CONVERGENCE of FOURIER SERIES in Lp SPACE Contents 1. Fourier Series, Partial Sums, and Dirichlet Kernel 1 2. Convolution 4 3. F

CONVERGENCE OF FOURIER SERIES IN Lp SPACE

JING MIAO

Abstract. The convergence of Fourier series of trigonometric functions is easy to see, but the same question for general functions is not simple to answer. We study the convergence of Fourier series in Lp spaces. This result gives us a criterion that determines whether certain partial diﬀerential equations have solutions or not.We will follow closely the ideas from Schlag and Muscalu’s Classical and Multilinear Harmonic Analysis.

Contents 1. Fourier Series, Partial Sums, and Dirichlet Kernel 1 2. Convolution 4 3. Fej´erkernel and Approximate identity 6 4. Lp convergence of partial sums 9 Appendix A. Proofs of Theorems and Lemma 16 Acknowledgments 18 References 18

1. Fourier Series, Partial Sums, and Dirichlet Kernel Let T = R/Z be the one-dimensional torus (in other words, the circle). We consider various function spaces on the torus T, namely the space of continuous functions C(T), the space of H¨oldercontinuous functions Cα(T) where 0 < α ≤ 1, and the Lebesgue spaces Lp(T) where 1 ≤ p ≤ ∞. Let f be an L1 function. Then the associated Fourier series of f is ∞ X (1.1) f(x) ∼ fˆ(n)e2πinx n=−∞ and Z 1 (1.2) fˆ(n) = f(x)e−2πinx dx. 0 The symbol ∼ in (1.1) is formal, and simply means that the series on the right-hand side is associated with f. One interesting and important question is when f equals the right-hand side in (1.1). Note that if we start from a trigonometric polynomial ∞ X 2πinx (1.3) f(x) = ane , n=−∞

Date: DEADLINES: Draft AUGUST 18 and Final version AUGUST 29, 2013. 1 2 JING MIAO where all but finitely many an are zero, then (1.4) Z 1 ∞ ! ∞ Z 1 ˆ X 2πimx −2πinx X 2πimx −2πinx f(n) = ame e dx = ame e dx. 0 m=−∞ m=−∞ 0 We can interchange summation and integration here because only finitely many an’s are nonzero. Note that the integral of the last term of (1.4) is 0 if n 6= m and ˆ 1 if n = m. Thus, we get f(n) = an for all n ∈ Z. The {an}n∈Z are also called Fourier coefficients. ∞ Therefore, we see that f(n) = P fˆ(n)e2πinx when only finite many Fourier n=−∞ coefficents of f are nonzero. In general, we want to know when the Fourier series of a function will converge to the original function. We will specify the sense of convergence of the infinite series later in this paper. It is natural to start from the most basic notion of convergence, namely that of pointwise convergence. The partial sums of f ∈ L1(T) are defined as N N Z X ˆ 2πinx X −2πiny 2πinx SN f(x) = f(n)e = e f(y) dy e n=−N n=−N T (1.5) Z N Z X 2πin(x−y) = e f(y) dy = DN (x − y)f(y) dy, T n=−N T N P 2πinx where DN (x) = e is called the Dirichlet kernel. If we simplify and bound n=−N the Dirichlet kernel, we get (1.6) N X e2πi(N+1)x − e−2πiNx D (x) = e2πinx = N e2πix − 1 n=−N cos(2π(N + 1)x) − cos(2πNx) + i[sin(2π(N + 1)x) + sin(2πNx)] = cos(2πx) − 1 + i sin(2πx) (cos(2πx) − 1)(cos(2π(N + 1)x) − cos(2πNx)) + sin(2πx)(sin(2π(N + 1)x) + sin(2πNx)) = 2 − 2 cos(2πx) i[(cos(2πx) − 1)(sin(2π(N + 1)x) + sin(2πNx)) − sin(2πx)(cos(2π(N + 1)x) − cos(2πNx))] + 2 − 2 cos(2πx) −2 sin((2N + 1)πx) sin(πx)(−2) sin2(πx) + sin(2πx)2 sin((2N + 1)πx) cos(πx) = 4 sin2(πx) sin((2N + 1)πx) cos2(πx) = sin((2N + 1)πx) sin(πx) + sin(πx) sin((2N + 1)πx) = . sin(πx)

Looking at Figure 1, we see that DN is very large at x = 0 and as N increases, the value at x = 0 will become larger. We have DN→∞(0) = ∞. Also DN is π 1 bounded by 1 away from zero. When πx ≤ , x ≤ . Our previously deﬁned |x| 2 2 domain is T = R/Z, which is the same as periodic function with period 1. Thus, CONVERGENCE OF FOURIER SERIES IN Lp SPACE 3

−5 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5

Figure 1. The Dirichlet kernel DN for N = 9

−1 1 we can translate the domain to [ 2 , 2 ]. Now let’s bound the Dirichlet kernel for each N. Since | sin nα| ≤ n| sin α|, we know that |DN (x)| ≤ 2N + 1 ≤ CN where C π 2|x| is some ﬁnite constant. Also note that when 0 < x ≤ , | sin x| ≥ . Therefore, 2 π | sin((2N + 1)πx)| 1 1 we have |D (x)| = ≤ = . To sum up, we have N | sin(πx)| 2π|x|/π 2|x| 1 (1.7) |D (x)| ≤ C min N, , N |x| for all N ≥ 1 and some absolute constant C. Next, we try to ﬁnd a bound of the 1 L norm of DN . From (1.7) we know that Z 1 Z 1 Z |DN (x)| dx ≤ dx + N dx 1 |x| 1 0 |x|> N |x|≤ N (1.8) 1 1 = 2 log − log + 1 2 N

2. Convolution The convolution of functions arises naturally in the discussion of the Dirichlet kernel. We deﬁne and discuss further properties of this operation here.

Deﬁnition 2.1. If f, g are two functions on T, and both f, g are continuous, then Z Z (2.2) (f ∗ g)(x) := f(x − y)g(y) dy = g(x − y)f(y) dy T T is the convolution of f and g. It is helpful to think of g or f as dirac delta functions which are generalized functions, or distributions, on the real number line that are zero everywhere except at zero, with an integral of one over the entire real line. Here we list and prove some useful lemmas about convolution.

1 Lemma 2.3. (1) Let f, g ∈ L (T). Then kf ∗ gkr ≤ kfkpkgkq for all 1 ≤ r, 1 1 1 p q p, q ≤ ∞, 1 + r = p + q , f ∈ L , g ∈ L . This is called the Young’s inequality. (2) For f, g ∈ L1(T) one has that for all n ∈ Z, f[∗ g(n) = fˆ(n)ˆg(n). Proof. (1) By Fubini’s theorem, since f(x − y)g(y) is jointly measurable on T × T 1 and belongs to L (T × T), we know that kf ∗ gk1 ≤ kfk1kgk1 < ∞. The details of the proof of the Young’s inequality are shown in the appendix.

(2) is a consequence of Fubini’s theorem and the homomorphism property of the exponentials e2πin(x+y) = e2πinxe2πiny: Z Z f[∗ g(n) = f(x − y)g(y) dy e−2πinx dx T T (2.4) Z Z = f(w)e−2πinw dw g(t)e−2πint dt T T = fˆ(n)ˆg(n).

One of the most basic as well as oldest results on the pointwise convergence of Fourier series is the following theorem. Note that it requires a function to be H¨older continuous, instead of just continuous.

α Theorem 2.5. If f ∈ C (T) with 0 < α ≤ 1, then kSN f − fk∞ → 0 as N → ∞. 1 Proof. One has, with δ ∈ (0, 2 ) to be determined, Z 1 SN f(x) − f(x) = (f(x − y) − f(x))DN (y) dy 0 Z (2.6) = (f(x − y) − f(x))DN (y) dy |y|≤δ Z + (f(x − y) − f(x))DN (y) dy. 1 2 >|y|>δ CONVERGENCE OF FOURIER SERIES IN Lp SPACE 5

R 1 2πinx Equation (2.6) is true because 0 DN (y) dy = 1, since the integration of e over [0, 1] is 0 except when n = 0. Also since DN (x) is periodic with period 1, integration 1 1 over [0, 1] is the same as that over [− 2 , 2 ]. We now use the bound from (1.8), i.e., 1 (2.7) |D (y)| ≤ C min N, . N |y| Here and in what follows, C will denote a numerical constant that can change from line to line. The ﬁrst integral in (2.6), which we denote by A, can be estimated as follows Z Z 1 α−1 α (2.8) A ≤ C |f(x) − f(x − y)| dy ≤ C[f]α |y| dy ≤ C[f]αδ , |y|≤δ |y| |y|≤δ with the usual Cα semi-norm: |f(x) − f(x − y)| (2.9) [f]α = sup α . x,y |y| To bound the second integral in (2.6), which we denote by B, we need to use the oscillation of DN (y). We can rewrite the expression as Z B = (f(x − y) − f(x))DN (y) dy 1 2 >|y|>δ Z = hx(y) sin((2N + 1)πy) dy 1 >|y|>δ (2.10) 2 Z 1 = − hx(y) sin (2N + 1) π y + dy 1 2N + 1 2 >|y|>δ Z 1 = − hx y − sin((2N + 1)πy) dy, 1 2N + 1 |y− 2N+1 |>δ f(x−y)−f(x) where hx(y) = sin(πy) . Therefore, with all integrals being understood to be in 1 1 the interval (− 2 , 2 ), we get Z 2B = hx(y) sin((2N + 1)πy) dy |y|>δ Z 1 − hx y − sin((2N + 1)πy) dy 1 2N + 1 |y− 2N+1 |>δ Z 1 (2.11) = hx(y) − hx y − sin((2N + 1)πy) dy |y|>δ 2N + 1 Z 1 − hx y − sin((2N + 1)πy) dy 1 2N + 1 [−δ,−δ+ 2N+1 ] Z 1 + hx y − sin((2N + 1)πy) dy. 1 2N + 1 [δ,δ+ 2N+1 ] We bound ﬁrst the last two integrals. Note that we have

f(x − z) − f(x) 2kfk∞ Ckfk∞ (2.12) |hx(z)| = ≤ ≤ . sin(πz) | sin(πz)| |z| 2|z| The last inequality is true by Jordan’s inequality | sin z| ≥ π when |z| ≤ π 2 2 . Provided that δ > 2N+1 , it follows that |z| > C|δ| for all z in the intervals 6 JING MIAO

1 1 [−δ, −δ + 2N+1 ] and [δ, δ + 2N+1 ]. Therefore, we obtain the following inequality: Ckfk Ckfk |h (z)| ≤ ∞ ≤ ∞ . x z |δ| This inequality allows us to bound each of the last two inequalities in (2.11) by Ckfk∞ δ(2N+1) . Now we bound the remaining integral:

f(x − y) − f(x) f(x − y − τ) − f(x) |hx(y) − hx(y + τ)| = − sin(πy) sin(π(y + τ))

f(x − y) − f(x − y − τ) ≤ sin(πy) 1 1 (2.13) + (f(x − y − τ) − f(x)) − sin(πy) sin(π(y + τ))

(f(x − y) − f(x)) − (f(x − y − τ) − f(x)) ≤ sin(πy) 1 1 + (f(x − y − τ) − f(x)) − . sin(πy) sin(π(y + τ)) If |y| > δ > 2τ. Using similar ideas, we bound the ﬁrst term by [f] |τ|α [f] |τ|α α ≤ α . |y| |δ| Using similar arguments, we bound the second term by Ckfk πτ C0kfk πτ ∞ ≤ ∞ . sin(πy) sin(π(y + τ)) |δ|2 Letting τ < 1/N, we get

N −α[f] N −1kfk (2.14) |B| ≤ C( α + ∞ ). δ δ2 Choosing δ = N −α/2 we can conclude from (2.8) and (2.14) that

−α2/2 −α/2 −1+α (2.15) |(SN f)(x) − f(x)| ≤ C(N + N + N ).

Thus, as N → ∞ we have |(SN f)(x) − f(x)| → 0, which means that kSN f − fk∞ → 0. 3. Fejer´ kernel and Approximate identity The Dirichlet kernel is not ideal in the sense that its L1 norm is not uniformly bounded as shown in (1.10). Why this unboundedness of the Dirichlet kernel is a problem that will be seen later in the paper. Thus, we introduce another form of partial sum, namely the average of the partial sum operator: N−1 1 X (3.1) σ f := S f. N N n n=0 N−1 1 P Setting KN = N Dn, which is called the Fej´erkernel, we therefore have n=0 σN f = KN ∗ f. Here we list some useful properties of the Fej´erkernel. CONVERGENCE OF FOURIER SERIES IN Lp SPACE 7

0 −0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5

Figure 2. The Fej´erkernel KN for N = 10 (blue), and N = 30 (green)

|n|+ Proposition 3.2. (1) Kˆ (n) = 1 − , where f + means the positive part of N N the function f, 1 sin(Nπx)2 (2) K (x) = , N N sin(πx) −1 2 −2 (3) 0 ≤ KN (x) ≤ CN min(N , x ). Proof. (1) By deﬁnition N−1 Z 1 1 X Kˆ (n) = D e−2πinx dx N N m 0 m=0 (3.3) N−1 m Z 1 1 X X = e2πikxe−2πinx dx. N 0 m=0 k=−m Note that R e2πikxe−2πinx dx = 1 if and only if k = n. Property (1) follows from T this fact. Using formula for summation of geometric sequence, we can get (2). By 1 1 Jordan’s inequality for x ∈ [− 2 , 2 ] and the inequality | sin nx| ≤ n| sin x|, we get (3).

We can see from Figure 2 that as N increases, the graph of KN “squeezes up” around x = 0.The above property of KN ensures that KN are what we call an approximate identity. ∞ ∞ Deﬁnition 3.4. We say that the family {ΦN }N=1 ⊂ L (T) form an approximate identity provided R 1 A1) 0 ΦN (x) dx = 1 for all N; R 1 A2) supN |ΦN (x)| dx < ∞; 0 R A3) for all δ > 0 one has |x|>δ |ΦN (x)| dx → 0 as N → 0. 8 JING MIAO

The name “approximate identity” derives from the fact that ΦN ∗ f → f as N → ∞ in many reasonable senses, which we will prove soon. Note that {DN }N≥1 are not an approximate identity since A2) does not hold. Also note that Deﬁnition 3.4 has nothing to do with the torus T. It applies equally well to the line R, tori d d T , or Euclidean spaces R . Next, we verify that the function KN belong to this class.

∞ Lemma 3.5. The Fej´erkernels {KN }N=1 form an approximate identity.

R 1 2πinx R 1 Proof. Since 0 e = 1 if n = 0 and 0 otherwise, we clearly have 0 KN (x) dx = R 1 1. Since KN (x) ≥ 0 by Proposition 3.2, we have 0 |KN (x)| dx = 1 for all N, so A2) holds. By part (3) of Proposition 3.2 we have Z Z −1 −2 0 −1 (3.6) |KN (x)| dx ≤ CN x dx ≤ C (δN) , |x|>δ |x|>δ where C and C0 are constants. Then clearly as N → ∞, the integral goes to 0. All of these prove the result. Now we establish the basic property of an approximate identity.

∞ Proposition 3.7. For any approximate identity {ΦN }N=1 one has (1) If f ∈ C(T), then kΦN ∗ f − fk∞ → 0 as N → ∞. p (2) If f ∈ L (T) where 1 ≤ p < ∞, then kΦN ∗ f − fkp → 0 as N → ∞. Proof. We begin with the uniform convergence. Since T is compact, f is uniformly continuous. Given > 0, let δ > 0 be such that (3.8) sup |f(x − y) − f(x)| < , |y|<δ for all x ∈ T. Then, by the deﬁnition of approximate identity we have (3.9) Z

|(ΦN ∗ f)(x) − f(x)| = (f(x − y) − f(x))ΦN (y) dy T Z Z ≤ |f(x − y) − f(x)| |ΦN (y)| dy + |ΦN (y)|2kfk∞ dy T |y|≥δ < C when N is large enough so that the second term in the second to last line of (3.9) p is less than 2kfk∞. Thus, we proved (1). To prove (2), ﬁx any f ∈ L (T) and let g ∈ C(T) be such that kf − gkp < . Then

(3.10) kΦN ∗ f − fkp ≤ kΦN ∗ (f − g)kp + kf − gkp + kΦN ∗ g − gkp. By Young’s inequality (Theorem A.1 in the appendix) we have

(3.11) kΦN ∗ (f − g)kp ≤ kΦN k1kf − gkp.

Using the fact that supkΦN k1 < ∞ and (1) we get kΦN ∗f −fkp → 0 as N → ∞. N Here we also list some basic and useful theorems about Fourier transform that will be used to prove the Lp convergence of Fourier series later. CONVERGENCE OF FOURIER SERIES IN Lp SPACE 9

Corollary 3.12. (1) For any f ∈ L2(T), we have 2 X ˆ 2 (3.13) kfk2 = |f(n)| . n∈Z This is called Plancherel’s theorem. (2) For f, g ∈ L2(T) we have Parseval’s identity Z X (3.14) f(x)g(x) dx = fˆ(n)gˆ(n). T n∈Z Proof. For (1) we have Z 1 Z 1 2 2 X ˆ 2πinx 2 kfk2 = |f| dx = |f(n)e | dx 0 0 n∈ (3.15) Z Z 1 X X = |fˆ(n)|2 dx = |fˆ(n)|2. 0 n∈Z n∈Z 2πinx The second equality is due to the orthogonality of {e }n∈Z. The last equality is true because fˆ(n) is independent of x for all n. For (2) the proof is similar: ! Z Z X f(x)g(x) dx = fˆ(n)e2πinx (gˆ(m)e2πimx) dx T T n∈Z ! Z X (3.16) = fˆ(n)e2πinx (gˆ(m)e−2πimx) dx T n∈Z X = fˆ(n)gˆ(n). n∈Z The last equality results from the orthogonality of the basis.

4. Lp convergence of partial sums p We now turn our attention to the convergence of the partial sums SN f in L (T) or C(T) (substituted for L∞(T)). Proposition 4.1. The following statements are equivalent for any 1 ≤ p ≤ ∞:

(a) For every f ∈ Lp(T) (or f ∈ C(T) if p = ∞) we have

(4.2) kSN f − fkp → 0 as N → ∞

(b) We have supkSN kp→p < ∞ where kSN kp→p = sup kSN fkp. N kfkp=1

p Proof. “⇒” AssumeSN f converges to f in L . Then we have kSN fkp → kfkp p by the triangle inequality. Since f ∈ L (T), we know that kfkp < ∞. Thus, supN kSN fkp < ∞. By the Uniform Boundedness Principle (see the appendix), we have supN kSN kp→p < ∞. 10 JING MIAO

“⇐” Let KN (x) be the Fej´erkernel. We have

kSN f − fkp = kSN (f − KN ∗ f) + (SN (KN ∗ f) − f)kp

= kSN (f − KN ∗ f) + (KN ∗ f − f)kp

≤ kSN (f − KN ∗ f)kp + kKN ∗ f − fkp (4.3) ≤ kSN kp→pkf − KN ∗ fkp + kKN ∗ f − fkp

= (kSN kp→p + 1)kf − KN ∗ fkp ≤ (M + 1), where M = supN kSN kp→p < ∞ by assumption and can be made arbitrarily small as N → ∞ since the Fej´erkernel forms an approximate identity. This ﬁnishes the proof. With the above proposition, we can now settle the question of the convergence of Fourier series in Lp space for p = 1 and p = ∞.

Corollary 4.4. Fourier series do not converge in C(T) and L1(T), i.e., there exists 1 f ∈ C(T) so that kSN f − fk∞ 9 0 and g ∈ L (T) so that kSN g − gk1 9 0 as N → ∞. Proof. By Proposition 4.1 it suﬃces to verify the limits

supkSN k∞→∞ = ∞ N

supkSN k1→1 = ∞. N

Recall that kDN k1 → ∞ as N → ∞ by (1.9). Notice that

kSN k∞→∞ = sup kDN ∗ fk∞ kfk∞=1

≥ sup |(DN ∗ f)(0)| kfk∞=1 (4.5) Z ≥ sup DN (−y)f(y) dy kfk∞=1 T Z = |DN (y)| dy = kDN k1. T The second to last inequality is true because kfk∞= 1, so DN (−y)f(y) ≤ 1 |DN (−y)| for all y. We can then let f be a continuous approximation (in L ) of signDN (y). Thus, Fourier series do not converge in C(T). R Recall that we proved that for the Fej´erkernel, we have |KM (x)| dx = 1 for T all M. Similar to (4.5), we have

(4.6) kSN k1→1 ≥ kDN ∗ KM k1 → kDN k1 1 as M → ∞. Thus, Fourier series do not converge in L (T). Now our goal is to prove that for any p such that 1 < p < ∞ we have that the Fourier series do converge in Lp(T). The case p = 2 is relatively easy. Recall the Plancherel theorem we proved earlier which states 2 X ˆ 2 (4.7) kfk2 = |f(n)| . n∈Z CONVERGENCE OF FOURIER SERIES IN Lp SPACE 11

Thus, by the Plancherel theorem, we have  1/2

X ˆ 2πinx X ˆ 2 (4.8) kSN f − fk2 = f(n)e =  |f(n)|  ,

|n|>N |n|>N 2 for all f ∈ L2(T) which goes to 0 as N → ∞. For 1 < p < ∞ and p 6= 2, we prove it in the following way which is a bit tricky. We start by defining the conjugate function. Definition 4.9. For f a trigonometric function, define the conjugate function f˜ by X (4.10) f˜(x) = −i sgn(m)fˆ(m)e2πimx, m∈Z Also define the Riesz projections P+ and P− by ∞ X ˆ 2πimx P+(f)(x) = f(m)e m=1 (4.11) −1 X ˆ 2πimx P−(f)(x) = f(m)e m=−∞ ˆ ˜ Thus, we have that f = P+(f) + P−(f) + f(0), where f = −iP+(f) + iP−(f), when f is a trigonometric function. The following is a consequence of Proposition 4.1.

Proposition 4.12. Let 1 < p < ∞. Then the expression SN (f) = DN ∗f converges p to f in L (T) as N → ∞ if and only if there exists a constant Cp > 0 such that ˜ p p kfkL (T) ≤ CpkfkL (T) for all trigonometric polynomials. Proof. By the previous deﬁnition, it is easy to show that

1 1 (4.13) P (f) = f + if˜ − fˆ(0). + 2 2 Therefore, the Lp boundedness of the operation f 7→ f˜ is equivalent to that of the operator f 7→ P+(f). Next, we need to show that 2N N X X (4.14) e−2πiNx (f(x\)e2πiNx)(m)e2πimx = fˆ(m)e2πimx. m=0 m=−N To prove this, note that Z Z f(x\)e2πiNx (m) = f(x)e2πiNxe−2πimx dx = f(x)e−2πi(m−N)x dx (4.15) T T = fˆ(m − N)

2N N Thus, the left hand side of (4.14) is equal to P fˆ(m−N)e2πi(m−N)x = P fˆ(m)e2πimx, m=0 m=−N which is the right hand side of (4.14). 12 JING MIAO

Since multiplication by exponentials does not aﬀect Lp norms, (4.15) implies p p that the norm operator SN (f) = DN ∗ f from L to L is equal to that of the operator

2N 0 X 2πimx (4.16) SN (g)(x) = gˆ(m)e m=0 from Lp to Lp. Therefore, we have the following equivalence:

0 (4.17) supkSN kLp→Lp < ∞ ⇔ supkSN kLp→Lp < ∞. N≥0 N≥0 p “⇒ ”. Suppose now that for all f ∈ L (T), we have SN (f) → f as N → ∞. 0 Proposition 4.1 then yields supkSN kLp→Lp < ∞ and thus supkSN kLp→Lp < ∞ by ˆ p (4.17). Let A(f) = P+(f) + f(0). We know that A(f) is bounded on L (T). Hence ˜ so is P+. Thus, f is also bounded.

p “⇐”. Conversely, suppose that P+ extends to a bounded operator from L (T) to itself. For every trigonometric function f we then have ∞ ∞ 0 X ˆ 2πimx X ˆ 2πimx SN (f)(x) = f(m)e − f(m)e m=0 m=2N+1 ∞ ∞ (4.18) X X = fˆ(m)e2πimx − e2πi(2N)x fˆ(m + 2N)e2πimx m=0 m=1 2πi(2N)x −2πi(2N)x ˆ = P+(f)(x) − e P+(e f) + f(0). (4.18) implies that 0 (4.19) supkSN (f)kLp ≤ (2kP+kLp→Lp + 1)kfkLp N≥0 for all trigonometric polynomials, and, by density, for all f ∈ Lp(T). Thus, estimate p (4.19) also holds for SN . By Proposition 4.1, we have that SN (f) → f in L for all p f ∈ L (T). We now have that convergence of Fourier series in Lp is equivalent to the Lp boundedness of the conjugate function. We now establish that these operators are bounded on Lp.

Theorem 4.20. Given 1 < p < ∞, there is a constant Ap > 0 such that for all f in L∞(T) we have ˜ (4.21) kfkLp ≤ ApkfkLp . Consequently, the Fourier series of an Lp function converge to the function in Lp for 1 < p < ∞. Proof. There is a short proof for this due to Bochner. Let f(t) be a trigonometric polynomial on T with coeﬃcients cj. We write N X 2πijt (4.22) f(t) = cje j=−N CONVERGENCE OF FOURIER SERIES IN Lp SPACE 13

Assume f(t) is complex-valued. Then f(t) + f(t) Re(f) = 2 N N P 2πijt P 2πijt cje + cje j=−N j=−N = 2 (4.23) N N P 2πijt P −2πijt cje + cje j=−N j=−N = 2 N X cj + c−j = e2πijt. 2 j=−N Similarly, the imaginary part can be expressed as f(t) − f(t) Im(f) = 2i N N P 2πijt P 2πijt cje − cje j=−N j=−N = 2i (4.24) N N P 2πijt P −2πijt cje − cje j=−N j=−N = 2i N X cj − c−j = e2πijt. 2i j=−N Thus, we have     N N ^ N ^ X X cj + c−j X cj − c−j (4.25) f˜(t) = c e2πijt =  e2πijt + i  e2πijt . j  2   2i  j=−N j=−N j=−N

Note that the expressions inside the square brackets are real-valued trigonometric ˆ ˆ polynomials. Since kfkP ≤ kRefkp + kImfkp ≤ 2kfkp, and kf(0)kp = |f(0)| ≤ ˆ kfkp, we can assume that f is real-valued and f(0) = 0. Since f is real-valued, we have that fˆ(−m) = fˆ(m) for all m, and since fˆ(0) = 0, we have (4.26) X X X fˆ(t) = −i fˆ(m)e2πimt + i fˆ(−m)e−2πimt = 2 Re [−i fˆ(m)e2πimt], m>0 m>0 m>0 which implies that f˜ is also real-valued. Note that X X f(t) + if˜(t) = fˆ(n)e2πint + (−i2) sgn (n)fˆ(n)e2πinx n∈Z n∈Z X ˆ 2πint X ˆ 2πinx X ˆ 2πinx (4.27) = f(n)e + f(n)e − f(n)e n∈Z n>0 n<0 X = fˆ(n)e2πint. n>0 14 JING MIAO

Thus, f + if˜ only has non-zero Fourier coeﬃcients for n > 0, which means the polynomial f + if˜ contains only positive frequencies. Then we claim: Z (4.28) (f(t) + if˜(t))2k dt = 0. T To prove this, note that since f +if˜is a sum of trigonometric functions, (f +if˜)2k can also be a sum of trigonometric functions. Thus, we have X (4.29) (f(t) + if˜(t))2k = (f\+ if˜)2k(n)e2πint. n≥0 The summation is over non-negative integers because f + if˜ only has positive frequencies. Thus, we have   Z Z 2k X 2πint 2k (4.30) (f(t) + if˜(t)) dt =  (f \+ if˜)2k(n)e  dt = (f\+ if˜) (0). T T n≥0

By our assumption we know that f\+ if˜(0) = 0. Since f + if˜ only has non- 2k negative frequencies, we have f\+ if˜ (0) = ((f\+ if˜)(0))2k = 0. Therefore, we know that (4.28) is true. Expanding the 2k power and taking real parts, we obtain k Z X 2k 2k−2j 2j (4.31) (−1)k−j f˜(t) f(t) dt = 0, 2j j=0 T where we used that f is real-valued. Separating the j = 0 term from the rest in (4.31), we get Z k Z 2k X 2k 2k−2j 2j (4.32) (−1)k f˜(t) dt = − (−1)k−j f˜(t) f(t) dt. 2j T j=1 T Then, we have Z k Z 2k X 2k 2k−2j 2j (4.33) f˜(t) dt ≤ f˜(t) f(t) dt. 2j T j=1 T ˜ 2k Notice that the left hand side of (4.33) is just kfkL2k . Now apply H¨older’s 2k 2k inequality to the right hand side of (4.33) with exponents 2k−2j and 2j . (The statement and proof of the H¨older’sinequality can be found in the Appendix.) We have (4.34) k Z k X 2k 2k−2j 2j X 2k 2k−2j 2j f˜(t) f(t) dt ≤ kf˜(t) k 2k kf(t) k 2k 2j 2j 2k−2j 2j j=1 T j=1 k Z Z X 2k 2k 2k−2j 2k 2j = ( f˜(t) dt) 2k ( f(t) dt) 2k 2j j=1 T T k X 2k 2k−2j 2j = kf˜k kfk . 2j L2k L2k j=1 CONVERGENCE OF FOURIER SERIES IN Lp SPACE 15

Thus, by (4.34) we have

k 2k X 2k 2k−2j 2j (4.35) kf˜k ≤ kf˜k kfk . L2k 2j L2k L2k j=1

2k Dividing (4.35) by kfkL2k we get

2k X (4.36) R2k ≤ R2k−2j, j=1

˜ where R = kfkL2k /kfkL2k . It is easy to see from (4.36) that there exists a positive C2k such that R ≤ C2k. Therefore, we conclude that

˜ (4.37) kfkLp ≤ CpkfkLp when p = 2k.

Now we want to generalize (4.37) to functions where fˆ(0) 6= 0. Apply (4.37) to f − fˆ(0), and we know that the conjugate function of a constant is zero (since ˆ constant functions only have Fourier terms for n = 0). Since |f(0)| ≤ kfk1 ≤ ˜ kfkp (the second inequality is true by H¨older’sinequality), we have that kfkp = ˆ^ ˆ ^ˆ ˆ ˆ k(f − f(0)) + f(0)kp ≤ kf − f(0)kp +kfg(0)kp ≤ Cpkfkp +kf(0)kp ≤ (Cp +1)kfkp. Thus, (4.37) is true for p = 2k and f real-valued trigonometric polynomial. Since a general trigonometric polynomial can be written as P + iQ where P and Q are ˜ real-valued trigonometric polynomials, we have kfkLp ≤ 2(Cp + 1)kfkp. Since trigonometric polynomials are dense in Lp, (4.37) holds for all smooth functions when p = 2k. Riesz-Thorin Interpolation implies that if the conjugate operator is bounded on Lp and Lq, where p < q, then it is bounded on Lr where p < r < q. Since every real number lies in an interval of the form [2k, 2k + 2], for some k ∈ Z+, we have that (4.37) is true for all p such that 2 ≤ p < ∞ when f is a trigonometric function. By density the same result is valid for all Lp functions when p ≥ 2. Finally, we need to prove that the result is valid for 1 < p < 2. The Lp norm of a polynomial can also be computed in another way:

Z 1 1 (4.38) kfkp = sup f(t)g(t) dt : kg(t)kq = 1, + = 1 . q p q g(t)∈L T 16 JING MIAO

Let p ∈ (1, 2), so that q > 2. Then by Corollary 3.12, we have ∞ Z X f˜(t)g(t) dt = f˜d(n)gˆ(n) T n=−∞ ∞ X = fˆ(n)(−i sgnn)gˆ(n) n=−∞ ∞ X (4.39) = fˆ(n)i sgn (n)ˆg(n) n=−∞ ∞ X = fˆ(n)−\˜g(n) n=−∞ Z = f(t)−˜g(t) dt. T By (4.38) we have Z ˜ ˜ 1 1 (4.40) kf(t)kp = sup f(t)g(t) dt : kg(t)kq = 1, + = 1 . q p q g(t)∈L T Thus, by (4.39) we have

Z ˜ 1 1 kfkLp = sup |f(t)g(t) dt : kg(t)kq = 1, + = 1 q b p q g(t)∈L T (4.41) ≤ sup kfkLp kg˜kLq ≤ sup kfkLp CpkgkLq g∈Lq g∈Lq

≤ kfkLp Cp. The second inequality in (4.41) is true by H¨older’sinequality. The third inequality is true because q > 2 and we proved the boundedness of the conjugate operator in Lq spaces for q > 2. The last inequality is true because we chose the g such that p kgkLq = 1. Therefore, we proved the boundedness of the conjugate operator on L for all 1 < p < ∞. By Proposition 4.12 and Theorem 4.20 we conclude that for all p such that 1 < p < ∞, the partial sum of the Fourier series of any Lp function converges to the original function.

Appendix A. Proofs of Theorems and Lemma Theorem A.1. (H¨older’sinequality) Let (S, Σ, µ) be a measure space and let p, q ∈ [1, ∞] with 1/p + 1/q = 1. Then, for all measurable real- or complex- value functions f and g on S, we have kfgk1 ≤ kfkpkgkq.

Proof. If kfkp = 0, then f is zero µ-almost everywhere, and the product fg is zero µ- almost everywhere. Hence the left hand side of H¨older’sinequality is zero. The same is true if kfkq = 0. Therefore, we may assume kfkp > 0 and kgkq > 0 in the following.

If kfkp = ∞ or kgkq = ∞, then the right hand side of H¨older’sinequality is inﬁnite. Therefore, we may assume that kfkp and kgkq are in (0, ∞). CONVERGENCE OF FOURIER SERIES IN Lp SPACE 17

If p = ∞ and q = 1, then |fg| ≤ kfk∞|g| almost everywhere and H¨older’s inequality follows from the monotonicity of the Lebesgue integral. Similarly for p = 1 and q = ∞. Therefore, we may also assume p, q ∈ (1, ∞).

Dividing f and g by kfkp and kgkq respectively, we can assume that kfkp = kgkq = 1. We now use Young’s inequality which states that ap bq (A.2) ab ≤ + p q for all nonnegative a and b, where equality is achieved if and only if ap = bq. Hence |f(s)|p |g(s)|q (A.3) |f(s)g(s)| ≤ + p q for all s ∈ S. Integrating both sides gives 1 1 (A.4) kfgk ≤ + = 1, 1 p q which proves the theorem. 1 1 1 p q Theorem A.5. If 1 ≤ p, q, r ≤ ∞, p + q = 1 + r , f ∈ L (T) and g ∈ L (T), then

(A.6) kf ∗ gkr ≤ kfkpkgkq. 1 1 Proof. if r = ∞, then p + q = 1. Hence by H¨older’sinequality and the translation invariance of Lebesgue measure, we have Z (A.7) |f(x − y)g(y)| dy ≤ kfkpkgkq.

Thus, f ∗ g(x) exists for each x and kf ∗ gk∞ ≤ kfkpkgkq. Next suppose 1 ≤ r < ∞. Note that p ≤ r and q ≤ r. Let s = p(1−1/q) = 1−p/r and note 0 ≤ s < 1. Let t = r/q and note that 1 ≤ t < ∞. Deﬁne q0 by 1/q + 1/q0 = 1 and note 1 < q0 ≤ ∞. Now let Z Z (A.8) h(x) = |f(x − y)g(y)| dy = |f(x − y)|1−s|g(y)||f(x − y)|s dy.

By H¨older’sinequality we have Z (1−s)q q 1/q s (A.9) h(x) ≤ ( |f(x − y)| |g(y)| dy) k|f| kq0 .

If s = 0 then q = 1. If s 6= 0 then sq0 = p. In either case taking the qth powers we obtain Z q (1−s)q q sq (A.10) h(x) ≤ |f(x − y)| |g(y)| dykfkq0 .

Thus, by Minkowski’s inequality we have t q khkqt = kh kt Z Z sq (1−s)q q t 1/t ≤ kfkp ( ( |f(x − y)| |g(y)| dy) dx) (A.11) Z Z sq (1−s)qt qt 1/t ≤ kfkp ( |f(x − y)| |g(y)| dx) dy

sq q (1−s)q = kfkp kgkq kfk(1−s)qt . 18 JING MIAO

But qt = r and (1 − s)r = p. Taking the qth roots we obtain the inequality in the conclusion of the theorem. Theorem A.12. (Uniform Boundedness Principle). Let X be a Banach space and Y be a normed vector space. Suppose that F is a collection of continuous linear operator from X to Y . If for all x in X we have

(A.13) supkT (x)kY < ∞, T ∈F then

(A.14) supkT kB(X,Y ) < ∞. T ∈F Proof. Suppose that for every x in the Banach space X, we have:

(A.15) supkT (x)kY < ∞. T ∈F For every integer n ∈ N, let

(A.16) Xn = {x ∈ X : supkT (x)kY ≤ n}. T ∈F

The set Xn is a closed set and by the assumption,

(A.17) ∪n∈NXn = X 6= ∅. By the Baire category theorem for a non-empty complete metric space X, there exist m such that Xm has non-empty interior, i.e., there exists x0 ∈ Xm and > 0 such that

(A.18) B(x0) := x ∈ X : kx − x0k ≤ ⊂ Xm. Let u ∈ X with kuk ≤ 1 and T ∈ F . we have: −1 kT (u)kY = kT (x0 + u) − T (x0)k −1 (A.19) ≤ (kT (x0 + u)kY + kT (x0)kY ) ≤ −1(m + m). Taking the supremum over u in the unit ball of X, it follows that

−1 (A.20) supkT kB(X,Y ) ≤ 2 m < ∞. T ∈F Acknowledgments. It is a pleasure to thank my mentors, Casey Rodriguez and Daniel Campos Salas, for leading me through the reading, answering all my ques- tions, and giving advice on my talk. I would also like to thank Professor Peter May for giving me the opportunity for a talk and reading my paper.

References [1] W. Schlag, C. Muscalu. Classical and Multilinear Harmonic Analysis. Cambridge University Press. 2013. [2] L. Grafakos. Classical Fourier Analysis. Springer Science + Business Media. 2008.