f, x) = [f(x) f(x t)]d N (t)dt. Theorem 4.. (Dirichlet-Dini criterion) Let f L (T ). If f stisfies f(x) f(x t) t dt < t point x [, π], then S N (f, x) f(x). If ( ) hods uniformly for x [, π], then the convergence {S N (f, x)} to f(x) is uniform. f(x) f(x t) Remrk 4.2. In the bove theorem, by, we men the function t { f(x) f(x t), t 0 ϕ(t) = t 0, t = 0. Proof of Theorem 4.. We observe tht f(x) S N (f, x) = [f(x) f(x t)]d N (t)dt = = π [f(x) f(x t)] sin(n + )t 2 sin( t ) dt 2 { } { f(x) f(x t) t/2 t sin( t ) 2 } sin(n + 2 )tdt Since (t/2)/[sin(t/2)] is bounded, in view of Riemnn Lebesgue lemm, we hve the following. The following corollries re immedite from Theorem 4.. Corollry 4.3. Suppose f is Lipschitz t point 2 x [, π]. Then S N (f, x) f(x) s N. 2 A function ϕ : I C is sid to be Lipschitz t point x 0 I if there exists K 0 > 0 such tht ϕ(x) ϕ(x 0 ) K 0 x x 0 for ll x I. ( )

TOPICS IN FOURIER ANALYSIS 9 Corollry 4.4. Suppose f is Lipschitz 3 on [, π]. Then uniformly on [, π]. S N (f, x) f(x) s N Nottion: We denote by C(T ) the spce of ll -periodic continuous functions on R, nd by C k (T ) for k N {0}, the spce of ll -periodic functions on R which re k-times continuously differentible on R. Corollry 4.5. If f C (T ), then uniformly on R. Now obtin more generl result. S N (f, x) f(x) s N Theorem 4.6. Suppose f is -periodic function such tht the following limits exist t point x R: Then f(x+) := lim t 0+ f(x + t), f(x ) := lim t 0+ f(x t), f f(x + t) f(x+) (x+) := lim, f f(x ) f(x t) (x ) := lim. t 0+ t t 0+ t S N (f, x) f(x+) + f(x ) 2 s N. Proof. Since D N (t) = D N ( t), we hve nd S N (f, x) = = = = 0 0 0 f(x t)d N (t)dt f(x t)d N (t)dt + f(x + t)d N (t)dt + 0 [f(x + t) + f(x t)]d N (t)dt π D N (t)dt = 2 D N (t)dt. 0 0 f(x t)d N (t)dt f(x t)d N (t)dt 3 A function ϕ : I C is sid to be Lipschitz on I if there exists K > 0 such tht ϕ(x) ϕ(x 0 ) K 0 x x 0 for ll x I.

0 M.T. NAIR Hence, for ny β R, S N (f, x) β = Tking β = f(x+)+f(x ) 2, we hve Thus, where A N = Note tht 0 [f(x + t) + f(x t) 2β]D N (t)dt. f(x + t) + f(x t) 2β = [f(x + t) f(x+)] [f(x ) f(x t)]. 0 S N (f, x) β = A N + B N, [f(x + t) f(x+)]d N (t)dt, B N = A = = = π 0 0 0 [f(x + t) f(x+)]d N (t)dt [f(x + t) f(x+)] sin(n + )t 2 sin( t ) dt 2 { } { f(x + t) f(x+) t/2 t sin( t ) 2 0 [f(x ) f(x t)]d N (t)dt. } sin(n + 2 )tdt Since f(x+t) f(x+) t f (x+) s t 0+, there exists δ > 0 such tht 0 < t < δ = f(x + t) f(x+) f (x+) t = f(x + t) f(x+) t + f (x+). Hence, the function { } { } f(x + t) f(x+) t/2 t t sin( t ), t 0, 2 is bounded on (0, δ), nd hence, belongs to L (T ). Therefore,by Riemnn Lebesgue lemm, A N 0 s N. Similrly, we see tht, B N 0 s N. An immedite corollry: Corollry 4.7. If f C(T ) nd hs left nd right derivtive t point x, then S N (f, x) f(x) s N. The following result is known s locliztion lemm.

TOPICS IN FOURIER ANALYSIS LEMMA 4.8. For 0 < r < π nd x [, π], f(x t)d N (t)dt 0 s N. r t π Proof. Observe tht where r t π f(x t)d N (t)dt = g(x, t) = r t π g(x, t) sin(n + /2)tdt, { f(x t)/ sin(t/2), r t π, 0, t r. Since g(x, ) is integrble, by Riemnn Lebesgue lemm, g(x, t) sin(n + /2)tdt 0 s N. r t π Proof of Corollry 4.4 using locliztion lemm. Suppose f is Lipschitz t point x [, π] with Lipschitz constnt K x, i.e., there exists δ > 0 such tht f(x) f(x t) K x t whenever t < δ. Now, f(x) S N (f, x) = [f(x) f(x t)]d N (t)dt = [f(x) f(x t)]d N (t)dt 0 t <δ + [f(x) f(x t)]d N (t)dt δ t π By Lemm 4.8, δ t π [f(x) f(x t)]d N (t)dt 0 s N. Hence, for given ε > 0, there exists N 0 N such tht for ll N N 0, [f(x) f(x t)]d N (t)dt < ε/2. Also, But, 0 t <δ δ t π [f(x) f(x t)]d N (t)dt 0 t <δ f(x) f(x t) D N (t) dt,

2 M.T. NAIR f(x) f(x t) D N (t) dt K x t D N (t) dt, 0 t <δ 0 t <δ t D N (t) = t sin(n + )t 2 sin( t ) = 2 t/2 sin( t ) sin(n + )t 2M, 2 2 2 where M is bound for t/2 sin( t ) on 0 < t δ. Hence, 2 [f(x) f(x t)]d N (t)dt 4MK xδ = 2MK xδ. π 0 t <δ We my tke δ such tht 2MK xδ π f(x) S N (f, x) < ε/2. Hence, [f(x) f(x t)]d N (t)dt 0 t <δ + [f(x) f(x t)]d N (t)dt δ t π < ε for ll N N 0. Exercise 4.9. Suppose f is -periodic nd Hölder continuous t x, i.e., there exist M > 0 nd α > 0 such tht f(x) f(y) M x y α for ll y [, π]. Then show tht S N (f, x) f(x) s N. Exercise 4.0. Suppose f is -periodic nd Hölder continuous on [, π], i.e., there exist M > 0 nd α > 0 such tht f(x) f(y) M x y α for ll x, y [, π]. Then show tht S N (f, x) f(x) uniformly. 5. Ce`sro summblity of Fourier series Theorem 5.. (Fejér s theorem) If f C(T ), then the Fourier series of f is uniformly Ce`sro summble on [, π], tht is, uniformly on [, π]. σ N (f, x) := N + S k (f, x) f(x) s N k=0 Recll tht S k (f, x) := k n= k ˆf(n)e inx = f(x t)d k (t)dt.

Hence, Thus, where σ N (f, x) = N + TOPICS IN FOURIER ANALYSIS 3 S k (f, x) = k=0 σ N (f, x) = K N (t) := N + { } f(x t) D k (t). N + f(x t)k N (t)dt, D k (t). Definition 5.2. The function K N (t) defined bove is clled the Fejér kernel. We observe tht Hence, k=0 π K N (t)dt =. f(x) σ N (f, x) = [f(x) f(x t)]k N (t)dt. For the proof of Theorem 5., we shll mke use of the following lemm. LEMMA 5.3. The following results hold. () For t 0, K N (t) = cos(n + )t N + cos t = N + k=0 sin 2 [(N + )t/2] sin 2. (t/2) (2) K N (t) is n even function nd K N (t) 0 for ll t [, π]. (3) For 0 < δ π, K N (t) ( ) N + sin 2. (δ/2) In prticulr, K N is positive nd K N (t) 0 s N uniformly on 0 < δ t π. Proof of Theorem 5.. Sine K N (t) is non-negtive function (see Lemm 5.3), we hve f(x) σ N (f, x) f(x) f(x t) K N (t)dt. Let ε > 0 be given. Since f is uniformly continuous, there exists δ (0, π] such tht Hence, f(x) f(y) < ε whenever x y < δ. f(x) f(x t) K N (t)dt < ε K N (t)dt = ε. t <δ t <δ

4 M.T. NAIR Also, since f is uniformly bounded there exists M > 0 such tht f(y) M for ll x [, π]. f(x) f(x t) K N (t)dt 2M K N (t)dt. t δ t δ We hve observed in Lemm 5.3 tht K N (t) is n even function nd K N (t) 0 s N uniformly on [δ, π]. Hence, there exists N 0 such tht f(x) f(x t) K N (t)dt 4M K N (t)dt < ε for ll N N 0. t δ Hence, f(x) σ N (f, x) f(x) f(x t) K N (t)dt < 2ε for ll N N 0. Note tht N 0 is independent of the point x. Thus, we hve proved tht S N (f, x) f(x) s N uniformly for x [, π]. Remrk 5.4. The proof of Theorem 5. revels more: Nottion: If f is pece-wise continuous nd -periodic, nd continuous t x, then σ N (f, x) f(x) s N. u n (x) := e inx, n Z. AC(T ) denotes the vector spce of ll -periodic complex vlued functions defined on R which re bsolutely continuous. spn{u n : n Z} is the spce (over C) of ll trigonometric polynomils. Corollry 5.5. The spce of ll trigonometric polynomils is dense in C(T ) with respect to the uniform norm, nd hence dense in L p (T ) w.r.t. p for p <. Proof. By Theorem 5., spce of ll trigonometric polynomils is dense in C(T ) with respect to the uniform norm. Hence, for ny f C(T ), there exists sequence (f n ) of trigonometric polynomils such tht s n. f f n p p = δ f(x) f n (x) p dx f f n p 0 Corollry 5.6. If f L 2 (T ) for some p < nd ˆf(n) = 0 for ll n Z, then f = 0.e.

TOPICS IN FOURIER ANALYSIS 5 Proof. Suppose f L 2 (T ) for some p < nd ˆf(n) = 0 for ll n Z, i.e., f, u n = 0 for ll n Z. By Corollry 5.5, it follows tht f L 2 = 0. Hence, f = 0.e. Corollry 5.7. If f C(T ) such tht ˆf(n) = 0 for ll n Z, then f = 0. prticulr, if f, g C(T ) such tht ˆf(n) = ĝ(n) for ll n Z, then f = g. In Proof. Suppose f C(T ) such tht ˆf(n) = 0 for ll n Z. Thus, f, u n L 2 = 0 for ll n Z. Since C(T ) L 2 [, π], f L 2 [, π]. Hence by Corollry, f = 0.e. Since f is continuous, f = 0. The bove corollry shows: The Fourier coefficients of f C(T ) determines f uniquely. Corollry 5.8. If f C 2 (T ), then f (n) = (in) 2 ˆf(n) for ll n Z. In prticulr, ˆf(n) = o( ), nd the Fourier series of f converges uniformly to f. n 2 Proof. Let f C 2 (T ). Then, using integrtion by prts, we obtin, ˆf(n) = = f(x)e inx dx ] [f(x) e inx π in = f (x)e inx dx in = ] [f (x) e inx π in in = (in) 2 f (x)e inx dx. [ e f inx ] (x) dx in f (x) in [ e inx Hence, f (n) = (in) 2 ˆf(n) for ll n Z. In prticulr, ˆf(n) = o(/n 2 ). Therefore, n Z ˆf(n) converges, nd hence the Fourier series converges uniformly. Suppose S N (f, x) g(x) uniformly. Then it follows tht g C(T ) nd ĝ(n) = ˆf(n) for ll n Z. Therefore, by Corollry 5.7, g = f. in ] dx Following the sme rguments s in the proof of Corollry 5.8, we obtin:

6 M.T. NAIR Corollry 5.9. If f C (T ) nd f is bsolutely continuous, then f exists lmost everywhere, f L [, π] nd f (n) = (in) 2 ˆf(n) for ll n Z, nd the Fourier series of f converges uniformly to f. More generlly, Theorem 5.0. If f C k (T ) nd f (k ) is bsolutely continuous for some k N, then f (k) exists lmost everywhere f (k) L (T ) nd f (k) (n) = (in) k ˆf(n) for ll n Z. Proof of Lemm 5.3. We hve K N (t) := D k (t) N + where D k (t) = Hence, But, Therefore, (N + )K N (t) = Subtrcting the (2) from (), Thus, k=0 k=0 sin(k + /2)t sin t/2 e i(k+/2)t e i(k+/2)t e it/2 e it/2 e i(k+/2)t e i(k+/2)t e it/2 e it/2 [e it ](N + )K N (t) = [ e it ](N + )K N (t) = [2 cos t 2](N + )K N (t) = 2 = sin(k + /2)t sin t/2 e i(k+/2)t e i(k+/2)t k=0 = ei(k+)t e ikt, e it = eikt e i(k+)t e it, e it/2 e it/2 [e i(k+)t e ikt ], () k=0 [e ikt e i(k+)t ] (2) k=0 [cos(k + )t cos kt] = 2[cos(N + )t ] k=0 K N (t) = cos(n + )t N + cos t = N + sin 2 [(N + )t/2] sin 2. (t/2)

TOPICS IN FOURIER ANALYSIS 7 Thus, we hve proved (). It is cler tht K N (t) is even nd non-negtive. Now, for 0 < δ π, sin 2 (t/2) sin δ/2, so tht Thus, δ K N (t)dt = N + δ K N (t)dt δ sin 2 [(N + )t/2] sin 2 dt (t/2) N + π δ δ sin 2 (δ/2) dt. (N + ) sin 2 (δ/2) 0 s N. Exercise 5.. Suppose f is piecewise continuous nd -periodic. If ˆf(n) = 0 for ll n Z, then f(x) = 0 for ll x t which f is continuous. Exercise 5.2. If f C (T ), then ˆf(n) = O(/n). More generlly, f C k (T ) implies ˆf(n) = O(/n k ). Exmple 5.3. Let f(x) = x 2, x π. Note tht ˆf(0) = so tht ˆf(0) = π 2 /3, nd for n 0, Hence, for n 0, Thus, ˆf(n) = = ] [x 2 e inx π x 2 e inx dx in ] [x 2 e inx π x 2 dx = 2 π3 3 2x e inx [2x e inx in dx ] π = in ( in) 2 ] = [2x e inx π = [2x e inx ( in) 2 n 2 = 4π ( )n n 2 x 2 π2 3 + 2 n 0 ( ) n n 2 ˆf(n) = 2 ( )n n 2. e inx = π2 3 + 4 n= Since the series of coefficients converges bsolutely, we hve f(x) = π3 3 + 4 n= ] π = 4π einx n 2 ( ) n cos nx. n 2 ( ) n cos nx. n 2

8 M.T. NAIR Tking x = 0, Thus, Tking x = π, Thus, 0 = π3 3 + 4 n= ( ) n+ n= π 2 = π3 3 + 4 n= n 2 ( ) n n 2 n= ( ) n n 2. = π2 2. ( ) n = π3 3 + 4 n. 2 n 2 = π2 6. n= Exmple 5.4. Let f(x) = x, x [, π]. Note tht ˆf(0) = 0 nd for n 0, ˆf(n) ] = xe inx dx = [x e inx π e inx [ ] in in dx = x e inx π. in Thus, so tht Hence, x = ˆf(n) = ] [x e inx π = in in [πe inπ + πe inπ ] = einπ in ˆf(n) = ( )n inπ = ( )n+ inπ. ( ) n+ e inx = ( ) n+ [e inx e inx ] = 2 ( ) n+ sin nx in in n n= n= n 0 Tking x = π/2 we obtin the Mdhv-Nīlkṅth series π 4 = n= ( ) n+ sin nπ n 2 = n=0 ( ) n 2n +. 6. Divergence of Fourier series Theorem 6.. There exists f C(T ) such tht {S N (f, 0)} is unbounded; in prticulr, the Fourier series of f does not converge to f t 0. For this we shll mke use of the Uniform Boundedness Principle from Functionl Anlysis:

TOPICS IN FOURIER ANALYSIS 9 Theorem 6.2. (Uniform Boundedness Principle) Let (T n ) be sequence of continuous liner trnsformtions from Bnch spce X to normed liner spce Y. If for ech u X, the set { T n u : n N} is bounded, then there exists M > 0 such tht Let sup T n u M n N. u ϕ N (f) := S N (f, 0), f C(T ). We see tht ϕ N : C(T ) C is liner functionl for ech N N nd ϕ N (f) = S N (f, 0) = π ( f( t)d N (t)dt π ) f D N (t) dt. Hence, ech ϕ N is continuous liner functionl on C(T ) nd In fct, sup ϕ N (f) u D N (t) dt. Theorem 6.3. nd sup ϕ N (f) = D N (t) dt u D N (t) dt 8 π k= k. Proof of Theorem 6.. By Theorem 6.3, there does not exist M > 0 such tht sup u ϕ N (f) M for ll n N. Hence, by Theorem 6.2, there exists f C(T ) such tht { ϕ n (f) : n N} is unbounded. Hence, there exists f C(T ) such tht Fourier series of f diverges t 0. Remrk 6.4. Let D := {f C(T ) : {S N (f, 0)} does not converge}. Then C(T ) \ D is subspce of C(T ), nd by Theorem 6., C(T ) \ D is proper subspce. Hence, C(T ) \ D is nowhere dense, nd hence D is dense in C(T ). Thus, we hve proved the following: There exists dense subset D of C(T ) such tht for ech f D, the Fourier series of f diverges t 0. In plce of 0, we cn tke ny point in [, π] nd obtin similr divergence result t tht point.

20 M.T. NAIR 7. Uniqueness Theorem 7.. (Uniqueness of Fourier series) Let f L (T ). If ˆf(n) = 0 for ll n N, then f = 0.e. Proof. Let g(t) = t f(x)dx, t [, π]. Then, by Fundmentl Theorem of Lebesgue Integrtion (FTLI), g is bsolutely continuous, g exists.e. nd g = f.e. Note tht g(t + ) g(t) = Hence g is -periodic. Let Then we see tht Tking we hve t+ t h(t) = h(t + ) h(t) = G(t) = t G(t + ) G(t) = t t+ t f(x)dx = g(x)dx, g(x)dx = [g(x) ĝ(0)]dx, Thus, G is -periodic, nd G = f.e.hence, f(x)dx = ˆf(0) = 0. t [, π]. g(x)dx = ĝ(0). t [, π], [g(x) ĝ(0)]dx = [ĝ(0) ĝ(0)] = 0. ˆf(n) = Ĝ (n) = (in) 2 Ĝ(n) for ll n 0. Therefore, Ĝ(n) = 0 for ll n 0. Hence, by Corollry 5.9, G(x) = Ĝ(0), nd hence G = 0, so tht f = 0.e. Recll tht for ech f L (T ), ˆf(n) 0 s n. Thus, ( ˆf(n)) c 0 (Z) for every f L [, π]. Nottion: c 0 (Z) is the set of ll sequences ϕ : Z C such tht ϕ(n) 0 s n.

TOPICS IN FOURIER ANALYSIS 2 Theorem 7.2. The mp F : L (T ) c 0 (Z) be defined by F(f) = ( ˆf(n)), f L (T ) is n injective continuous liner opertor which is not onto. Proof. For f, g L (T ) nd α C, we hve (f + g)(n)) = ˆf(n) + ĝ(n) for ll n Z, αf(n)) = α ˆf(n). for ll n Z, Thus, F is liner opertor. Note tht ˆf(n) = π f(x)e inx dx Thus, if we endow L [, π] with the norm f L := f(x) dx, f L (T ), f(x) dx. then we see tht F is continuous liner opertor. By Theorem 7., F is injective. So, it remins to show tht F is not onto. If it is onto, then my Bounded Inverse Theorem, its inverse is lso continuous. Note tht F(D N ) = { D N (n)} nd D N (n) = for n N so tht (F(D N )) = for ll N N. If F is onto, then, by Bounded Inverse Theorem 4 its inverse F is continuous so tht ( D N ) = { F (F(D N ) } is bounded, which is not true. By the bove theorem there exists (c n ) c 0 (Z) such tht there is no f L ( T ) stisfying c n = ˆf(n) for ll n N. It is nturl urge to hve n exmple of such sequence c n ). We shll show tht c n ) with { / log(n), n 2, c n = 0, n, is such sequence. This is consequence of the first prt of the following theorem. 4 If X nd Y re Bnch spces nd T : X Y is continuous bijective liner opertor, then T is lso continuous.

22 M.T. NAIR Theorem 7.3. Let f L (T ). Then n 0 ˆf(n) n einx converges t every x R nd b b f(x)dx = n Z ˆf(n)e inx dx. For proving the bove theorem we shll mke use of the following theorem: Theorem 7.4. (Jordn) If f L (T ) is of bounded vrition 5, then for every x R, In prticulr, if f AC(T ), then for every x R. It cn be esily shown tht: S N (f, x) (f(x+) + f(x ) s N. 2 S N (f, x) f(x) s N Every bsolutely continuous function is of bounded vrition. Proof of Theorem 7.3. Let g(t) = t [f(x) ˆf(0)]dx. Then g is bsolutely continuous nd g is -periodic, i.e., g AC(T ), g L (T ) nd g = f ˆf(0).e. Therefore, ĝ (n) = inĝ(n) for ll n 0 so tht By Jordn s theorem, g(x) = ĝ(0) + n 0 ĝ(n) = ˆf(n) in, n 0. ĝ(n)e inx = ĝ(0) + n 0 ˆf(n) in einx. In prticulr, n 0 ˆf(n) n einx converges. Also, g(x) g(y) = n 0 ˆf(n) in [einx e inx ] = n 0 ˆf(n) x y e int dt. 5 A function f : [, b] C is of bounded vrition if there exits κ > 0 such tht for every prtition x 0 < x < < x n = b, n k= f(x k+) f(x k κ.

TOPICS IN FOURIER ANALYSIS 23 But, g(x) g(y) = x g (t)dt = x [f(t) ˆf(0)]dt = x y y y f(t)dt ˆf(0)(x y)dt. Hence, x b y f(t)dt = n Z ˆf(n)e int dt. This competes the theorem. Corollry 7.5. Let (c n ) be with c n = { / log(n), n 2, 0, n, Then there is no f L (T ) stisfying c n = ˆf(n) for ll n N. Proof. Suppose f L (T ) stisfying c n = ˆf(n) for ll n N. Then by the first prt of Theorem 7.3, the series e inx n=2 converges. In prticulr, tking x = 0, n log n n=2 n log n converges, which is not true (e.g., by integrl test). 8. Convolution Given f, g L (T ), it cn be shown tht (x, y) f(x y)g(y) is mesurble on R R, nd hence, for ech x [, π], the integrl converges. f(x y)g(y)dy Definition 8.. The convolution of f, g L (T ) is defined by (f g)(x) = f(x y)g(y)dy, x [, π]. We observe the following:

24 M.T. NAIR () f g L (T ) nd f g f g : f(x y) g(y) dydx = = [ ] f(x y) dx g(y) dy f g(y) dy = () 2 f g. (2) f g = g f: f(x y)g(y)dy = = = x+π x f(τ)g(x τ)dy f(τ)g(x τ)dy f(τ)g(x τ)dy. (3) f g(n) = ĝ(n ˆf(n) for ll n Z: f g(n) = (f g)(x)e inx dx, (f g)(x)e inx = = f(x y)g(y)e inx dy f(x y)g(y)e in(x y) e iny dy, (f g)(x)e inx dx = = = [ [ = () 2 ˆf(n)ĝ(n). ] f(x y)g(y)e in(x y) e iny dy dx ] f(x y)e in(x y) dx g(y)e iny dy ˆf(n)g(y)e iny dy

(4) (f g) h = f (g h): (f g)(x y)h(y)dy = TOPICS IN FOURIER ANALYSIS 25 = = = [ [ f(x τ) ] f(x y t)g(t)dt h(y)dy ] f(x τ)g(τ y)dτ h(y)dy [ ] g(τ y)h(y)dy dτ f(x τ)(g h)(τ)dτ = () 2 [f (g h)](x). Theorem 8.2. With respect to convolution s multipliction, L (T ) is Bnch lgebr. However, The Bnch lgebr L (T ) does not hve multiplictive identity: Suppose there exists ϕ L ( T ) such tht f ϕ = f for ll f L (T ). Then ˆf(n) ˆϕ(n) = ˆf(n) for ll f L (T ). Hence, ˆϕ(n) = whenever ˆϕ(n) 0. But, ˆϕ(n) 0 s n. Hence, there exists N N such tht ˆϕ(n) = 0 for ll n N. Let f L (T ) be such tht ˆf(n) 0 for some n N. Then for such n, we obtin which is contrdiction. 0 = ˆf(n) ˆϕ(n) = ˆf(n) 0, There exists (ϕ n ) in L (T ) such tht f ϕ n f 0. In fct, we hve the following. Theorem 8.3. Let K n be the Fejér kernel. Then, for every f L (T ), f K n f 0 s N. Proof. Recll tht if g C(T ), then g ϕ n g 0. Let f L (T ) nd ε > 0 be given. Let g C(T ) be such tht f g < ε, nd let N N be such tht

26 M.T. NAIR g ϕ n g < ε for ll n N. Then, for n N, we hve f K n f f K n g K n + g K n g + g f 3ε. (f g) K n + ε + ε (f g) K n + 2ε f g K n + 2ε The lst inequlity is due the fct tht K n(t) dt = π K n(t)dt =. The norm on L 2 (T ) is given by Observe: f 2 = 9. L 2 -Theory ( π /2 f(x) dx) 2. () If u n (x) := e inx, n Z, then the set {u n : n Z} is n orthonorml set nd spn{u n : n Z}, the spce of ll trigonometric polynomils, is dense in L 2 (T ). (2) Let f L 2 (T ) nd S N (f) := S N (f, ). Then () S N (f) = f, u n u n. n= N (b) S N (f) 2 2 = n= N ˆf(n) 2. (c) f S N (f) 2 2 = f 2 2 S N (f) 2 2 = f 2 2 (d) (Bessel s inequlity): n= N n= N ˆf(n) 2. ˆf(n) 2 f 2 2 N N. In prticulr, ˆf(n) 0 s n. (e) f S N (f), u n = 0 n N}. (f) f S N (f) 2 f g g spn{u n : n Z, n N}. Only (2)(f) requires some explntion. Note tht for every g spn{u n : n Z, n N}, f g 2 2 = f S N (f) 2 2 + S N (f) g 2 2, becuse, in view of (2)(e), f S N (f), S N (f) g = 0.

TOPICS IN FOURIER ANALYSIS 27 The result in (2)(d) gives nother proof for the Riemnn Lebesgue lemm, becuse L 2 (T ) is dense in L (T ). In view of (2)(f), f S N (f) 2 = inf{ f g 2 : g X N }, where X N := spn{u n : n Z, n N}. In other words, S N (f) is the (unique!) best pproximtion of f from X N. Uniqueness is due to the following: Suppose ϕ be in X N such tht Then, f ϕ 2 = inf{ f g 2 : g X N }. f ϕ 2 2 = f S N (f) 2 2 + S N (f) ϕ 2 2 since f S N (f), S N (f) ϕ = 0 so tht we obtin S N (f) ϕ 2 = 0. Theorem 9.. Let f L 2 (T ). Then we hve the following: () spn{u n : n Z} is n orthonorml bsis of L 2 (T ), i.e., mximl orthonorml set in L 2 (T ). (2) (Fourier expnsion) f = ˆf(n)u n in L 2 (T ). n Z (3) (Prsevl s formul) f 2 2 = n Z ˆf(n) 2. Proof. () It cn be seen tht f, u n = 0 for ll n Z implies f = 0 in L 2 (T ). Hence, spn{u n : n Z} is mximl orthonorml set in L 2 (T ). (3) We observe tht, for n > m, S n (f) S m (f) 2 n k m ˆf(n) 2. Hence, {S n (f)} is cuchy sequence in L 2 (T ). Therefore, it converges to some g L 2 (T ). It cn be seen tht ĝ(n) = ˆf(n) for ll n Z. Therefore, g = f in L 2 (T ). (3) Follows from (2). Now, we give nother proof for the following theorem: Theorem 9.2. If f C (T ), then the Fourier series of f converges bsolutely, nd uniformly to f. Further, ( ) f S N (f, ) = O. N

28 M.T. NAIR Proof. Hence, ˆf(n) = n 0 n 0 f C (T ) = ˆf (n) = in ˆf(n). n in ˆf(n) = ( n ˆf (n) n 0 n 0 ) /2 n ˆf 2 2 = π 3 ˆf 2. Hence the Fourier series of f converges bsolutely, nd uniformly to continuous function, sy g C(T ). Since ĝ(n) = ˆf(n) for ll n Z, we obtin g = f. We lso observe tht, for ll x R, f(x) S N (f, x) n >N This completes the proof. ˆf(n) = n 0 n ˆf (n) References n 2 n >N /2 ˆf 2 ˆf 2 N. [] R. Bhti, Fourier Series, TRIM, Hindustn Book Agency, 993 (Second Edition: 2003). [2] S. Kesvn, Lectures on Fourier Series, Notes: Third Annul Foundtion School, December, 2006. [3] M.T. Nir, Mesure nd Integrtion, Notes for the MSc. Course, Jn-my, 204. [4] R. Rdh & S. Thngvelu, Fourier Series, Web-Course, NPTEL, IIT Mdrs, 203. [5] B. O. Turesson, Fourier Anlysis, Distribution Thoery, nd Wvelets, Lecture Notes, Mrch, 202. Deprtment of Mthemtics, I.I.T. Mdrs, Chenni-600 036, INDIA E-mil ddress: mtnir@iitm.c.in