Solutions to some exercises and problems

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Solutions to some exercises and problems from Stein and Shakarchi's. Fourier Analysis. The book by Y. Ketznelson, ”An introduction of Har- monic Analysis” ...

Solutions to some exercises and problems Teck-Cheong Lim Department of Mathematical Sciences George Mason University 4400, University Drive Fairfax, VA 22030 U.S.A. e-mail address: [email protected] Abstract Solutions to some exercises and problems from Stein and Shakarchi’s Fourier Analysis. The book by Y. Ketznelson, ”An introduction of Harmonic Analysis” (2nd corrected edition) is referred to frequently. Chapter 1: The Genesis of Fourier Analysis Chapter 2: Basic Properties of Fourier Series Chapter 3: Convergence of Fourier Series Chapter 4: Some applications of Fourier Series Chapter 5: The Fourier transform on R Chapter 6: The Fourier transform on Rd Chapter 7: Finite Fourier Analysis Chapter 8: Dirichlet’s Theorem

Chapter 1 The Genesis of Fourier Analysis 1. (Exercise 8) Suppose F is a function on (a, b) with two continuous derivatives. Show that whenever x and x + h belong to (a, b), one may write h2 00 F (x) + h2 φ(h), 2

F (x + h) = F (x) + hF 0 (x) + where φ(h) → 0 as h → 0. Deduce that

F (x + h) + F (x − h) − 2F (x) → F 00 (x) as h → 0. h2 Proof. Firstly one has Z

x+h

F (x + h) − F (x) = x

1

F 0 (y) dy.

Then F 0 (y) = F 0 (x)+(y−x)F 00 (x)+(y−x)ψ(y−x), where ψ is continuous and ψ(h) → 0 as h → 0. Therefore F (x + h) − F (x) = F 0 (x)h + F 00 (x) By mean-value theorem Z h Z uψ(u) du = ψ(ξ) 0

h2 + 2

h

u du = h2

0

Z

h

uψ(u) du. 0

ψ(ξ) = h2 φ(h) 2

where ξ is between 0 and h, and φ(h) → 0 as h → 0 by the continuity of ψ. From above, one easily gets F (x + h) + F (x − h) − 2F (x) − F 00 (x) = φ(h) + φ(−h) → 0 as h → 0. h2 (x) Remark. Suppose F is periodic. The function ζ(x, h) = F (x+h)+F (x−h)−2F − h2 00 F (x) is continuous for all x, h. And there exists M such that

|F (x + h) + F (x − h) − 2F (x)| ≤ M h2 for all x, h (since F, F 00 , ζ are all periodic in x). 2. (Exercise 10) Show that the expression of the Laplacian 4=

∂2 ∂2 + ∂x2 ∂y 2

is given in polar coordinates by the formula 4=

∂2 1 ∂ 1 ∂2 + + 2 2. 2 ∂r r ∂r r ∂θ

Also prove that 2 2 2 2 ∂u + ∂u = ∂u + 1 ∂u . ∂x ∂y ∂r 2 r ∂θ Solution. From x = r cos θ, y = r sin θ, we get  ∂x ∂x   cos θ ∂r ∂θ = ∂y ∂y sin θ ∂r ∂θ

 −r sin θ , r cos θ

hence ∂r ∂x ∂θ ∂x

∂r ∂y ∂θ ∂y

!

 =

cos θ sin θ

−r sin θ r cos θ 2

−1

 =

cos θ − 1r sin θ

 sin θ . 1 r cos θ

By Chain Rule, ∂u ∂u ∂r ∂u ∂θ = + ∂x ∂r ∂x ∂θ ∂x ∂u 1 ∂u = cos θ − sin θ. ∂r r ∂θ By Chain Rule and Product Rule,   ∂ ∂u cos θ ∂x ∂r     ∂ ∂u ∂r ∂ ∂u ∂θ = cos θ + cos θ ∂r ∂r ∂x ∂θ ∂r ∂x ∂2u 1 ∂u 1 ∂2u 2 2 = cos θ + sin θ − cos θ sin θ ∂r2 r ∂r r ∂θ∂r Similarly,   1 ∂u ∂ − sin θ ∂x r ∂θ 1 ∂2u 2 2 ∂u 1 ∂2u 2 = sin θ + + sin θ cos θ − cos θ sin θ. r2 ∂θ2 r2 r2 ∂θ r ∂r∂θ Since

∂2u ∂r∂θ

=

∂2u ∂θ∂r ,

we get

∂2u 1 ∂u 1 ∂2u 2 ∂u 2 ∂2u ∂2u 2 2 2 = cos θ+ sin θ+ sin θ+ sin θ cos θ− cos θ sin θ. ∂x2 ∂r2 r ∂r r2 ∂θ2 r2 ∂θ r ∂r∂θ Similarly, ∂u ∂u 1 ∂u = sin θ + cos θ, ∂y ∂r r ∂θ ∂2u ∂2u 1 ∂u 1 ∂2u 2 ∂u 2 ∂2u = sin2 θ+ cos2 θ+ 2 2 cos2 θ− 2 sin θ cos θ+ cos θ sin θ. 2 2 ∂y ∂r r ∂r r ∂θ r ∂θ r ∂r∂θ Consequently, ∂2u ∂2u ∂ 2 u 1 ∂u 1 ∂2u + 2 = + + 2 2. 2 2 ∂x ∂y ∂r r ∂r r ∂θ Assume the general case that u is complex-valued. Then 2 2 2 ∂u = ∂u cos2 θ + 1 ∂u sin2 θ − 2 M |

N X

an bn | ≤ B(aN + aM +

n=M

N −1 X

(an − an+1 )) = 2BaM

n=M

proving that the series is Cauchy and hence converges. P einx 1 is the Fourier series of the 2π6. (Exercise 8) Verify that 2i n6=0 n periodic sawtooth function illustrated in Figure 6, defined by f (0) = 0, and  −π/2 − x/2 if − π < x < 0 f (x) = π/2 − x/2 if 0 < x < π. Note that this function is not continuous. Show that nevertheless, the series converges for every x (by which we mean, as usual, that the symmetric partial sums of the series converge). In particular, the value of the series at the origin, namely 0, is the average of the values of f (x) as x approaches the origin from the left and the right. Solution. Checked with Maple and found it correct. bn = sin nx, Bn =

n X k=1

bk =

sin(nx/2) sin((n + 1)x/2) sin(x/2)

8

1 (to prove write sin nx as 2i (einx − e−inx ))) and |Bn | ≤ csc(|x|/2) for |x| < π, x 6= 0. 1/n is decreasing to 0. So series converges for x 6= 0. At x = 0, all the symmetric sums are 0, so series converges to 0. It must converge to the function since its Cesaro means converge to the function.

7. (Exercise 9) Let f (x) = χ[a,b] (x) be the characteristic function of the interval [a, b] ⊂ [−π, π], that is,  1 if x ∈ [a, b] χ[a,b] (x) = 0 otherwise. (a) Show that the Fourier series of f is given by f (x) ∼

b − a X e−ina − e−inb inx + e . 2π 2πin n6=0

The sum extends over all positive and negative integers excluding 0. (b) Show that if a 6= −π or b 6= π and a 6= b, then the Fourier series does not converge absolutely for any x. (c) However, prove that Fourier series converges at every point x. What happens if a = −π and b = π? Solution. (a) Straightforward. (b) |e−ina − e−inb |2 = 1 + 1 − 2 cos n(b − a) = 4 sin2 (n(b − a)/2), thus |e−ina − e−inb | = 2| sin nθ0 |, where 0 < θ0 = The series

b−a π < . 2 2

∞ X | sin nθ0 | n n=1

diverges because ....... (c) an = 1/n, bn = sin(na) − sin(nb) which has bounded partial sums by the solution of Exercise 8. So series converges by Exercise 7. If a = −π and b = π, then the Fourier series is 1. 8. (Exercise 10) Suppose f is a periodic function of period 2π which belongs to the class C k . Show that fˆ(n) = O(1/|n|k ) as |n| → ∞. Solution. Integrate by parts k times as in Corollary 2.4. 9. (Exercise 11) Suppose that {fk }∞ k=1 is a sequence of Riemann integrable functions on the interval [0, 1] such that Z 1 |fk (x) − f (x)| dx → 0 as k → ∞. 0

9

Show that fˆk (n) → fˆ(n) uniformly in n as k → ∞. Proof. Use Z 1 |ˆ g (n)| ≤ |g(x)| dx 0

for all n. P 10. (Exercise cn converges to P12) Prove that if a series of complex numbers s, then cn is Cesaro summable to s. Proof. Let the sequence of partial sums be sn , n = 1, 2, · · · . First assume that s = 0. Let  > 0. Choose N1 such that |sn | < 2 for all n ≥ N1 . Choose N > N1 such that PN1  k=1 |sk | < n 2 for all n ≥ N . Then for all n ≥ N , s1 + · · · + sn | n |s1 | + · · · + |sn | n |s1 + · · · + |sN1 | |sN1 +1 + · · · + |sn | + n n  n/2 + 2 n  |

≤ =
0. Choose K such that |sk | <  for all k ≥ K. Then |

∞ X

ck rk | ≤ (1 − r)

k=1

K−1 X

M rk + (1 − r)

K−1 X rK ≤ (1 − r) M rk +  1−r k=1

k=1

It follows that lim sup | r→1−

∞ X

ck rk | ≤ .

k=1

P∞ Since  > 0 is arbitrary, lim supr→1− | k=1 ck rk | ≤ 0 which is P∞we have equivalent to limr→1− k=1 ck rk = 0. P∞ NextP assume that sP6= 0. Define dn = cn − s/2n . Then n=1 dn = 0, ∞ ∞ n n and r). Thus by our proof above, n=1 Pd∞n r =n n=1 cn r − sr/(2 − P ∞ limr→1− n=1 P cn r − s = 0, i.e. limr→1− n=1 cn rn = s. P ∞ ∞ n (b). The series n=1 (−1) Pdoes not converge, but limr→1− n=1 (−1)n rn = ∞ −r n limr→1− 1+r = −1/2, i.e. n=1 (−1) is Abel summable to −1/2. (c). Write s1 + · · · + sn σn = n and τn = nσn . Since τn − τn−1 = sn , by the same argument as in (a), we have ∞ ∞ X X (1 − r) τk r k = sk rk , k=1

hence (1 − r)2

∞ X

k=1

σk krk =

k=1

∞ X

ck rk

k=1

from the proof of (a). Assume that σ = 0. Let  > 0. Choose N such that |σk | <  for all n ≥ N , and B be a bound for |σk |, k = 1, 2, · · · . Then |

∞ X k=1

k

ck r | ≤ (1−r)

2

N −1 X k=1

N −1 X rN ((1 − r)N + r) 2 Bkr +(1−r)  ≤ (1−r) Bkrk +rN ((1−r)N +r) (1 − r)2 k

2

k=1

N P∞ +r) where we have use the fact that k=N krk = r ((1−r)N . As in the (1−r)2 P∞ k proof of (a), this implies that limr→1− k=1 ck r = 0. For the case σ 6= 0,

11

P∞ consider dn = cn − σ/2n . Since n=1 cn is Cesaro summable to σ and P∞ P∞ n σ/2 = σ (and hence Cesaro summable to σ), we have n=1 n=1 dn Cesaro summable to 0. Thus by our proof above and the same argument as in the proof of (a), the follows. Presult ∞ (d). Consider the series n=1 (−1)n n. It is Abel summable to −1/4 since ∞ X

(−1)n nrn =

n=1

−r . (1 + r)2

Note that

n−1 an σn−1 = . n n P∞ Thus for a Cesaro P∞ summable series n=1 an , limn→∞ proves that n=1 (−1)n n is not Cesaro summable. σn −

an n

must be 0. This

12. (Exercise 14) This exercise deals with a theorem of Tauber which says that under an additional condition on the coefficients cn , the above arrows can be reversed. P (a) IfP cn is Cesaro summable to σ and cn = o(1/n) (that is, ncn → 0), then cn converges to σ. (b) The above statement holds if we replace Cesaro summable by Abel summable. Proof. (a). Denote (n − 1)cn by tn . Since tn = n−1 n ncn and ncn → 0 we have tn → 0. Now sn −σn = sn −

s1 + · · · + sn (sn − s1 ) + · · · + (sn − sn ) t2 + · · · + tn = = n n n

and it is immediate that sn − σn → 0. (b). Let r = 1 − N1 . We have |

N X n=1

cn −

N X n=1

cn r n | ≤

 X  N n 1 |cn | |cn | 1 − (1 − )n ≤ N N n=1 n=1 N X

n where we have used (1 − N1 )n ≥ 1 − N for 1 ≤ n ≤ N , which can be proved

12

easily by induction on n. Also if |cn n| <  for all n ≥ N , then | ≤ ≤ ≤

∞ X

cn r n −

n=1 ∞ X

N X

cn r n |

n=1

|cn |(1 −

n=N +1 ∞ X n=N +1 ∞ X n=N +1

= (1 −

1 n ) N

1 n|cn | (1 − )n N N 1  (1 − )n N N

 1 N +1 ) → N e

as N → ∞. 13. (Exercise 15) Prove that the Fejer kernel is given by FN (x) =

1 sin2 (N x/2) . N sin2 (x/2)

Proof. Recall that N FN (x) = D0 (x)+· · ·+DN −1 (x) where Dn (x) is the Dirichlet kernel. Write ω = eix . Then Dn (x) =

ω −n + · · · + ω −1 + 1 + ω + · · · + ω n

(ω −n + · · · + ω −1 ) + (1 + ω + · · · + ω n ) ω −n − 1 1 − ω n+1 + = ω −1 −1 ω −1 1−ω n+1 −n ω −1 1−ω + = 1−ω 1−ω ω −n − ω n+1 = 1−ω =

13

So N FN (x) =

N −1 X n=0

=

ω −n − ω n+1 1−ω

1 1−ω

=

1 1−ω

=

1 1−ω

= ω

N −1 X

ω −n −

n=0  −N

N −1 X

! ω n+1

n=0

ω −1 1 − ωN −ω −1 ω −1 1−ω





ω −N +1 − ω 1 − ωN −ω 1−ω 1−ω



ω −N − 2 + ω N (1 − ω)2

=

(ω N/2 − ω −N/2 )2 (1 − ω)2 (ω −1/2 )2

=

(ω N/2 − ω −N/2 )2 (ω 1/2 − ω −1/2 )2

=

−4 sin2 (N x/2) −4 sin2 (x/2)

=

sin2 (N x/2) . sin2 (x/2)

1

Therefore FN (x) =

1 sin2 (N x/2) . N sin2 (x/2)

14. (Exercise 16) The Weierstrass approximation theorem states: Let f be a continuous function on the closed and bounded interval [a, b] ⊂ R. Then, for any  > 0, there exists a polynomial P such that sup |f (x) − P (x)| < . x∈[a,b]

Proof. Let  > 0. We may extend f to a continuous (c−a)-periodic function where b ≤ c. By Corollary PN 5.4 of Fejer’s theorem, there exists a trigonometric polynomial Q = n=M an einx such that |Q(x)−f (x)| < /2 for all x. For each n, M ≤ n ≤ N , there exists a polynomial pn (x) such that |an einx − pn (x)| < /2N for all x ∈ [a, c]. Then P = pM + · · · + pN is a polynomial in x that satisfies the requirement. 15. (Exercise 17) In Section 5.4 we proved that the Abel means of f converge to f at all points of continuity, that is, lim Ar (f )(θ) = lim (Pr ∗ f )(θ) = f (θ), with 0 < r < 1,

r→1

r→1

14

whenever f is continuous at θ. In this exercise, we will study the behavior of Ar (f )(θ) at certain points of discontinuity. An integrable function is said to have a jump discontinuity at θ if the two limits lim

h→0,h>0

f (θ + h) = f (θ+ ) and

lim

h→0,h 0 r 2π 0 be given. Choose δ > 0 so that 0 < h < δ implies |f (θ − h) − f (θ−)| <  and |f (θ + h) − f (θ+)| < . Let M be such that |f (y)| ≤ M for all y. Then + − (f ∗ Pr )(θ) − f (θ ) + f (θ ) 2   Z π 1 f (θ+ ) + f (θ− ) = Pr (y)f (θ − y) dy − 2π −π 2 Z 0 Z π 1 1 ≤ Pr (y)|f (θ − y) − f (θ+ )| dy + Pr (y)|f (θ − y) − f (θ− )| dy 2π −π 2π 0 Z Z 1 1 ≤ Pr (y)|f (θ − y) − f (θ+ )| dy + Pr (y)|f (θ − y) − f (θ− )| dy 2π −δ 0. Choose a δ, 0 < δ < min{1/2, /(4C)}. Choose η < 1 such that |f (x)| < (1/4)δ for all x > η. We claim that |(1 − x)f 0 (x)| <  for all x > η. Let x > η. With x0 = x + δ(1 − x) we have 1 f (x0 ) = f (x) + δ(1 − x)f 0 (x) + δ 2 (1 − x)2 f 00 (ζ) 2 for some x < ζ < x0 . Since δ < 1/2, we have 1 − x ≤ 2(1 − ζ), so that |(1 − x)2 f 00 (ζ)| ≤ 4|(1 − ζ)2 f 00 (ζ)| ≤ 4C. Therefore |(1 − x)f 0 (x)| f (x0 ) − f (x) 1 = | − δ(1 − x)2 f 00 (ζ)| δ 2 |f (x0 )| + |f (x)| 1 ≤ + δ(1 − x)2 |f 00 (ζ)| δ 2 1 1 + = < 2 2 P∞ n Lemma 2 Let n=0 an x be a real power series with an ≥ 0 for all n. Suppose ∞ X an xn = 1. lim− (1 − x) x→1

n=0

Then for any integrable function g(t) on [0, 1] lim− (1 − x)

x→1

∞ X

n

n

Z

an x g(x ) =

1

g(t) dt. 0

n=0

Proof. First prove for functions of the type xk . Then for any polynomial; then for any continuous function; then for any integral function. P∞ Corollary 1 Let n=0 an xn be a real power series with an ≥ 0 for all n. Suppose ∞ X lim− (1 − x) an xn = 1. x→1

n=0

Then

PN

n=0

lim

N

N →∞

20

an

= 1.

Proof. Apply the above lemma to the function g(t) = 0 for t < e−1 , 1/t otherwise, and let x = e−1/N . Finishing the solution of Problem 3 (x) We shall write f (x) ∼ g(x) to mean fg(x) → 1 as x → 1, and f (n) ∼ g(n) (n) to mean fg(n) → 1 as n → ∞. (a) We may assume P∞ that s = 0 (see Exercise 13, Chapter 2), i.e. we assume that f (x) = n=0 an xn → 0 as x → 1. Then f 00 (x) = O(1/(1 − x)2 ) because

f 00 (x) =

∞ X

n(n − 1)an xn−2 = O(

n=2

∞ X

(n − 1)xn−2 ) = O(1/(1 − x)2 ).

n=2

So by Lemma 1, 

0

f (x) = o

1 1 − x)

 .

Suppose |nan | ≤ c. Then ∞ X

(1 −

n=1

Since 1 −

nan c

nan n−1 1 f 0 (x) 1 )x = − ∼ c 1−x c 1−x

≥ 0, Corollary 1 implies that n X

(1 −

k=1

or what is the same

n X

kak )∼n c

kak = o(n).

k=1

21

Write wn =

Pn

k=1

= = = = = = =

kak , w0 = 0. So wn /n → 0 as n → ∞. Then f (x) − a0 ∞ X an xn n=1 ∞ X

wn − wn−1 n x n n=1

∞ ∞ X wn n X wn−1 n x − x n n n=1 n=1 ∞ ∞ X wn n X wn n+1 x − x n n+1 n=1 n=1   n ∞ X xn+1 x − wn n n+1 n=1   ∞ X xn xn − xn+1 + wn n+1 n(n + 1) n=1 ∞ X

(1 − x)



wn

n=1

X wn xn + xn n + 1 n=1 n(n + 1)

Since f (x) → 0 and the first term in the last sum approaches 0 as x → 1, we get ∞ X wn lim xn = −a0 x→1 n(n + 1) n=1 Since

wn n(n+1)

= o(1/n), by the regular Tauberian theorem ∞ X

wn = −a0 . n(n + 1) n=1 Now N X

=

=

=

wn n(n + 1) n=1   N X 1 1 wn − n n+1 n=1 N X wn − wn−1 wN − n N +1 n=1 N X

an −

n=1

Letting N → ∞ we get

P∞

n=1

wN . N +1

an = −a0 , i.e. 22

P∞

n=0

an = 0.

Chapter 3 1. (Exercise 1) Show that the first two examples of inner product spaces, namely Rd and Cd , are complete. 2. (Exercise 2) Prove that the vector space `2 (Z) is complete. 3. (Exercise 3) Construct a sequence of integrable functions {fk } on [0, 2π] such that Z 2π 1 lim |fk (t)|2 dt = 0 k→∞ 2π 0 but limk→∞ fk (t) fails to exist for any t. 4. (Exercise 4) (In (c), use the fact that f is continuous except possibly on a set of measure 0.) 5. (Exercise 5) Let  f (t) =

0 log(1/t)

for t = 0 for 0 < t ≤ 2π,

and define a sequence of functions in R by  0 for 0 ≤ t ≤ 1/n fn (t) = f (t) for 1/n < t ≤ 2π. Prove that {fn }∞ n=1 is a Cauchy sequence in R. However, f does not belong to R. Solution. Z (log t)2 dt = t(log t)2 − 2t log t + 2t and limt→0 t log t = 0, limt→0 t log2 t = 0. 6. (Exercise 6) Consider the sequence {ak }∞ k=−∞ defined by  1/k if k ≥ 1 ak = 0 if k ≤ 0. Note that {ak } ∈ l2 (Z), but that no Riemann integrable function has k th Fourier coefficient equal to ak for all k. Solution. Let M = sup−π≤tπ |f (t)|. Recall that Z π 1 Ar (f )(θ) = Pr (θ − t)f (t) dt. 2π −π So |Ar (f )(θ)| ≤ M

1 2π 23

Z

π

Pr (θ − t) dt = M −π

Also Ar (f )(θ) =

∞ X

fˆ(n)einθ r|n| .

n=−∞

So Ar (f )(0) =

∞ X

fˆ(n)r|n| .

n=−∞

It follows that if f is a Riemann integrable function that has k th Fourier coefficient equal to ak for all k, then ∞ X rn n n=1

is bounded for 0 ≤ r < 1. This is a contradiction since the sum of the series is − log(1 − r). 7. (Exercise 7) Show that the trigonometric series X sin nx n≥2

log n

converges for every x, yet it is not the Fourier series of a Riemann integrable function. P sin nx The same is true for for 0 < α < 1, but the case 1/2 < α < 1 is nα more difficult. See Problem 1. Solution. Apply Parseval’s identity.(A Riemann integrable function is in L2 .) Series P converges because the partial sums of sin nx is bounded; see Exercise 9. Postpone the case 1/2 < α < 1 to Problem 1. 8. (Exercise 8) Exercise 6 in Chapter 2 dealt with the sums X n

odd

≥1

∞ X 1 π2 1 π2 = and = . n2 8 n2 6 n=1

Similar sums can be derived using the methods of this chapter. (a) Let f be the function defined on [−π, π] by f (θ) = |θ|. Use Parseval’s identity to find the sums of the following two series: ∞ X

∞ X 1 1 and . 4 4 (2n + 1) n n=0 n=1 4

4

In fact, they are π96 and π90 , respectively. (b) Consider the 2π-periodic odd function defined on [0, π] by f (θ) = θ(π − θ). Show that ∞ X

∞ X π6 1 π6 1 = and = . (2n + 1)6 960 n6 945 n=1 n=0

24

P∞ Remark. The general expression when k is even for n=1 1/nk in terms of π k is given in Problem 4. P However, finding a formula for the sum P∞ ∞ 3 k 1/n , or more generally n=1 n=1 1/n with k odd, is a famous unresolved question. Solution. The Fourier series for f (θ) = |θ| is π 2 − 2 π

einx + e−inx . n2

X n

odd

≥1

By Parseval’s identity, π2 4 + 2 4 π

1 2 = n4 2π

X n

from which we get

odd

Z

≥1

π

|θ|2 dθ =

−π

π2 3

∞ X

π4 1 = . (2n + 1)4 96 n=0 The sum

P∞

1 n=1 n4

is obtained from solving for x in the equation x=

π4 x + 96 16

4

which yields x = π90 . The Fourier series for the function f (θ) = θ(π − θ) is 8 π

4 sin kθ = −i k3 π

X k

odd

≥1

eikθ − e−ikθ . k3

X k

odd

≥1

By Parseval’s identity, Z π ∞ 16 X 2 1 1 4 = π , θ2 (1 − θ)2 dθ = 2 6 π n=0 (2n + 1) 2π −π 30 from which we get

∞ X

1 π6 . = 6 (2n + 1) 960 n=0 Solving x from x= we get x =

1 π6 x+ 64 960

π6 945 .

9. (Exercise 9) Show that for α not an integer, the Fourier series of π ei(π−x)α sin πα 25

on [0, 2π) is given by ∞ X einx . n+α −∞

Apply Parseval’s formula to show that ∞ X

π2 1 = . 2 (n + α) (sin πα)2 −∞

Solution Straightforward checking. 10. (Exercise 10) Consider the example of a vibrating string which we analyzed in Chapter 1. The displacement u(x, t) of the string at time t satisfies the wave equation 1 ∂2u ∂2u 2 = , c = τ /ρ. 2 2 c ∂t ∂x2 The string is subject to the initial conditions u(x, 0) = f (x) and

∂u (x, 0) = g(x), ∂t

where we assume that f ∈ C 1 and g is continuous. We define the total energy of the string by 1 E(t) = ρ 2

L

Z



0

∂u ∂t

2

Z

1 dx + τ 2

L



0

∂u ∂x

2 dx.

The first term corresponds to the ”kinetic energy” of the string (in analogy with (1/2)mv 2 , the kinetic energy of a particle of mass m and velocity v), and the second term corresponds to its ”potential energy.” Show that the total energy of the string is conserved, in the sense that E(t) is constant. Therefore, E(t) = E(0) =

1 ρ 2

Z 0

L

1 g(x)2 dx + ρ 2

Z

L

f 0 (x)2 dx.

0

Solution. We have E 0 (t) = ρ

Z

L

∂u ∂ 2 u dx + τ ∂t ∂t2

Z

L

∂u ∂ 2 u dx − τ ∂t ∂t2

Z

0

Z =ρ 0

L

∂u ∂ 2 u dx ∂x ∂x∂t

L

∂ 2 u ∂u dx ∂ 2 x ∂t

0

0

=0 where we have used integration by parts, and that for all t. 26

∂u ∂t (0, t)

=

∂u ∂t (L, t)

=0

11. (Exercise 11) The inequalities of Wirtinger and Poincare establish a relationship between the norm of a function and that of its derivative. RT (a) If f is T -periodic, continuous, and piecewise C 1 with 0 f (t) dt = 0, show that Z T Z T T2 2 |f (t)| dt ≤ |f 0 (t)|2 dt, 4π 2 0 0 with equality if and only if f (t) = A sin(2πt/T ) + B cos(2πt/T ). (b) If f is aa above and g is just C 1 and T -periodic, prove that Z 2 Z T Z T T T2 2 ≤ |f (t)| f (t)g(t) dt dt |g 0 (t)|2 dt. 0 4π 2 0 0 (c) For any compact interval [a, b] and any continuously differentiable function f with f (a) = f (b) = 0, show that Z b Z (b − a)2 b 0 2 |f (t)|2 dt ≤ |f (t)| dt. π2 a a Discuss the case of equality, and prove that the constant (b−a)2 /π 2 cannot be improved. Solution. RT (a) The condition 0 f (t) dt = 0 implies that fˆ(0) = 0. The continuity 2π ˆ of f guarantees that fˆ0 (n) = 2πin T f (n). Indeed, write τ for T . Then for n 6= 0, Z 1 T fˆ(n) = f (t)e−inτ t dt T 0 Z 1 1 1 1 T 0 = (f (0+ ) − f (T − )) + f (t)e−inτ t dt T inτ inτ T 0 T ˆ0 f (n) = 2πin Therefore, by Parseval’s identity,recalling that fˆ(0) = 0, Z

T

|f (t)|2 dt = T

0

=

X |n|>0

T 3 X |fˆ0 (n)|2 4π 2 n2 |n|>0

3



= =

T X ˆ0 |f (n)|2 4π 2 |n|>0 Z 3 T 1 T 0 2 |f (t)| dt 4π 2 T 0 Z T T2 |f 0 (t)|2 dt 4π 2 0 27

|fˆ(n)|2

From the above inequalities, we see that equality holds if and only if fˆ(n) = 0 for all n ≥ 2. This means that, writing an for fˆ(n), f (x) = a1 eiτ x + a−1 e−iτ x which simplifies to A sin(τ x) + B cos(τ x). Remark. It is clear from the proof above that in the absence of the RT condition 0 f (t) dt = 0, the inequality in (a) is X

|an |2 ≤

|n|>0

T2 X |bn |2 4π 2 |n|>0

where an = fˆ(n), bn = fˆ0 (n). (b) Let an = fˆ(n), bn = gˆ(n), cn = gˆ0 (n). Then 2 Z T f (t)g(t) dt 0 X = T| an bn |2 |n|≥0

X

= T|

an bn |2

|n|>0

X

≤ (T

|an |2 )(

|n|>0 T

Z

X |n|>0

T2 |f (t)| dt 2 4π 2

≤ 0

|bn |2 )

Z

T

|g 0 (t)| dt

0

(c) Extend f to a function on [a, 2b − a] such that f (b + h) = −f (b − h) for 0 ≤ h ≤ b − a and then extend it so that it is T = 2(b − a)periodic. It is easy to see that now f (b + h) = −f (b − h) and f 0 (b + h) = RT R 2b−a RT f 0 (b − h) for all h. Then 0 f (x) dx = a f (x) dx = 0, 0 |f (x)| dx = R 2b−a Rb RT Rb |f (x)| dx = 2 a |f (x)| dx, and 0 |f 0 (x)| dx = 2 a |f 0 (x)| dx. Check a that f so extended is also C 1 on R. Therefore by (a), we have Z b |f (x)| dx a

= ≤ = =

1 2

Z

T

|f (x)| dx 0 2

Z 1 2 (b − a)2 T 0 |f (x)| dx 2 4π 2 0 Z (b − a)2 1 T 0 |f (x)| dx π2 2 0 Z (b − a)2 b 0 |f (x)| dx π2 a

For the function f (x) that we define, its translation g(x) = f (x + b) is an odd function. So gˆ(n) = einb fˆ(n), and the Fourier series for g is of the 28

form

∞ X

∞ X 2πnx πnx An sin = An sin . 2(b − a) b −a n=1 n=1

Thus the Fourier series for f is of the form ∞ X

An sin

n=1

πn(x − b) . b−a

According the inequality in (c) is an equality iff f (x) is of the form π(x−a) π(x−b) A sin π(x−b) b−a . (The book says A sin b−a ; this is equivalent since sin b−a = − sin π(x−a) b−a ). R∞ 12. (Exercise 12) Prove that 0 sinx x dx = Proof. R π We have −π DN (t) dt = 2π. So Z

π

−π

π 2.

sin(N + 1/2)x dx = 2π sin(x/2)

Write csc(x/2) as csc(x/2) − 2/x + 2/x. limx→0 csc(x/2) − 2/x = 0, so it has a removable discontinuity at 0 on the interval [−π, π]. By LebesgueRiemann lemma we get Z π 2 sin(N + 1/2)x dx → 2π as N → ∞. x −π This yields Z

π

0

sin(N + 1/2)x π dx → as N → ∞. x 2

By change of variable, we get Z

(N +1/2)π

0

Since get

R (N +1/2)π Nπ

sin x x

sin x π dx → as N → ∞. x 2

dx → 0 as N → ∞, (use Mean-Value Theorem), we Z



0

sin x π dx → as N → ∞. x 2

By MVT, Z

(N +1)π



| sin x| 1 dx ≤ , x N

so for any t > π there exists N > 0 such that Z t sin x 1 dx ≤ . x N Nπ 29

It follows that Z 0



sin x dx = lim N →∞ x

Z 0



sin x π dx = . x 2

13. (Exercise 13) Suppose that f is periodic and of class C k . Show that fˆ(n) = o(1/|n|k ), that is, |n|k fˆ(n) goes to 0 as |n| → ∞. This is an improvement over Exercise 10 in Chapter 2. Solution. We have (see p.43) Z 2π 1 f (k) (θ)e−inθ dθ. 2π fˆ(n) = (in)k 0 So 2π(in) fˆ(n) = k

Z



f (k) (θ)e−inθ dθ.

0

Now use Lebesgue-Riemann lemma. 14. (Exercise 14) Prove that the Fourier series of a continuously differentiable function f on the circle is absolutely convergent. Proof. 1 ˆ0 f (n), we get, by Cauchy-Schwarz inequality and ParseFrom fˆ(n) = in val’s identity, ∞ X

=

|fˆ(n)|

n=−∞ ∞ X

1 ˆ0 |f (n)| |n| n=−∞ ∞ π 2 1/2 X ˆ0 ) ( |f (n)|2 )1/2 6 n=−∞ Z π 2 1 π |f 0 (t)|2 dt < ∞. (2 )1/2 6 2π −π

≤ (2 =

15. (Add in) Prove that the Fourier series of a 2π periodic absolutely continuous function whose derivative (exists a.e.) in [0, 2π] is square integrable (in particular, Riemann integrable), is absolutely convergent. (Note that Exercise 16 below shows that Lipschitz condition alone is enough. But derivative of a Lipschitz function is bounded. And Lipschitz functions are precisely functions representable as integral of a bounded measurable function.) Proof. Use the proof in Exercise 14. Note that the integration by parts formula is valid for absolutely continuous functions. 30

16. (Exercise 15) Let f be a 2π-periodic and Riemann integrable on [−π, π]. (a) Show that Z π 1 π ˆ f (n) = − f (x + )e−inx dx 2π −π n hence

1 fˆ(n) = 4π

Z

π

[f (x) − f (x + −π

π −inx )]e dx. n

(b) Now assume that f satisfies a H¨older condition of order α, namely |f (x + h) − f (x)| ≤ C|h|α for some 0 < α ≤ 1, some C > 0, and all x, h. Use part (a) to show that fˆ(n) = O(1/|n|α ). (c) Prove that the above result cannot be improved by showing that the function ∞ X k f (x) = 2−kα ei2 x , k=0

where 0 < α < 1, satisfies |f (x + h) − f (x)| ≤ C|h|α , and fˆ(N ) = 1/N α whenever N = 2k . Solution. Note that for any real x, |1 − eix | ≤ |x| (to prove, just note that LHS is 2| sin(x/2)| and | sin a| ≤ |a| for all real number a). (a). Z π π 1 f (x + )e−inx dx 2π −π n Z π+ nπ 1 f (u)e−inu eiπ du = 2π −π+ nπ Z π 1 = − f (u)e−inu du 2π −π = −fˆ(n) (b) |fˆ(n)| Z π 1 π ≤ |f (x) − f (x + )| dx 4π −π n 1 πα ≤ 2πC α 4π |n| C1 = |n|α 31

(c).

=

|f (x + h) − f (x)| ∞ ∞ X X k k | 2−kα ei2 (x+h) − 2−kα ei2 x | k=0



∞ X

k=0 k

2−kα |ei2

h

− 1|

k=0

X



X

2−kα 2k |h| +

2k ≤1/|h|

2−kα 2

2k >1/|h|

The second sum is easily seen to be less than 2|h|α . The first sum is 0 if |h| > 1. So assume |h| ≤ 1. Let l be the unique nonnegative integer such that 2−l−1 < |h| ≤ 2−l . Then the first sum is l X

k

1−α

(2 |h|)

α

α

|h| ≤ |h|

k=0

l X

(2k−l )1−α ≤

k=0

1 |h|α . 1 − 2α−1

Since the series converges uniformly, the coefficient of einx is fˆ(n) for all n. 17. (Exercise 16) Let f be a 2π-periodic function which satisfies a Lipschitz condition with constant K; that is |f (x) − f (y)| ≤ K|x − y| for all x, y. This is simply the H¨ older condition with α = 1, so by thePprevious exercise, we see that fˆ(n) = O(1/|n|). Since the harmonic series 1/n diverges, we cannot say anything (yet) about the absolute convergence of the Fourier series of f . The outline below actually proves that the Fourier series of f converges absolutely and uniformly. (a) For every positive h we define gh (x) = f (x + h) − f (x − h). Prove that Z 2π ∞ X 1 |gh (x)|2 dx = 4| sin nh|2 |fˆ(n)|2 , 2π 0 n=−∞ and show that

∞ X

| sin nh|2 |fˆ(n)|2 ≤ K 2 h2 .

n=−∞

(b) Let p be a positive integer. By choosing h = π/2p+1 , show that X

|fˆ(n)|2 ≤

2p−1 0 so that fˆ(ξ) = O(1/|ξ|1+ ) as |ξ| → ∞. Solution. By remarks in section 1.7, since both f and fˆ are of moderate decrease, the inverse Fourier transform of fˆ is f . Thus Z ∞ f (x + h) − f (x) = fˆ(ξ)e2πiξx (e2πiξh − 1) dξ. −∞

The condition on fˆ is equivalent to |fˆ(ξ)| ≤

A for all ξ ∈ R 1 + |ξ|1+α

52

for some A > 0. Thus (note that |e2πiξh − 1| = 2| sin(πξh)|) Z ∞ f (x + h) − f (x) A|e2πiξh − 1| ≤ 1 dξ α α h |h| −∞ 1 + |ξ|1+α Z ∞ 4A | sin(πξh)| ≤ dξ |h|α 0 1 + ξ 1+α Z 4A ∞ | sin(u)| = du π 0 |h|1+α + u1+α ! Z 1 sin(u) Z ∞ | u | 1 4A ≤ du + du π uα u1+α 0 1 Z 1  Z ∞ 4A 1 1 ≤ du + du α π u1+α 0 u 1 0. So by (a), fˆ is not of moderate decrease. 4. (Exercise 4) Examples of compactly supported functions in S(R) are very handy in many applications in analysis. Some examples are: (a) Suppose a < b, and f is the function such that f (x) = 0 if x ≤ a or x ≥ b and f (x) = e−1/(x−a) e−1/(b−x) if a < x < b. Show that f is indefinitely differentiable on R. (b) Prove that there exists an indefinitely differentiable function F on R such that F (x) = 0 if x ≤ a, F (x) = 1 if x ≥ b, and F is strictly increasing on [a, b]. (c) Let δ > 0 be so small that a + δ < b − δ. Show that there exists an indefinitely differentiable function g such that g is 0 if x ≤ a or x ≥ b, g is 1 on [a + δ, b − δ], and g is strictly monotonic on [a, a + δ] and [b − δ, b]. Solution. Note that the graph of −1/(x − a) − 1/(b − x) is symmetric about the line x = (a + b)/2 with maximum value of −4/(b − a) at the point x = (a + b)/2. R∞ Rx For (b) consider F (x) = c −∞ f (t) dt where c is the reciprocal of −∞ f (t) dt. 5. (Exercise 5) Suppose f is continuous and of moderate decrease. (a) Prove that fˆ is continuous and fˆ(ξ) → 0 as |ξ| → ∞. (b) Show that if fˆ(ξ) = 0 for all ξ, then f is identically 0. Solution.

53

Z ∞ f (x)e−2πiξx (e−2πihx − 1) dx |fˆ(ξ + h) − fˆ(ξ)| = −∞ Z ∞ |f (x)||e−2πihx − 1| dx ≤ −∞ Z ∞ = |f (x)|2| sin 2πhx| dx −∞ Z ∞ | sin 2πhx| ≤ 2A dx 1 + x2 −∞ R 1 Let  > 0. Choose N > 0 such that 2A |x|>N 1+x Since 2 dx < /2. | sin 2πhx| → 0 uniformly on [−N, N ] as h → 1+x2 R N | sin 2πhx| 2A −N 1+x2 dx < /2 for |h| < δ. This

0, there exists δ > 0 such that proves that fˆ is continuous. Note: Lebesgue dominated convergence theorem could be applied to yield a simpler proof. (b). Z ∞  Z ∞ 1 −2πiξx −2πi(ξx+1/2) ˆ f (ξ) = f (x)e − f (x)e dx 2 −∞ −∞ Z ∞ 1 1 = (f (x) − f (x − ))e−2πiξx dx 2 −∞ 2ξ

Now the result follows from applying Lebesgue dominated convergence theorem; without it, the proof is harder. (b). If f, g are of moderate decrease, then the function f (x)g(y)e−2πixy is integrable over R2 . Then it follows from Fubini’s theorem that Z Z f (x)ˆ g (x) dx = g(x)fˆ(x) dx. Suppose fˆ(x) = 0 for all x. For any t and any δ > 0, Kδ (t−x) as a function of x is in S(R), so by Corollary 1.10 it is equal to the Fourier transform gˆδ R R for some gδ in S(R). So f (x)gˆδ (x) dx = f (x)Kδ (t − x) dx = 0. Letting δ → 0, we get f (t) = 0. 6. (Exercise 8) Prove that if f is continuous, of moderate decrease, and R∞ 2 f (y)e−y e2xy dy = 0 for all x ∈ R, then f = 0. −∞ Proof. R∞ 2 2 2 R∞ 2 Let g(x) = e−x . Then (f ∗g)(x) = −∞ f (y)e−(x−y) dy = e−x −∞ f (y)e−y e2xy dy = √ 2 2 0 for all x. This implies that f[ ∗ g(ξ) = fˆ(ξ)ˆ g (ξ) = fˆ(ξ) πe−π ξ = 0 for all ξ. So fˆ = 0. By Theorem 1.9, f = 0. 7. (Exercise 12) Show that the function defined u(x, t) = 54

x Ht (x) t

satisfies the heat equation for t > 0 and limt→0 u(x, t) = 0 for every x, but u is not continuous at the origin. Proof. Maple shows that x(x2 − 6t) x2 /4t ∂u ∂2u √ = e . = ∂t ∂x2 8 πt7/2 It is clear that limt→0+ u(x, t) = 0 for all x. If we approach the origin along the parabola x2 = t, we get lim

x→0 x2

1 √



e−1/4 = ∞.

8. (Add in) Suppose f is an even function such that both f and fˆ are of moderate decrease. Then fˆ is also even and the Fourier transform of fˆ(x) is f (ξ). Proof. R R∞ The below denotes −∞ . We have Z Z Z fˆ(−ξ) = f (x)e2πixξ dx = f (−x)e−2πixξ dx = f (x)e−2πixξ dx = fˆ(ξ) so fˆ is even. Since

Z f (x) =

we have Z

fˆ(t)e−2πiξt dt =

Z

fˆ(t)e2πixt dt

fˆ(−t)e2πiξt dt =

Z

fˆ(t)e2πiξt dt = f (ξ)

proving that the Fourier transform of fˆ(x) is f (ξ). 9. (Exercise 15) This exercise provides another example of periodization. (a) Apply the Poisson summation formula to the function gˆ(x) and its Fourier transform in Exercise 2 to obtain ∞ X

1 π2 = 2 (n + α) (sin πα)2 n=−∞ whenever α is real, but not equal to an integer. (b) Prove as a consequence that ∞ X

1 π = n+α tan πα n=−∞ whenever α is real but not equal to an integer. Solution. 55

(1)

By the last item, the Fourier transform of h(x) = gˆ(x) is g(x) in Exercise 2. By Poisson summation formula ∞ ∞ X X sin2 (π(n + α)) = g(n)e2πinα = 1. 2 (n + α)2 π n=−∞ n=−∞

(If α is an integer, then the equality is reduced to trivial 1 = 1.) Since sin(nπ + πα) = (−1)n sin(πα), the result follows. (b). Assume that 0 < α < 1. We have from (a), for n 6= 0,   Z α 1 1 1 1 + dx = − + 2 (n + x)2 −n + α n + α 0 (−n + x) 2

and since limx→0 (sinππx)2 − x12 = π 2 /3, it has a removable discontinuity at x = 0 and limx→0 − tanππx + x1 = 0, we have Z α π2 1 π 1 − 2 dx = − + . 2 (sin πx) x tan πα α 0 Therefore



1 X + α n=1



1 1 + n + α −n + α

 =

π . tan πα

Evaluating at α = 1/2 yields 0; this is correct since ∞ X

∞ X 4 2 2 = − 2 4n − 1 n=1 2n − 1 2n + 1 n=1

is a telescoping series whose sum is 2. Now we have to prove for any α, a noninteger. Let n1 = bαc. Then α = n1 + α1 for some 0 < α1 < 1. Then ∞ X

∞ X 1 1 π π = . = = n + α n=−∞ n + α1 tan πα1 tan πα n=−∞

10. P (Exercise 19) The following is a variant of the calculation of ζ(2m) = ∞ 2m found in Problem 4, Chapter 3. n=1 1/n (a) Apply the Poisson summation formula to f (x) = t/(π(x2 + t2 )) and fˆ(ξ) = e−2πt|ξ| where t > 0 in order to get ∞ ∞ X 1 X t = e−2πt|n| . π n=−∞ t2 + n2 n=−∞

(b) Prove the following identity: ∞ ∞ t 1 2 X 1 X = + (−1)m+1 ζ(2m)t2m−1 , 0 < t < 1 π n=−∞ t2 + n2 πt π m=1

56

as well as

∞ X

e−2πt|n| =

n=−∞

2 − 1, 0 < t < 1. 1 − e−2πt

(c) Use the fact that ∞ X z z B2m 2m = 1 − + z , z e −1 2 m=1 (2m)!

where Bk are the Bernoulli numbers to deduce from the above formula, 2ζ(2m) = (−1)m+1

(2π)2m B2m . (2m)!

Solution. (a) By Lemma 2.4, the Fourier transform of f (x) = t/(π(x2 +t2 )) is fˆ(ξ) = e−2πt|ξ| . Then by Poisson summation formula with x = 0 ((x + n)2 = n2 ), we get (a). (b). ∞ t 1 1X t 1 X = + π n=−∞ t2 + n2 πt π t2 + n2 n6=0 ∞ X

=

1 2 + πt π

t 2 + n2 t n=1

=

∞ t/n2 2X 1 + πt π n=1 1 + (t/n)2

=

∞ ∞ 2m 1 2X t X m t + (−1) πt π n=1 n2 m=0 n2m

=

∞ ∞ 2XX t2m+1 1 + (−1)m 2(m+1) πt π n=1 m=0 n

∞ ∞ 1 2XX t2m−1 (−1)m+1 2m = + πt π n=1 m=1 n

=

∞ ∞ 1 2 XX t2m−1 + (−1)m+1 2m πt π m=1 n=1 n

=

∞ 1 2 X + (−1)m+1 ζ(2m)t2m−1 πt π m=1

57

Next, ∞ X

e−2πt|n| = 1 + 2

n=−∞

∞ X

e−2πtn

n=1 −2πt

=1+2

e 1 − e−2πt

1 + e−2πt 1 − e−2πt 2 − (1 − e−2πt ) = 1 − e−2πt 2 = −1 1 − e−2πt =

Dividing by 2 on both sides, we get ∞ 1 1 X 1 1 + (−1)m+1 ζ(2m)t2m−1 = − . 2πt π m=1 1 − e−2πt 2

Letting z = −2πt and multiplying both sides by −z, we get 1+

∞ 1 X 2πz 2m z z (−1)m+1 ζ(2m) = z + . 2m π m=1 (2π) e −1 2

Comparing with ∞ X z B2m 2m z = 1 − + z , ez − 1 2 m=1 (2m)!

we get 2ζ(2m) = (−1)m+1

(2π)2m B2m . (2m)!

11. (Exercise 23) The Heisenberg uncertainty principle can be formulated in d2 2 terms of the operator L = − dx 2 + x , which acts on Schwartz functions by the formula d2 f L(f ) = − 2 + x2 f. dx This operator, sometimes called the Hermite operator, is the quantum analogue of the harmonic oscillator. Consider the usual inner product on S given by Z ∞ (f, g) = f (x)g(x) dx whenever f, g ∈ S. −∞

(a) Prove that the Heisenberg uncertainty principle implies (Lf, f ) ≥ (f, f ) for all f ∈ S. 58

This is usually denoted by L ≥ I. (b) Consider the operators A and A∗ defined on S by A(f ) =

df df + xf and A∗ (f ) = − + xf. dx dx

The operators A and A∗ are sometimes called the annihilation and creation operators, respectively. Prove that for all f, g ∈ S we have (i) (Af, g) = (f, A∗ g), (ii) (Af, Af ) = (A∗ Af, f ) ≥ 0, (iii) A∗ A = L − I. In particular, this again shows that L ≥ I. (c) Now for t ∈ R, let At (f ) =

df df + txf and A∗t (f ) = − + txf. dx dx

Use the fact that (A∗t At f, f ) ≥ 0 to give anotherR proof of the Heisenberg ∞ uncertainty principle which says that whenever −∞ |f (x)|2 = 1 then Z ∞  Z ∞ 2 ! df 2 2 dx ≥ 1 . x |f (x)| 4 −∞ −∞ dx Solution. p (a) By dividing f by (f, f ) if necessary, we may assume that (f, f ) = 1. We need to prove that (Lf, f ) ≥ 1. Now Z Z Z Z (Lf, f ) = −(f 00 , f ) + (x2 f, f ) = (f 0 , f 0 ) + (x2 f, f ) = |f 0 |2 + x2 |f |2 = 4π 2 ξ 2 |fˆ|2 + x2 |f |2 Z Z 1 2 ˆ 2 1/2 =1 ≥ 4π( ξ |f | ) )( x2 |f |2 )1/2 ≥ 4π 4π where we have used integration by parts and theorem 4.1 (Heisenberg uncertainty principle). (b) Direct checking. Use integration by parts. Note: AA∗ = L + I. (c) Similar to (b), we also have (A∗t At f, f ) ≥ 0 for all t. When this is written out, we get t2 (x2 f, f )−t(f, f )−(f 00 , f ) = t2 (x2 f, f )−t+(f 0 , f 0 ) ≥ 0 for all t. It follows that 1 − 4(f 0 , f 0 )(x2 f, f ) ≤ 0 from which the result follows. 12. (Problem 1) The equation x2

∂2u ∂u ∂u + ax = ∂x2 ∂x ∂t

with u(x, 0) = f (x) for 0 < x < ∞ and t > 0 is a variant of the heat equation which occurs in a number of applications. To solve it, make the 59

change of variables x = e−y so that −∞ < y < ∞. Set U (y, t) = u(e−y , t) and F (y) = f (e−y ). Then the problem reduces to the equation ∂U ∂2U ∂U + (1 − a) = , ∂y 2 ∂y ∂t with U (y, 0) = F (y). This can be solved like the usual heat equation (the case a = 1) by taking the Fourier transform in the y variable. One must then compute the integral Z ∞ 2 2 e(−4π ξ +(1−a)2πiξ)t e2πiξν dξ. −∞

Show that the solution of the original problem is then given by Z ∞ 2 dν 1 e(− log(ν/x)+(1−a)t) /(4t) f (ν) . u(x, t) = ν (4πt)1/2 0 Solution. I have checked that what it says up to the new differential equation is correct. Taking the Fourier transform w.r.t. y of the new equation, we get ˆ ∂U ˆ. = (−4π 2 ξ 2 + (1 − a)2πiξ)U ∂t ˆ (ξ, 0) = Fˆ (ξ), to get Solve, using U ˆ (ξ, t) = Fˆ (ξ)e(−4π2 ξ2 +(1−a)2πiξ)t . U 2 2

It remains to find the inverse Fourier transform of e(−4π ξ +(1−a)2πiξ)t and then write U (y, t) as a convolution. We have Z ∞ Z ∞ 2 2 2 2 e(−4π ξ +(1−a)2πiξ)t e2πiξy dξ = e−4π ξ t e((1−a)t+y)2πiξ dξ −∞

−∞ 2

2

which is equal to the inverse transform of e−4π tξ evaluated at (1−a)t+y. √ 2 2 Using F(e−πy ) = e−πξ and F(δf (δy)) = fˆ(δ −1 ξ) with δ = 1/ 4πt, we find that the required inverse transform is √

1 −y2 /4t e 4πt

evaluating at (1 − a)t + y yields √

1 −(y+(1−a)t)2 /(4t) e 4πt

Therefore Z



U (y, t) = −∞

F (µ) √

1 −(y−µ+(1−a)t)2 /(4t) e dµ. 4πt

Make a substitution µ = − log ν, and substitute y = − log x, we get the result. 60

13. (Problem 2) The Black-Scholes equation from finance theory is ∂V ∂V σ 2 s2 ∂ 2 V − rV = 0, 0 < t < T, + rs + ∂t ∂s 2 ∂s2

(2)

subject to the ”final” boundary condition V (s, T ) = F (s). An appropriate change of variables reduces this to the equation in Problem 1. Alternatively, the substitution V (s, t) = eax+bτ U (x, τ ) where x = log s, τ =  σ2 1 r 1 r 2 reduces (2) to the one2 (T − t), a = 2 − σ 2 , and b = − 2 + σ 2 dimensional heat equation with the initial condition U (x, 0) = e−ax F (ex ). Thus a solution to the Black-Scholes equation is Z ∞ (log(s/s∗ )+(r−σ2 /2)(T −t))2 ds∗ e−r(T −t) − 2σ 2 (T −t) e F (s∗ ) ∗ . V (s, t) = p 2 s 2πσ (T − t) 0 Solution. Multiplying the equation by e−rt . Define V1 = e−rt V . Then the equation becomes ∂V1 ∂V1 σ 2 s2 ∂ 2 V1 + rs + = 0. ∂t ∂s 2 ∂s2 Let t1 = −(σ 2 /2)t. Then on both sides we get

∂V1 ∂t

=

∂V1 2 ∂t1 (−σ /2),

and upon dividing by −σ 2 /2

∂V1 ∂ 2 V1 2r ∂V1 + s2 . = 2s ∂t1 σ ∂s ∂s2 This is the equation in Problem 1. For the alternative way, compute and canceling out eax+bτ , we get −

∂V ∂t

, ∂V ∂s and

∂2V ∂s2

, replacing 1/s by ex

∂U ∂U σ2 ∂U ∂ 2 U σ2 (bU + )+r(aU + )+ (a(a−1)U +(2a−1) + )−rU = 0. 2 ∂τ ∂x 2 ∂x ∂x2 2

2

2

Since r + σ2 (2a − 1) = 0, − σ2 b + ra + σ2 a(a − 1) − r = 0, the coefficients σ2 of U and ∂U ∂x are zero. Upon canceling out 2 , we get the heat equation ∂U ∂2U x ax ∂τ = ∂x2 . Since F (e ) = F (s) = V (s, T ) = e U (x, 0), the initial −ax F (ex ). From the formula condition for the heat equation is U (x, 0) = e for solution of heat equation, we get Z ∞ 2 1 U (x, τ ) = √ e−ay F (ey )e−(x−y) /(4τ ) dy. 4πτ −∞ Making a change of variable y = log s∗ and substitute x = log s, we find Z ∞ ∗ ∗ 2 ds∗ 1 e−a log s F (s∗ )e−(log(s/s ) /(4τ ) ∗ . U (log s, τ ) = √ s 4πτ 0

61

and V (s, t) = √

1 4πτ

Z





∗ 2

ea log s+bτ e−a log s F (s∗ )e−(log(s/s

) /(4τ )

0

ds∗ . s∗

Adding up the exponents of e, we get a log(s/s∗ ) + bτ − (log(s/s∗ )2 /(4τ ) 1 = − [log2 (s/s∗ ) − 4aτ log(s/s∗ ) − 4bτ 2 ] 4τ 1 = − (log(s/s∗ ) − 2aτ )2 + (a2 + b)τ 4τ 2

Since 4τ = 2σ 2 (T − t), −2aτ = (r − σ2 )(T − t) and (a2 + b)τ = −r(T − t), we are done. 14. (Problem 4) If g is a smooth function on R, define the formal power series u(x, t) =

∞ X n=0

g (n) (t)

x2n (2n)!

(3)

(a) Check formally that u solves the heat equation. (b) For a > 0, consider the function defined by  −a e−t if t > 0 g(t) = 0 if t ≤ 0. One can show that there exists 0 < θ < 1 depending on a so that |g (k) (t)| ≤

k! − 1 t−a e 2 for t > 0. (θt)k

(c) As a result, for each x and t the series (3) converges; u solves the heat equation; u vanishes for t = 0; and u satisfies the estimate |u(x, t)| ≤ 2a/(a−1) Cec|x| for some constants C, c > 0. (d) Conclude that for every  > 0 there exists a nonzero solution to the heat equation which is continuous for x ∈ R and t ≥ 0, which satisfies 2+ u(x, 0) = 0 and |u(x, t)| ≤ Cec|x| . 15. (Problem 7) The Hermite functions hk (x) are defined by the generating identity ∞ X 2 2 tk hk (x) = e−(x /2−2tx+t ) . (4) k! k=0

(a) Show that an alternate definition of the Hermite functions is given by the formula  k 2 2 d hk (x) = (−1)k ex /2 e−x . (5) dx 62

Conclude from the above expression that each hk (x) is of the form Pk (x)e−x where Pk is a polynomial of degree k. In particular, the Hermite functions 2 2 belong to the Schwartz space and h0 (x) = e−x /2 , h1 (x) = 2xe−x /2 . ∞ (b) Prove that the family {hk }k=0 is complete in the sense that if f is a Schwartz function, and Z ∞ (f, hk ) = f (x)hk (x) dx = 0 for all k ≥ 0, −∞

then f = 0. (c) Define h∗k (x) = hk ((2π)1/2 x). Then c∗ (ξ) = (−i)k h∗ (ξ). h k k Therefore, each h∗k is an eigenfunction for the Fourier transform. (d) Show that hk is an eigenfunction for the operator defined in Exercise 23, and in fact, prove that Lhk = (2k + 1)hk . In particular, we conclude that the functions hk are mutually orthogonal for the L2 inner product R ∞on the Schwartz space. (e) Finally, show that −∞ hk (x)2 dx = π 1/2 2k k!. Solution. 2 2 2 2 (a) Note that e−(x /2−2tx+t ) = ex /2 e−(t−x) . Note that 

d dx

k

2

e−x = pk (x)e−x

2

for some degree k polynomial pk (x) which is odd if k is odd, and even if k 2 is even. Therefore the k partial derivative of e−(t−x) w.r.t. t, evaluated 2 2 at t = 0 is pk (−x)e−x = (−1)k pk (x)e−x . The formula then follows from Taylor expansion. (b) Under the hypothesis, we have by equation (4) Z ∞ Z ∞ 2 2 2 2 f (x)e−(x /2−2tx+t ) dx = e−t f (x)e−x /2+2tx dx −∞

−∞

for all t. By Exercise 8, f = 0. (c) From equation (4), we get ∞ X

√ tk hk ( 2πx) k! k=0 √ = exp(−(πx2 − 2t 2πx + t2 )) p = exp(t2 ) exp(−π(x − 2/πt)2 )

63

2

/2

,

Taking Fourier transform of both sides, we get ∞ X

k √ c∗ (ξ) t = exp(t2 ) exp(−2 2πitξ) exp(−πξ 2 ) h k k! k=0 √ = exp(−(πξ 2 − 2(−it) 2πξ + (−it)2 )) ∞ X (−it)k = h∗k (ξ) k! k=0

Comparing both sides, we get the result. (d) From the RHS of (4), we have ∞ X

L(hk (x))

k=0

2 2 tk = L(e−(x /2−2tx+t ) ) k!

2 2 d2 −(x2 /2−2tx+t2 ) e + x2 e−(x /2−2tx+t ) dx2 2 2 2 2 d = − e−(x /2−2tx+t ) (−x + 2t) + x2 e−(x /2−2tx+t ) dx 2 2 2 2 = −e−(x /2−2tx+t ) ((−x + 2t)2 − 1) + x2 e−(x /2−2tx+t )

= −

= e−(x

2

/2−2tx+t2 )

2

(1 − 4t2 + 4tx)

2

= e−(x /2−2tx+t ) + (−4t2 + 4tx)e−(x ∞ X 2 2 d tk = hk (x) + 2t e−(x /2−2tx+t ) k! dt = =

k=0 ∞ X

2

/2−2tx+t2 )



hk (x)

k=0 ∞ X

tk X tk + 2khk (x) k! k! k=0

(2k + 1)hk (x)

k=0

tk k!

from which we get Lhk = (2k + 1)hk . By Exercise 23, L = I + A∗ A, so it is Hermitian. Therefore the functions hk are mutually orthogonal. (e) Multiply both sides of equation (4) by hk (x) and integrate. On the LHS, using the orthogonality property, we get Z tk ∞ [hk (x)]2 dx. k! −∞ On the RHS, using equation (5) and integration by parts k times, we get (−1)k (−1)k

Z



−∞

2

e−x



d dx

k

2

e2tx−t = (2t)k

Z



−∞

Therefore result. 64

√ 2 e−(x−t) dx = 2k tk π.

Chapter 6 The Fourier Transform on Rd 1. (Exercise 1) Suppose that R is a rotation in the plane R2 , and let   a b c d denote its matrix with respect to the standard basis vectors e1 = (1, 0) and e2 = (0, 1). (a) Write the conditions Rt = R−1 and det(R) = ±1 in terms of equations in a, b, c, d. (b) Show that there exists φ ∈ R such that a + ib = eiφ . (c) Conclude that if R is proper, then it can be expressed as z 7→ ze−iφ , and if R is improper, then it takes the form z 7→ z¯eiφ , where z¯ = x − iy. Solution. (a) Write D = det(R). Then D = ±1 and the conditions Rt = R−1 implies     a c d/D −b/D = , b d −c/D a/D i.e. we have D = ad − bc = ±1, a = dD, b = −cD, c = −bD, d = aD. (b) From (a) we have D = ad − bc = a2 D − (−b2 D) = (a2 + b2 )D which implies that a2 + b2 = 1. Thus (b) follows. (c). Suppose D = 1. Then ze−iφ = (x + iy)(a − ib) = (ax + by) + i(−bx + ay) = (ax + by) + i(cx + dy). If D = −1, then z¯eiφ = (x − iy)(a + ib) = ax + by + i(bx − ay) = ax + by + i(cx + dy). This proves (c). 2. (Exercise 2) Suppose R : R3 → R3 is a proper rotation. (a) Show that p(t) = det(R − tI) is a polynomial of degree 3, and prove that there exists γ ∈ S 2 (where S 2 denotes the unit sphere in R3 ) with R(γ) = γ. (b) If P denotes the plane perpendicular to γ and passing through the origin, show that R : P → P, and that this linear map is a rotation. Solution. (a). p(t) is clearly a polynomial of degree 3. (The coefficient of t3 is −1.). p(0) = det(R) > 0. Since limt→∞ p(t) = −∞, we see that there exists λ > 0 such that p(λ) = 0. So R − λI is singular, and its kernel is nontrivial. Chapter 7 Finite Fourier Analysis 1. (Exercise 1.) Let f be a function on the circle. For each N ≥ 1 the discrete Fourier coefficients of f are defined by aN (n) =

N −1 N −1 1 X 1 X f (e2πik/N )e−2πikn/N = f (e2πik/N )en (k), for n ∈ Z. N N k=0

k=0

65

We also let Z a(n) =

1

f (e2πix )e−2πinx dx

0

denote the ordinary Fourier coefficients of f . (a) Show that aN (n) = aN (n + N ). (b) Prove that if f is continuous, then aN (n) → a(n) as N → ∞. Proof. (a) is easy since e−2πi = 1. (b) Note that aN (n) is the Riemann sum of the integral a(n) with partition: k/N, k = 0, · · · , N , and choice of points: k/N, k = 0, · · · , N − 1. 2. (Exercise 2) If f is a C 1 function on the circle, prove that |aN (n)| ≤ c/|n| whenever 0 < |n| ≤ N/2. Proof. Let l be an integer. Then aN (n)[1 − e2πiln/N ] =

=

=

=

N −1 N −1 1 X 1 X f (e2πik/N )e−2πikn/N − f (e2πik/N )e−2πi(k−l)n/N N N

1 N 1 N 1 N

k=0

k=0

N −1 X

NX −l−1

f (e2πik/N )e−2πikn/N −

k=0 N −1 X

f (e2πik/N )e−2πikn/N −

k=0

1 N 1 N

f (e2πi(k+l)/N )e−2πikn/N

k=−l N −1 X

f (e2πi(k+l)/N )e−2πikn/N

k=0

N −1 X

[f (e2πik/N ) − f (e2πi(k+l)/N )]e−2πikn/N

k=0

f ∈ C 1 implies f is Lipschitz. Let M be its Lipschitz constant. Then from the above equations, |aN (n)||1 − e2πiln/N | ≤ M |1 − e2πil/N | N Let l be the integer closest to 2n (in case of ambiguity, choose either integer). Then |l − N/2n| ≤ 1/2, from which we get ln 1 − ≤ |n| ≤ 1 . N 2 2N 4

Thus |1 − e2πil/N | ≤ |2πl/N | = |2πln/N |/|n| ≤ On the other hand, since ln 1 1 − ≤ N 2 4

66

3π . 2|n|

implies π 2πln 3π ≤ ≤ , 2 N 2 √ |1 − e2πiln/N | ≥ 2.

we have

Therefore for 0 < |n| ≤ N/2, 3πM |aN (n)| ≤ √ . 2 2|n| 3. (Exercise 3) By a similar method, show that if f is a C 2 function on the circle, then c |aN (n)| ≤ , whenever 0 < |n| ≤ N/2. |n|2 As a result, prove the inversion formula for f ∈ C 2 , ∞ X

f (e2πix ) =

a(n)e2πinx

n=−∞

from its finite version. Solution. We have aN (n)[2 − e2πiln/N − e−2πiln/N ] =

N −1 1 X [2f (e2πik/N ) − f (e2πi(k+l)/N ) − f (e2πi(k−l)/N )]e−2πikn/N N k=0



1 N

N −1 X

M |e2πi(k+l)/N − e2πik/N |2

k=0 2πil/N

= M |e

− 1|2

Then do as we did in Exercise 2 to choose l to reach the conclusion. For the second part, let N be odd, then for a fixed k, X aN (n)e2πikn/N |n|