be the set of complex valued 2π-periodic functions f on R such that

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1 . Fourier series. Definition.. Given a real number P, we say a complex valued function f on R is P -periodic if f(x + P ) f(x) for all x R. We let be the set of complex valued -periodic functions f on R such that P I f Leb whenever I is a bounded interval. (Replace Leb by Riem if Leb makes you nervous. A great deal of what follows will still go through.) It follows from our previous work that P is a vector space over C with respect to pointwise addition and scalar multiplication. Here is a Corollary of Hölder s Inequality. Theorem.. Suppose p < q. Then In particular, f p () /p /q f q whenever f P. P q P p. Proof. If q the inequality holds trivially (Why?) so suppose q <. Let p q/p and q p/( p ) so p and q are conjugate. From the Hölder s Inequality we infer that f p p f p R f p p R q f p q() p/q. Of particular interest is the case p ; we will frequently write for f. f Proposition.. For any f P and any a R we have π+a +a f(x) dx Exercise.. Prove this Proposition. Lebesgue integrable on R n then I(τ a f) I(f) π f(x) dx. To start, show that if f is Riemann or for a R n where I is either R or L, respectively. Here τ a (x) x + a for x R n and τ a f f (τ a ) f τ a. Definition.. Suppose f, g P. We say f and g are complementary if fg P in which case we set (f, g) π f(x)g(x) dx. If f P p and g P q for some p, q such that p, q and p and q are conjugate then f and g are complementary by virtue of Hölders Inequality.

2 In case p q one easily verifies from the linearity of the integral that (, ) is a pseudo-hermitian inner product on P. The pseudo is necessary because f (f, f) 0 only implies that {x R : f(x) 0} is a set of measure zero and not that f 0 which means, by definition that f(x) 0 for all x R. (Note that if 0 < η < then η L ({x (, π] : f(x) η}) π Definition.3. For each n Z we let E n (x) e inx for x R; evidently, E n P. f(x) dx f.).. Discussion and more definitions. Suppose A C {0}. Then d e Ax dx A e Ax, x R. So if < a < b < we may use the Fundamental Theorem of Calculus to obtain Thus b a e Ax dx eax A (E m, E n ) b a eab e Aa. A { if m n, 0 else. That is, the set {E n : n Z} is orthonormal with respect to (, ). For each N N we let T N be the linear subspace of P spanned by {E n : n N} and we call the members of T N trigonometric polynomials of degree N. For each f P we define ˆf : Z C, the Fourier transform of f, by letting ˆf(n) (f, E n ) for n Z. One of our goals is to reconstruct f from its Fourier transform. As a first step in this direction, for each nonnegative integer N and each f P we set S N f (f, E n )E n ˆf(n)E n. In particular, if f P then S N f is the orthogonal projection with respect to (, ) of f onto T N. Theorem.. (Bessel s Inequality.) For any f P we have ˆf(n) f. Remark.. Plancherel s Theorem, which comes later, will give the opposite inequality.

3 3 Proof. This follows from our work on orthogonal projections. Here is a recap of what we did there. Let N be a positive integer N. Keeping in mind the orthogonality of the E n s we obtain N N 0 (f ˆf(n)E n, f ˆf(n)E n ) n N n N (f, f) ( ˆf(n)E n, f) (f, ˆf(n)E n ) + ( (f, f) f ˆf(n)(E n, f) ˆf(n). ˆf(n)(f, E n ) + ˆf(n)E n, ˆf(n) ˆf(n)E n The Fourier transform behaves nicely with respect to translations. h R and f P. Recall that τ h f(x) f(x h) for x R. ˆf(n)E n ) Suppose By the by the translation invariance of integration and the first Proposition in this section we have π π h π τ h f(x) dx f(x) dx f(x) dx < h so that τ h f P. It follows that τ h f P p whenever p and f P p. Proposition.. We have (i) τ h is linear for each h R; (ii) τ h τ h τ h+h whenever h, h R; (iii) (τ h f, τ h g) (f, g) whenever f, g P and h R. Exercise.. Prove this. It s very straightforward. Proposition.3. Suppose f P and h R. Then Exercise.3. Exercise. τ h f(n) e inh ˆf(n). Definition.4. Suppose f, g P. For each x R we set { π f g(x) f(x y)g(y) dy if π f(x y)g(y) dy < 0 else and we call f g the convolution of f and g. Recalling Tonelli s Theorem we note that f g is a complex valued -periodic Lebesgue measurable function on R. Recalling Young s Inequality we note that if p, q, r [, ], p + q r +, f P p and g P q then f g r f p g q.

4 4 By Tonelli s Theorem the first of the above cases holds for almost all x R. Evidently, f g is -periodic. Our previously developed theory of convolutions, with natural modifications, applies here. Proposition.4. Suppose f, g P. Then Exercise.4. Prove this. f g ˆfĝ. Definition.5. For each nonnegative integer N we define the Dirichlet kernel D N by letting D N E n. Proposition.5. Let N be a no-negative integer. Then (i) D N (x) { sin(n+ )x sin x N + (ii) D N is even and π D N (x) dx and (iii) S N f D N f for any f P. Proof. Suppose N and x R {0}. Then D N (x) N n N e inx if x 0, else; e inx (eix ) N+ e ix e i(n+ )x e i(n+ )x e i x e i x sin(n + )x sin x, and it is evident that S N (0) N+ so (i) holds. To prove (ii), note that { π e inx 0 if n 0, dx (E n, E 0 ) else for any integer n. To prove (iii), suppose f P and x R and observe that S N f(x) (f, E n )E n π ( D N f(x). ( ) π f(t)e int dt e inx e in(x t) )f(t) dt

5 5 Lemma.. (The Riemann Lebesgue Lemma.) Suppose f is Lebesgue integrable on R. (This means that f Leb.) Then f(x) sin tx dx 0. lim t Proof. Suppose < a < b <. Then It follows that () b a sin tx dx cos tb cos ta t 0 as t. s(x) sin tx dx 0 as t. for any elementary function s such that {x R : s(x) 0} is bounded.. Let η > 0. Choose an elementary function s such that {x R : s(x) 0} is bounded and f s η; (Remember this is practically the definition of what it means for f to be Lebesgue integrable.) Then f(x) sin tx dx [f(x) s(x)] sin tx dx + s(x) sin tx dx f(x) s(x) dx + s(x) sin tx dx η + s(x) sin tx dx ; for any t R. Now use () to complete the proof. Corollary.. Suppose f P. Then (x δ,x) lim ˆf(n) 0. n Theorem.3. Suppose f P, x R, L +, L C and ( f(t) ) L f(t) L+ () lim dt + dt 0. δ 0 t x t x Then (x,x+δ) lim S Nf(x) L + L +. N Remark.. (Very important) For example, if f is differentiable at x the hypothesis holds with L ± f(x). Proof. Part One. Suppose x 0. Set K [, 0) and K + (0, π). For each positive integer N let g ± N (t) D N( t)(f(t) L ± ) for t K ±.

6 6 Recall that D N is even and π D N ; this implies D N K D N. K +. Thus S N f(0) L + L + π 0 π D N ( t)f(t) dt L D N ( t) dt L + D N ( t) dt 0 (t) dt + g + N (t) dt. K + K g N For any δ (0, π) we set Let η > 0. Since J ± δ K ± ( δ, δ), I ± δ K ± J ± δ. g ± N (t) f(t) L ± t t sin t for any t (, π) and any N we may choose δ (0, π) such that g ± N η. Moreover, I ± δ J ± δ t f(t) L± sin t is Lebesgue integrable so, by the Riemann-Lebesgue Lemma, g ± N 0; lim N the Theorem follows in case x 0. Part Two. From Part One we infer that But S N (τ x f)(0) I ± δ lim S N (τ x f)(0) L + L +. N τ x f(n) ˆf(n)e inx S N f(x). Proposition.6. Suppose m is a positive integer, f P and f is m times continuously differentiable. Then f (m) (n) (in) m ˆf(n).

7 7 Proof. We use integration by parts to obtain π ˆf (n) f (x) e inx dx π e inx d(f(x)) π e inx f(x) xπ x π in f(x) e inx dx in ˆf(n). f(x) d(e inx ) Thus the Proposition holds if m and follows for arbitrary m by induction. Corollary.. Suppose m is a positive integer, f P and f is m times continuously differentiable, N is a nonnegative integer and x R. Then Proof. Note that f(x) S N f(x) f (m). π N m ( ) / ( ) / a n b n a n b n whenever a and b are complex valued functions on Z; that n m dx x m N m ; nn+ and that, by Bessel s Inequality, n >N N f (m) (n) f (m). Let O N and x R. From the Fourier Inversion Formula we have that f(x) S N f(x) lim ˆf(n)E n (x). Thus if O N and N < O we have ˆf(n)E n (x) N< n <O N< n <O O N<n O f (in) m (m) (n)e n (x) / f (m) (n) n m N< n <O N m f (m). n >N /

8 8.. Approximate identities. Let φ be a smooth nonnegative real valued function on R which has integral one and which is supported in [, ]. For each ɛ > 0 let φ ɛ (x) ɛ φ( x ɛ ) for x R. Theorem.4. Suppose f P. Then lim f S N f 0. N Proof. Let η > 0. Choose ɛ > 0 such that if g φ ɛ f then f g < η. Then for any nonnegative integer N we have f S N f f g + S N (f g) + S N g g η + S N g g because S N (f g) f g by Bessel s Inequality. But S N g g 0 as N because, by the previous Theorem, S N g converges uniformly to g as N since g is smooth. Corollary.3. (Parseval s?) Theorem.) If f P then ( ) / f ˆf(n). Proof. This follows from the preceding Corollary and the Pythagorean identity f S N f + S N f f. Corollary.4. n n π 6. Proof. Apply Parseval s Theorem to that member f of P such that f(x) x for x [, π). Corollary.5. Suppose f P and ˆf 0. Then f equals zero almost everywhere. Proof. We have φ ɛ f 0 for any ɛ > 0 so, as φ ɛ f is smooth, φ ɛ f 0 by the previous Theorem. Since f φ ɛ f 0 as ɛ 0 we infer that f 0. Corollary.6. Suppose f P. Then f ˆf(n). Proof. In view of Bessel s Inequality and the foregoing we need to show that if the right hand side is finite so is the left hand side, so suppose the right hand side is finite. Now φ ɛ f φ ɛ ˆf and ˆφɛ so φ ɛ f is bounded independently of ɛ by Plancherel s Theorem. Do you know what to do now? Definition.6. (The Fejér kernel.) For each nonnegative integer N let F N N + N D N. n0

9 9 Proposition.7. Proof. N e inx n0 m n e ix F N (x) N + ( sin( N+ )x sin x N e inx e i(n+)x n0 ) if x 0, else. [ e i(n+)x e ix e ix e ix e i(n+)x over e ix] [ ( e ix )( e i(n+)x ) ( e ix )e ix ( e i(n+)x ) ] e ix ( e ix )( e ix ) ( e i(n+)x ) + ( e i(n+)x ) ( e ix )( e ix ) ( ) N+ i( e )x N+ i( e )x ( ) e i x e x ( sin( N+ )x ). sin x Theorem.5. Suppose f P and f is continuous. uniformly to f as N. Proof. Exercise for the reader. Then F N f converges Theorem.6. The Weierstrass Approximation Theorem. Polynomial functions are uniformly dense in the continuous functions on a compact rectangle. Proof. Exercise for the reader.