Fourier Analysis, Stein and Shakarchi
Chapter 5 The Fourier Transform on R
Yung-Hsiang Huang*
2018.07.10
Abstract
Problem 3(c) need the theory of residues from complex analysis. Here we assume it and
modify the computations given in page 83 of Book II.
After Problem 6, we contain a proof for Widder’s uniqueness theorem in the class of
nonnegative solutions of heat equation in the strip ST := {(x, t) ∈ Rd × (0, T )}. It’s taken
from [4, Section 5.14]. In this proof, We need an interior gradient estimate for heat equation,
that is, Proposition 25 and refer its proof to [4, Section 5.12 and 5.12c].
We also combine Problem 7 and Book III’s Problem 4.11 on Hermite functions here.
I finish this solution file when I am a teaching assistant of the course “Analysis II” in
NTU 2018 Spring. The following students contribute some answers:
Problem 3(b)(e): Ge-Cheng Cheng.
Problem 3(c): Sam Wang.
Problem 8 is discussed with Shuang-Yan Li and Yi-Heng Tsai.
Exercises
1. Corollary 2.3 in Chapter 2 leads to the following simplified version of the Fourier
inversion formula. Suppose f is a continuous function supported on [−M, M ], whose
Fourier transform fˆ is of moderate decrease.
(a) Fix L with L/2 > M , and show that f (x) =
1
an (L) =
L
* Department
Z
P
an (L)e2πinx/L where
L/2
f (x)e−2πinx/L dx =
−L/2
1 ˆ
f (n/L).
L
of Math., National Taiwan University. Email: d04221001@ntu.edu.tw
1
P∞
Alternatively, we may write f (x) = δ
n=−∞
fˆ(nδ)e2πnδx with δ = 1/L.
(b) Prove that if F is continuous and of moderate decrease, then
Z ∞
∞
X
F (ξ) dξ = lim+ δ
F (δn).
δ→0
−∞
(c) Conclude that f (x) =
R∞
−∞
n=−∞
fˆ(ξ)e2πxξ dξ.
Proof. (c) is proved by taking F (ξ) = fb(ξ)e2πiξx in (b) and then apply (a).
P
For (a), we note that the Fourier series of f on [− L2 , L2 ] is
an (L)e2πinx/L . which converges
absolutely and hence uniformly since we have, by the moderate decreasing property of fb,
1
L
|an (L)| = | fˆ(n/L)| ≤ 2
.
L
L + n2
The uniqueness theorem of Fourier series and continuity of f imply the desired equality at
every point x.
Rn
F (ξ) dξ −
−n
(b) Given > 0, we have N = N () such that
R∞
−∞
F (ξ) dξ < whenever n ≥ N .
Note that there is a L > 0 such that for each L ≥ L and for each 0 < δ < 1,
Z ∞
X
X
δ
1
π
≤ 2A
dx = 2A( − arctan L) < .
|δ
F (δn)| ≤ Aδ
2
2
2
2
L
1+δ n
1+δ x
2
L+1
L
δ
|n|>
|n|> δ +1
δ
Therefore, for M =: max{N, L + 1} we have
Z
Z ∞
Z ∞
X
F (ξ) dξ −
F (ξ) dξ − δ
F (δn) ≤
−∞
+ δ
Z
X
F (δn) − δ
|n|≤ M
δ
M
F (ξ) dξ − δ
F (ξ) dξ +
−M
−M
−∞
n∈Z
M
X
X
F (δn)
|n|≤ M
δ
F (δn)
n∈Z
Z
M
≤ 2 +
F (ξ) dξ − δ
−M
X
F (δn) .
|n|≤ M
δ
Then by the uniform continuity of F on [−M, M ], there is a δ > 0 such that the last term is
less than whenever δ ∈ (0, δ ).
2. Let f and g be the functions defined by
f (x) = χ[−1,1] (x) and g(x) = χ[−1,1] (x) 1 − |x| .
Although f is not continuous, the integral defining its Fourier transform still make sense. It’s
easy to see that
sin πξ 2
sin 2πξ
b
f (ξ) =
and gb(ξ) =
πξ
πξ
with the understanding that fb(0) = 2 and gb(0) = 1. Note that one can interpret g as a
convolution of cfδ with itself, for some c, δ ∈ R and fδ (x) := f (δx).
2
3. The following exercise illustrates the principle that the decay of fb is related to the
continuity properties of f .
(a) Suppose that f is a function of moderate decrease on R whose Fourier transform
fb is continuous and satisfies
fb(ξ) = O
1 as |ξ| → ∞
|ξ|1+α
for some 0 < α < 1. Prove that f satisfies a Hölder condition of order α.
(b) Let f be a continuous function on R which vanishes for |x| ≥ 1, with f (0) = 0,
and which is equal to 1/ log(1/|x|) for all x in a neighborhood of the origin. Prove
that fb is not of moderate decrease. In fact, there is no > 0 so that fb(ξ) = O(1/|ξ|1+ )
as |ξ| → ∞.
Proof. (a) Since fb ∈ L1 (R) ∩ C 0 (R), we have, for each |h| > 0
Z
fb(ξ)[e2πi(x+h)ξ − e2πixξ ] dξ
|f (x + h) − f (x)| =
R
Z
Z
2πi(x+h)ξ
2πixξ
b
≤
f (ξ)[e
−e
] dξ +
fb(ξ)[e2πi(x+h)ξ − e2πixξ ] dξ
−1
−1
|ξ|>|h|
|ξ|≤|h|
Z ∞
Z
≤2
2C|ξ|−1−α dξ +
|fb(ξ)|2 sin(π|hξ|) dξ.
|h|−1
|ξ|≤|h|−1
By the hypotheses, there is for some M > 0 such that |fb| ≤ M/(1 + |ξ|1+α ) for all ξ ∈ R
Z |h|−1
sin(π|hξ|)
α
|f (x + h) − f (x)| ≤ 4Cα|h| + 2M
dξ
1 + |ξ|1+α
0
Z 1
sin(πu)
α
α
≤ 4Cα|h| + 2M |h|
du
u1+α
0
Z 1
1
α
α
0
≤ 4Cα|h| + 2M |h| M
du = K|h|α ,
α
u
0
where K is independent of h.
(b) The continuity of fb is a general fact. Suppose fb satisfies the decay condition for some
1 > > 0, then f is -Hölder, especially near the origin. However one find that as |h| → 0,
|
f (h) − f (0)
|h|−
|
=
→ ∞.
h
log |h|−1
This is a contradiction.
4. Examples of compactly supported functions in S(R) are very handy in many applications in analysis. Some examples are:
3
(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].
Proof. We may assume a > 0 (why?). (a) is easy (by mathematical induction).
Rx
R∞
For (b) consider F (x) = c −∞ f (t) dt where c is the reciprocal of −∞ f (t) dt.
√ √
For (c) we first observe that e
h(x) = 1 − F (x2 ) is a function that equals to 1 on [− a, a],
√
√
√
decreasing on { a ≤ |x| ≤ b} and vanishes outside {|x| ≥ b}. After a suitable scaling and
translation, we have a desired one.
5. Suppose f is continuous on R and of moderate decrease.
(a) Prove that fb is uniformly continuous and vanishes at infinity.
(b) Show that if fb(ξ) = 0 for all ξ, then f is identically 0.
Proof. (a) (1st proof, without LDCT) Given > 0, pick large N such that 2A
R
1
|x|>N 1+x2
dx <
/2 where A > 0 is the moderate constant. Then
Z
Z
| sin πhx|
−2πixh
b
b
|f (ξ + h) − f (ξ)| ≤
|f (x)||e
− 1| dx ≤ 2A
dx
2
R
R 1+x
Z N | sin πhx| Z
1
+
dx
<
≤ 2A
2
2
−N 1 + x
|x|>N 1 + x
provided h is small enough such that | sin πhx| <
8AN
for all x ∈ [−π, π]. Note that the
smallness for h is independent of ξ, so the continuity is uniform.
(2nd proof, with LDCT.) The continuity is directly from LDCT. Note that
Z
Z
1
−2πixξ
b
f (ξ) =
f (x)e
dx =
−f (y − )e−2πiyξ dy.
2ξ
R
R
R So fb(ξ) = 21 R f (x) − f (x − 2ξ1 ) e−2πixξ dx and hence the LDCT implies the decay at ∞.
(b) Given g ∈ S(R), then by Fubini’s theorem
Z
Z
Z
Z
Z
Z
−2πiyx
−2πiyx
f (x)b
g (x) dx =
f (x) g(y)e
dy dx =
g(y) f (x)e
dx dy =
g(y)fb(y) dy = 0.
R
R
R
R
4
R
R
2
In particular, for each y ∈ R, we take gy (ξ) = e−πξ δ e2πiξy whose Fourier transform is gby (x) =
2 /δ
e−π(x−y)
δ −1/2 . Letting δ → 0, we have f (y) = 0.
Remark 1. See Exercise 2.22 (Riemann-Lebesgue lemma) and 2.25 (the decay rate of fb may
be slow as |ξ|− for any ) in Book III. Also consult [18, Section I.4.1] or [15, Exercise 1.6] for
showing the Fourier transform map from L1 (Rd ) to C0 (Rd ) is not surjective. For L1 (T) case,
see Section 3.1 of Chapter 4 in Book IV.
Remark 2. For functions on T, one can construct functions with arbitrary slow decay for their
Fourier transform, see [9, Section 3.3.1].
2
6. The function e−πx is its own Fourier transform. Generate other functions that (up
to a constant multiple) are their own Fourier transforms. What must the constant
multiples be? To decide this, prove that F 4 = I. Here F(f ) = fb is the Fourier
transform, F 4 = F ◦ F ◦ F ◦ F, and I is the identity operator (If )(x) = f (x) (see also
Problem 7).
Proof. If F(g) = cg, then c4 = 1 since F 4 = I. So c ∈ {1, −1, i, −i}. The case c = 1 is solved
2
in the statement with g(x) = e−πx . Note that g 0 = −2πxg(x) and hence gb0 (x) = 2πixb
g (x) =
2
2πixg(x) = −ig 0 (x), that is, the case c = −i has an eigenfunction g 0 (x) = −2πxe−πx .
Furthermore, we differentiate g 0 again and see that g 00 (x) = −2πg(x) + 4π 2 x2 g(x) and hence
gb00 (x) = −4π 2 x2 gb(x) = −4π 2 x2 g(x) = −2πg(x) − g 00 (x), that is, the case c = −1 has an
00 + πg(x) = −(πg(x) + g 00 (x)).
eigenfunction since g\
Finally, we differentiate the g 00 once again and see that g 000 (x) = −2πg 0 (x) + 8π 2 xg(x) +
4π 2 x2 g 0 (x) = −6πg 0 + 4π 2 x2 g 0 (x). Taking Fourier transform to both sides, we have
g 000\
+ 3πg 0 (ξ)
d2 b0
0
b
= −3π g (ξ) − 2 g = i(3πg 0 + g 000 )(ξ),
dξ
that is, g 000 + 3πg 0 is an eigenfunction for c = i.
Remark 3. Note that these four eigenspaces E1 , E−1 , Ei , E−i span all of L2 (Rd ), that is, given
f ∈ L2 (Rd ) we are searching fk ∈ Ek for k = ±1, ±i such that f = f1 + f−1 + fi + f−i . To find
these fk , one can solve this equation with the following three equations through linear algebra
(matrix): Ff = f1 − f−1 + ifi − if−i , F 2 f = f1 + f−1 − fi − f−i , F 3 f = f1 − f−1 − ifi + if−i .
This decomposition of L2 can be used to give new examples of eigenfunctions for F, e.g. the
previous f1 should be (f +Ff +F 2 f +F 3 f )/4 and now we take f (x) = e−2π|x| (so Ff (x) =
5
1 1
π 1+x2
and F 2 f = f . Therefore 2f1 (x) = e−2π|x| +
1 1
π 1+x2
is also an eigenfunction for c = 1 different
from the one given in the exercise. I learned this perspective from [12, page 338-339].
7. Prove that the convolution of two functions of moderate decrease is a function of
moderate decrease.
Proof. Let f, g be functions of moderate decrease. Then the continuity of f ∗ g can be proved
by LDCT if one realize the integral in Lebesgue’s sense, or using an standard decomposition of
the integral as ”local” part and ”far-away” part like the following proof for moderate decrease
if one realize the integral in Riemann’s sense, we leave the details to the readers.
On the other hand, the moderate decreasing property can be proved as follows:
Z
Z
|f ∗ g(x)| ≤
|f (t)||g(x − t)| dt +
|f (t)||g(x − t)| dt
|x|
|x|
|t|≤ 2
|t|> 2
Z
Z
cg
cf
M
≤
|f (t)| dt
+ |g(x − t)| dt
≤
.
|x|2
|x|2
1 + |x|2
1+ 4
1+ 4
R
R
8. Prove that if f is continuous, of moderate decrease, and
R
R
2
f (y)e−y e2xy dy = 0 for all
x ∈ R, then f ≡ 0.
2
2 R
2
Proof. We know [f ∗ e−x ](z) = e−z R f (y)e−y e2zy dy = 0 for all z ∈ R. So fb ≡ 0 since
−x2 (ξ) = f\
fb(ξ)ed
∗ e−x2 (ξ) = 0 for all ξ ∈ R. Thus f ≡ 0 by Fourier inversion theorem.
Remark 4. This is true for all L1 (Rd ) or L2 (Rd ) functions.
9. If f is of moderate decrease, then
Z R
|ξ| b
1−
f (ξ)e2πixξ dξ = (f ∗ FR )(x),
R
−R
where the Fej́er kernel on the real line is defined by FR (t) = R
sin πtR
πtR
2
if t 6= 0 and
FR (0) = R. Show that {FR } is a family of good kernels as R → ∞, and therefore
(f ∗ FR ) tends uniformly to f (x) as R → ∞. This is the analogue of Fej́er’s theorem
for Fourier series in the context of the Fourier transform.
Proof. Note that for all R > 0
Z
Z
Z
Z
2 ∞ sin2 x
2 ∞ sin z
|FR (t)| dt =
FR (t) dt =
dx =
dz = 1
π 0
x2
π 0
z
R
R
6
and hence for all δ > 0
Z
2
FR (t) dt =
π
[−δ,δ]c
Z
∞
πRδ
sin2 x
dx → 0 as R → ∞
x2
10. Below is an outline of a different proof of the Weierstrass approximation theorem.
Dfine the Landau kernels by
Ln (x) =
(1 − x2 )n
χ|x|≤1 .
cn
where cn is chosen so that inf ∞
−∞ Ln (x) dx = 1. Prove that {Ln }n≥0 is a family of good
kernels as n → ∞. As a result, show that if f is a continuous function supported
in [−1/2, 1/2], then (f ∗ Ln )(x) is a sequence of polynomials on [−1/2, 1/2] which
converges uniformly to f .
R1
R1
R1
Proof. Note that cn = −1 (1 − x2 )n dx = 2 0 (1 − x2 ) dx ≥ 2 0 x(1 − x2 )n dx =
R
each δ > 0, δ≤|x| Ln (x) dx ≤ (1 − δ 2 )n (n + 1) → 0 as n → ∞.
1
.
n+1
So for
11. Suppose that u is the solution to the heat equation given by u = f ∗ Ht where
f ∈ S(R). If we also set u(x, 0) = f (x), prove that u is continuous on the closure of
the upper half-plane, and vanishes at infinity, that is, u(x, t) → 0 as |x| + t → ∞.
Remark 5. We should not only consider the tangential limit, i.e. (x0 , t) → (x0 , 0), but also
the nontangential limit (x, t) → (x0 , 0). However, the proof is almost the same idea. Also, the
condition on f can be weaken to f ∈ C(R) and of moderate decreasing.
Proof. By the even weaker hypothesis stated in the remark, f is uniformly continuous. So for
every > 0, there is δ > 0 such that |f (y) − f (x0 )| < whenever |y − x0 | < δ. Note that for
each x with |x − x0 | < 2δ , we have {y : |x − y| < 4δ } ⊂ {|x0 − y| < δ}, and hence
Z h
i
|u(x, t) − f (x0 )| =
f (y) − f (x0 ) Ht (x − y) dy
R
Z
Z
2Ht (x − y) dy + 2kf k∞
Ht (x − y) dy.
≤
|x−y|< 4δ
|x−y|≥ 4δ
And hence there is there is t > 0 independent of x such that |u(x, t) − f (x0 )| ≤ 3 whenever
t ∈ (0, t ) and |x − x0 | < 4δ .
7
Note that |u(x, t)| ≤ kf kL1 (R) (4πt)−1 for all x ∈ R, t > 0. So it tends to zero as |x| + t → ∞
and |x| ≤ 2t. For the case |x| > t, we note that
Z
1
2
|u(x, t)| ≤
|f (y)|(4πt)− 2 e−|x−y| /4t dy
ZR
Z
1
2
− 12 −|x−y|2 /4t
=
|f (y)|(4πt) e
dy +
|f (y)|(4πt)− 2 e−|x−y| /4t dy
|x|
|x|
|x−y|> 2
|x−y|≤ 2
Z
Z
1
M
2
− 21 −|x−y|2 /4t
≤ (4πt)− 2 e−|x| /16t
|f (y)| dy +
(4πt)
e
dy
2
|x|
|x|
|x−y|≤ 2 1 + |y|
|x−y|> 2
1
M
2
≤ (4πt)− 2 e−|x| /16t kf kL1 (R) +
2 .
1 + |x|4
If |x| + t → ∞ with t ≥ for some > 0, then |x| → ∞ (since we assume |x| > t) and
1
|u(x, t)| ≤ (4π)− 2 e−|x|/16 +
4M
→ 0 as |x| → ∞.
4 + |x|2
If |x| + t → ∞ with t → 0+ , then |x| → ∞ and
1
|u(x, t)| ≤ (4πt)− 2 e−1/16t +
4M
→ 0 as |x| → ∞, t → 0+ .
4 + |x|2
12. Show that the function
u(x, t) =
1
x
−x2 /4t
e
t (4πt)1/2
(1) satisfies the 1-d heat equation for t > 0, (2) limt→0 u(x, t) = 0 for every x ∈ R,
(3) lim(x,t)→(x0 ,0) u(x, t) = 0 for every x0 6= 0 and (4) u is not continuous at the origin.
Remark 6. Hence, it can not serve as an example to the non-uniqueness phenomenon of the
heat equation in R × (0, T ). An exact counterexample is given in Problem 4.
Proof. (1)(2) are trivial. (3) is a consequence of Lopital’s rule and standard squeeze theorem.
(4) u(x, x2 ) =
√ 1 e−1/4
4πx2
→ ∞ as x → 0.
13. Prove the following uniqueness theorem for harmonic functions in the strip S =
{(x, y) : 0 < y < 1, −∞ < x < ∞}: if u is harmonic in S, continuous on S with
u(x, 0) = u(x, 1) = 0 for all x ∈ R, and u vanishes at infinity, then u = 0.
Proof. For each > 0, one can use maximum modulus principle (Mean-Value theorem is proved
in the textbook) to conclude |u| ≤ on S. So u ≡ 0.
We sketch another way which is similar to the proof for Theorem 2.7. Suppose not, we may
assume M = sup u > 0. Hence u(x1 , y1 ) = M for some (x1 , y1 ) ∈ S. By the uniform continuity
8
of u and mean-value property with ball Br min{|y1 |,1−|y1 |} (x1 , y1 ), we have a contradiction that
M ≤ M − δ for some δ > 0 if r is close to 1.
14. Prove that the periodization of the Fejér kernel FN on the real line (Exercise 9) is
equal to the Fejér kernel for periodic functions of period 1. In other words,
∞
X
FN (x + n) = FN (x),
n=−∞
when N ≥ 1 is an integer, and where
FN (x) =
N
X
n=−N
cR (ξ) = (1 −
Proof. Since F
|n| 2πinx
1 sin2 (N πx)
1−
e
=
.
N
N sin2 (πx)
|ξ|
)χ[−R,R] (ξ)
R
for all R > 0, ξ ∈ R. The desired result follows from
Poisson summation formula.
15. This exercise provides another example of periodization.
(a) Apply the Poisson summation formula to the function g in Exercise 2 to obtain
∞
X
1
π2
=
(n + α)2
(sin πα)2
n=−∞
whenever α is real, but not equal to an integer.
(b) Let α ∈ R \ Z. Prove as a consequence that
∞
X
1
π
=
n+α
tan πα
n=−∞
where the series is defined through the symmetric partial sum about bαc, the
largest integer ≤ α. (Note that this series is not absolutely convergent, so we need
to assign the order of summation here.)
Remark 7. Other proofs are given in Exercise 3.9 and Book II’s Exercise 3.12.
Proof. (a) Poisson summation formula applied to f = gb and x = α implies that
∞
X
∞
∞
X
X
sin2 πα
sin2 (π(n + α))
=
=
g(−n)e2πinα = 1.
2 (n + α)2
2 (n + α)2
π
π
n=−∞
n=−∞
n=−∞
(b) Assume that 0 < α < 1 first. Note that for n ∈ N,
Z α
1
1
1
1
+
=−
+
dx
2
n + α −n + α
(−n + x)2
0 (n + x)
9
Note that
P
1
n6=0 (n+x)2
converges uniformly on (0, 1) by M -test with supx∈(0,1) (n + x)−2 = n−2
and supx∈(0,1) (−n + x)−2 = (n − 1)−2 for all n ≥ 2. Therefore
∞
α
Z α
1
π2
1
1
−
lim
−
dx
=
dx
2
2
α →0+ sin πx x2
0 n6=0 (n + x)
π
1
π
1
π
1 +
+ lim+ −
+
=
.
= − −
→0
α
tan πα α
tan π tan πα
1 X 1
1
1
+
+
= −
α n=1 n + α −n + α
α
Z
X
Finally, for arbitrary α ∈ R, α = bαc + α1 , where 0 < α1 < 1. So by the definition of this series
and previous result
∞
X
1
π
π
1
=
=
=
n + α n=−∞ n + α1
tan πα1
tan πα
n=−∞
∞
X
16. The Dirichlet kernel on the real line is defined by
Z R
sin(2πRx)
fb(ξ)e2πixξ dξ = (f ∗ DR )(x) so that DR (x) = χ\
.
[−R,R] (x) =
πx
−R
Also, the modified Dirichlet kernel for periodic functions of period 1 is defined by
X
∗
DN
(x) =
|n|≤N −1
1
e2πinx + (e−2πiN x + e2πiN x ).
2
Show that the result in Exercise 15 gives
∞
X
∗
DN (x + n) = DN
(x),
n=−∞
where N ≥ 1 is an integer, and the infinite series must be summed symmetrically.
∗
In other words, the periodization of DN is the modified Dirichlet kernel DN
. Also
show that if R ∈ R+ \ N, then
∞
X
DR (x + n) = D[R] (x),
n=−∞
where [R] is the largest integer ≤ R.
Remark 8. Corresponding to Problem 3.1, one can define modified conjugate Dirichlet kernel
e ∗ similarly ,that is,
D
N
∗
eN
D
(x) =
X
|n|≤N −1
1
sign(n)e2πinx + (−e−2πiN x + e2πiN x ).
2
10
Proof. Note that we can’t apply Poisson summation formula since DR is not in L1 (R) (so the
inversion formula breaks down). So we go back to the symmetric partial sum:
Z R
Z R
L
L
L Z R
X
X
X
−2πiξx
−2πiξk
−2πiξ(x+k)
e−2πiξx DL (ξ) dξ
e
e
dξ =
DR (x + k) =
e
dξ =
k=−L
X
=
−R
−R
k=−L
e−2πiξm
+e
−R
k=−L
fx (0 − ξ)DL (ξ) dξ + e−2πi[R]x
Z
− 12
m≤[R]−1
2πi[R]x
1
2
Z
Z
1
2
R−[R]
fx (0 − ξ)DL (ξ) dξ
− 12
fx (0 − ξ)DL (ξ) dξ
−R+[R]
where fx (s) := e2πiξs . So the Tauberian theorem for Cesaro or Abel sum (see Exercise 2.14 and
Problem 2.3) of Fourier series of fx (s), fx (s)χ(− 1 ,δ) (s) and fx (s)χ(−δ, 1 ) (s) (δ ≥ 0) implies that
2
2
as L → ∞,
Z 1
Z
2
fx (0−ξ)DL (ξ) dξ → f (0) = 1 if δ > 0 and
δ
1
2
fx (0−ξ)DL (ξ) dξ →
0
f (0+) + f (0−)
1
= .
2
2
Another case is similar, so we complete the proof.
17. The gamma function is defined for s > 0 by
Z
Γ(s) =
∞
e−x xs−1 dx.
0
(a) One can easily show that for s > 0 the above integral makes sense, that is, the following
two limits exist:
Z
lim+
δ→0
1
Z
−x s−1
e x
dx and lim
A→∞
δ
A
e−x xs−1 dx.
1
(b) One then can use integration by parts to prove Γ(s + 1) = sΓ(s) whenever s > 0, and
√
conclude that (1) for every integer n ≥ 1 we have Γ(n + 1) = n!; (2) Γ( 12 ) = π and Γ( 32 ) = π2
easily.
P
1
18. The zeta function is defined for s > 1 by ζ(s) = ∞
n=1 ns . Verify the identity
Z
1 ∞ s −1
−s/2
π
Γ(s/2)ζ(s) =
t 2 (ϑ(t) − 1) dt whenever s > 1
2 0
P
−πn2 s
where Γ and ϑ are the gamma and theta functions, respectively. ϑ(s) := ∞
.
n=−∞ e
More about the zeta function and its relation to the prime number theorem can
be found in Book II.
Proof. For s > 1
Z ∞
Z
s
−1
t 2 (ϑ(t) − 1) dt =
0
∞
t
s
−1
2
(
0
=2
∞
X
n=1
∞
X
−πn2 t
e
− 1) dt = 2
n=−∞
s
π − 2 n−s
Z
0
11
∞ Z
X
n=1
∞
s
∞
2
s
e−πn t t 2 −1 dt
0
e−z z 2 −1 dz = 2π −s/2 Γ(s/2)ζ(s),
where the interchange of the order of sum and integration is permitted by Monotone convergence
theorem (if one use Lebesgue integral) or an careful argument to uniform convergence and
improper Riemann integral (interchange twice, one is for integral over [0, M ] and sum, another
one is for limM →∞ and sum)
19. The following is a variant of the calculation of ζ(2m) =
P∞
n=1
1/n2m found in Problem
4, Chapter 3.
(a) Applying the Poisson summation formula to f (x) = t/(π(x2 +t2 )) and fb(ξ) = e−2πt|ξ|
where t > 0, we get
∞
∞
X
1 X
t
=
e−2πt|n| .
π n=−∞ t2 + n2 n=−∞
(b) Prove the following identity valid for 0 < t < 1:
∞
∞
t
1 X
1
2X
(−1)m+1 ζ(2m)t2m−1 .
=
+
π n=−∞ t2 + n2
πt π m=1
Also note the following trivial fact
∞
X
e−2πt|n| =
n=−∞
2
− 1.
1 − e−2πt
(c) Use the fact that
∞
z X B2m 2m
z
=
1
−
+
z ,
ez − 1
2 m=1 (2m)!
where Bk are the Bernoulli numbers to deduce from the above formula,
2ζ(2m) = (−1)
m+1 (2π)
2m
(2m)!
B2m .
Proof. (b) For 0 < t < 1,
∞
X
n=1
=
∞
∞
∞
∞ X
∞
2m−1
X
X
X
t
t
1
t X
m t 2m
m+1 t
=
=
(−1)
(
)
=
(−1)
t2 + n2
n2 (t/n)2 + 1 n=1 n2 m=0
n
n2m
n=1
n=1 m=1
∞
X
(−1)m+1 t2m−1
m=1
∞
X 1
X
=
(−1)m+1 t2m−1 ζ(2m),
2m
n
n=1
m=1
P
P
where the Fubini’s theorem, that is, the interchange of the order of the sums n and m is
P
2m−1
permitted by the absolute convergence of the double series n,m (−1)m+1 t n2m which can be
proved easily.
12
(c) Note that for z = −2πt, (0 < t < 1)
∞
∞
X
X
B2m (2π)2m 2m
−2πt
t = −2πt
− 1 − πt = −1 + πt
e−2πt|n|
(2m)!
e
−
1
m=1
n=−∞
∞
∞
∞
X
X
1 X
t
m+1
2m−1
= −1 + πt
= −1 + 1 + 2t
(−1)
ζ(2m)t
=
2(−1)m+1 ζ(2m)t2m .
2
2
π n=−∞ t + n
m=1
m=1
By the uniqueness of power series (see Baby Rudin’s Theorem 8.5),
B2m (2π)2m
= 2(−1)m+1 ζ(2m).
(2m)!
20. The following results are relevant in information theory when one tries to recover
a signal from its samples.
Suppose f is of moderate decrease and that its Fourier transform fb is supported
in I = [−1/2, 1/2]. Then, f is entirely determined by its restriction to Z. This
means that if g is another function of moderate decrease whose Fourier transform
is supported in I and f (n) = g(n) for all n ∈ Z, then f = g. More precisely:
(a) Prove that the following reconstruction formula holds:
f (x) =
∞
X
f (n)K(x − n) where K(y) =
n=−∞
sin πy
πy
Note that K(y) = O(1/|y|) as |y| → ∞.
(b) If λ > 1, then
∞
X
1 n
n
cos πy − cosπλy
f (x) =
f ( )Kλ (x − ) where K(y) =
.
2 (λ − 1)y 2
λ
λ
λ
π
n=−∞
Thus, if one samples f ”more often,” the series in the reconstruction formula
converges faster since Kλ (y) = O(1/|y|2 ) as |y| → ∞. Note that Kλ (y) → K(y) as
λ → 1.
(c) Prove that
R∞
−∞
|f (x)|2 dx =
P∞
n=−∞
|f (n)|2 .
Remark 9. Compare with Exercise 7.8.
Remark 10. This is the well-known Shannon sampling theorem. See the survey article [1]. Its
relation to continuous wavelet transform can be found in [3, Chapter 2]
Proof. (a) Using Poisson summation formula for g(x) = fb(x), we have
fb(ξ) = χ[− 1 , 1 ] (ξ)
X
2 2
fb(ξ + n) = χ[− 1 , 1 ] (ξ)
X
2 2
n∈Z
n∈Z
13
f (n)e2πi(−n)ξ .
Hence
1
2
Z
fb(ξ)e2πiξx dξ =
f (x) =
− 12
Z
1
2
X
f (n)e2πiξ(x−n) dξ =
− 12 n∈Z
X
f (n)
n∈Z
sin πξ(x − n)
,
π(x − n)
where the interchange of sum and integration is promised by the uniform convergence with
upper bound |f (n)| ≤
M
1+n2
from the moderate decreasing.
(b) It’s the same idea: Using Poisson summation formula with g(x) = fb(λx) and the characteristic function χ[− 1 , 1 ] there is replaced by the function φ(ξ) describe in the Figure 2. We omit
2 2
Rλ
n
the brutal computation for −2λ φ(ξ)e2πiξ(x− λ ) dξ.
2
(c) By Planchel’s theorem and Poisson summation formula, we have
Z
2
Z
|f (x)| dx =
1
2
− 12
R
Z
1
2
X
=
− 12
|fb(ξ)|2 dξ =
Z
− 12
2πi(m−n)ξ
f (n)f (m)e
1
2
X
n∈Z
dξ =
(m,n)∈Z2
f (n)e2πi(−n)ξ
X
f (m)e2πimξ dξ
m∈Z
X
Z
1
2
f (n)f (m)
(m,n)∈Z2
e2πi(m−n)ξ dξ =
− 12
X
|f (n)|2 ,
n∈Z
where the third and fourth equalities are due to the absolute (and hence uniform) convergence
of the series with bound |f (n)f (m)| ≤
M2
.
(1+n2 )(1+m2 )
21. Suppose that f is continuous on R. Show that f and fb cannot both be compactly
supported unless f = 0. This can be viewed in the same spirit as the uncertainty
principle.
[Hint: Assume f is supported in [0, 1/2]. Expand f in a Fourier series in the interval
[0, 1], and note that as a result, f is a trigonometric polynomial.]
Proof. Suppose f is compactly supported in [0, 21 ] and fb is also compactly supported (say, in
[−M, M ] for some M > 0.
R
P
P
2πinx
2πinx
b
b
As hint, f (x) ∼
=
where fb(n) = R f (t)e−2πint dt =
n∈Z f (n)e
|n|≤M f (n)e
R1
f (t)e−2πint .
0
So the series is an trigonometric polynomial that converges on R.
The uniqueness of Fourier series implies f equals to this polynomial. Unless f ≡ 0, f can have
a finite number of roots in [0, 1] and hence cannot be compactly supported in [0, 21 ].
Remark 11. The assumption on widehatf can be weaken to the exponential decay, that is,
|widehatf (ξ)| . e−k|ξ| . The idea of the proof is: first, we use inversion formula to see the
smoothness of f and convergence of its power series with a fix radius of convergence that is at
14
least
k
2π
everywhere. Thus, it is real analytic over R, and cannot have compact support unless
it is identically 0.
22. The heuristic assertion stated before Theorem 4.1 can be made precise as follows.
If F is a function on R, then we say that the preponderance of its mass is contained
in an interval I (centered at the origin) if
Z
Z
1
2
2
x |F (x)| dx ≥
x2 |F (x)|2 dx
2
I
R
(1)
Now suppose f ∈ S, and (1) holds with F = f and I = I1 ; also with F = fb and I = I2 .
Then if Lj denotes the length of Ij , we have
L1 L2 ≥
1
.
2π
A similar conclusion holds if the intervals are not necessarily centered at the origin.
Proof. We may assume kf k2 = 1 and hence kfbk2 = 1 by Planchel’s theorem. So Theorem 1.4
implies the desired result as follows:
Z
Z
Z
L 2 L 2 Z
1
2
2
2
2
2
2
b
(L1 L2 ) = 16
|f (x)| dx |f (ξ)| dξ ≥ 16
x |f (x)| dx
|ξ|2 |fb(ξ)|2 dξ
2
2
R
I1
I2
Z
ZR
1
≥ 4 x2 |f (x)|2 dx |ξ|2 |fb(ξ)|2 ≥ 2 .
4π
R
R
23. The Heisenberg uncertainty principle can be formulated in terms of the operator
2
d
2
L = − dx
2 + x , which acts on Schwartz functions by the formula
L(f ) = −
d2 f
+ x2 f.
dx2
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 ) ∀f ∈ S.
This is usually denoted by L ≥ I.
15
(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) (A∗ Af, f ) = (Af, Af ) ≥ 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 another proof of the Heisenberg uncertainty
R∞
principle which says that whenever −∞ |f (x)|2 dx = 1 then
Z ∞ df 2 1
Z ∞
2
2
x |f (x)| dx
dx ≥ .
4
−∞ dx
−∞
Remark 12. See Problem 7 and [6, Chapter 6].
Proof. (a) By Theorem 1.4 and arithmetic-geometric mean inequality
Z
Z
df 2
2
2
4π 2 x2 |fb(x)|2 + x2 |f (x)|2
+ x |f (x)| =
(Lf, f ) =
dx
R
R
Z
1/2 Z
1/2
1
2 2 b
2
2
2
≥2
4π x |f (x)|
x |f (x)|
≥ 4π (f, f ).
4π
R
R
(b) (i)(ii) Use the integration by parts, (iii) is obvious.
R
R
R df
(c) Note that 0 ≤ (At f, At f ) = (A∗t At f, f ) = t2 R x2 |f (x)|2 dx − t R |f (x)|2 dx + R | dx
(x)|2 dx.
Then we complete the proof through the basic algebra that
Z df
Z
2
Z
2
2
2
| (x)|2 ≤ 0.
− |f (x)| dx − 4
x |f (x)| dx
R
R
R dx
Problems
1. The equation
x2
∂ 2u
∂u
∂u
+ ax
=
2
∂x
∂x
∂t
16
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 this equation, make the 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
∂ 2U
∂U
∂U
=
,
+ (1 − a)
2
∂y
∂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
R∞
2 2
integral −∞ e(−4π ξ +(1−a)2πiξ)t e2πiξv dξ. Show that the solution of the original problem
is then given by
1
u(x, t) =
(4πt)1/2
Z
∞
2 /(4t)
e−(log(v/x)+(1−a)t)
f (v)
0
dv
.
v
Proof. We omit the easy proof for equivalence of these two equation. Taking Fourier transform
in y to the equation, we have
b (ξ, t) = ∂t U
b (ξ, t)
(−4π 2 ξ 2 + (1 − a)2πiξ)U
b (ξ, t) = e(−4π2 ξ2 +2πi(1−a)ξ)t Fb(ξ) = where G(y, t) is determined by the inversion formula,
So U
that is,
Z
G(y, t) =
(−4π 2 ξ 2 +2πi(1−a)ξ)t 2πiξy
e
e
Z
R
Hence U (y, t) =
e−4π
dξ =
2 ξ2 t
e2πiξ(t(1−a)+y) dξ = √
R
R
R
1 −(t(1−a)+y)2 /(4t)
e
.
4πt
1
e−(t(1−a)+y−s)/(4t) ds and then (with s = − log v and y = − log x)
F (s) √4πt
Z ∞
1
dv
2
u(x, t) =
e−(log(v/x)+(1−a)t) /(4t) f (v) .
1/2
(4πt)
v
0
2. The Black-Scholes equation from finance theory is
∂V
∂V
σ 2 s2 ∂ 2 V
+ rs
+
− rV = 0, 0 < t < T,
∂t
∂s
2 ∂s2
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
(T
2
− t), a =
1
2
−
r
,
σ2
and b = −( 12 +
r 2
)
σ2
reduces the
equation to the one-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
e−r(T −t)
−
2σ 2 (T −t)
V (s, t) = p
e
F (s∗ ) ds∗ .
2πσ 2 (T − t) 0
17
3. The Dirichlet problem in a strip. Consider the equation ∆u = 0 in the horizontal
strip
{(x, y) : 0 < y < 1; −∞ < x < ∞}
with boundary conditions u(x, 0) = f0 (x) and u(x, 1) = f1 (x), where f0 and f1 are both
in the Schwartz space.
(a) Show (formally) that if u is a solution to this problem, then
u
b(ξ, y) = A(ξ)e2πξy + B(ξ)e−2πξy .
Express A and B in terms of fb0 and fb1 , and show that
u
b(ξ, y) =
sinh(2π(1 − y)ξ) b
sinh(2πyξ) d
f0 (ξ) +
f1 (ξ).
sinh(2πξ)
sinh(2πξ)
(b) Prove as a result that we have L2 -convergence to the boundary conditions, that
is,
Z
∞
|u(x, y) − f0 (x)|2 dx → 0 as y → 0
−∞
and
Z
∞
|u(x, y) − f1 (x)|2 dx → 0 as y → 1.
−∞
(c) If Φa (ξ) = (sinh 2πaξ)/(sinh 2πξ), with 0 ≤ a < 1, then Φa is the Fourier transform
of ϕa where
ϕa (x) =
1
sin πa
·
.
2
cosh πx + cos πa
This can be shown, for instance, by using contour integration and the residue
formula from complex analysis (see Book II, Chapter 3).
(d) Use this result to express u in terms of Poisson-like integrals involving f0 and
f1 as follows:
sin πy u(x, y) =
2
Z
∞
−∞
f0 (x − t)
dt +
cosh πt − cos πy
Z
∞
−∞
f1 (x − t)
dt .
cosh πt + cos πy
(e) Finally, one can check that the function u(x, y) defined by the above expression
is harmonic in the strip, and converges uniformly to f0 (x) as y → 0, and to f1 (x) as
y → 1 which are improvements of (b). Moreover, one sees that u(x, y) vanishes at
infinity, that is, lim|x|→∞ u(x, y) = 0, uniformly in y.
Remark 13. The identity in (c) can also be applied to prove the sum of two squares theorem,
see Section 3.1 of Chapter 10 in Book II.
18
Proof. We omit (a)(d) since they are easy.
(b) By Planchel’s theorem,
Z ∞
Z ∞
2
|b
u(ξ, y) − fb0 (ξ)|2 dξ
|u(x, y) − f1 (x)| dx =
−∞
−∞
Z ∞
sinh 2πyξ
sinh 2π(1 − y)ξ
fb1 (ξ)
=
− 1 + fb0 (ξ)
sinh 2πξ
sinh 2πξ
−∞
2
dξ.
(2)
Note that, by using a simple geometrical argument, one can show the radial function ξ 7→
sinh 2πaξ
sinh 2πξ
is decreasing in |ξ| for each 0 < a < 1 and hence bounded by a.
Hence the integrand in (2) is bounded by
2
2
2
2 fb1 (ξ)(y + 1) + 2 fb0 (ξ)(1 − y)
2
≤ 2 fb1 (ξ) · 2 + 2 fb0 (ξ)
∈ L1 (R).
Then the desired result follows from LDCT. Similar for y → 0.
(c) Our goal is to show for fixed ξ ∈ R,
Z ∞
e−2πixξ
2 sinh 2πaξ
I :=
dx =
.
sin πa sinh 2πξ
−∞ cosh πx + cos πa
Let γR be as the following figure (copy from Book II) whose width goes to infinity but whose
height is fixed.
Let f (z) =
e−2πizξ
cosh πz+cos πa
and note that the denominator of f vanishes precisely when eπz +e−πz =
−eπia − e−πia . So the only poles of f inside γR are α = i(1 − a) and β = i(1 + a). To find the
residue of f at α, denoted by Res(f, α), we note that, by L’Hôspital rule,
Res(f, α) = lim (z −α)f (z) = lim e−2πizξ
z→α
Similar for Res(f, β) =
z→α
2e−2πiαξ
e2π(1−a)ξ
2(z − α)
=
=
eπz + e−πz + eπia + e−πia
π(eπα − e−πα )
πi sin πa
e2π(1+a)ξ
−πi sin πa
Then the residue theorem implies
I
f (z) = 2πi Res(f, α) + Res(f, β) =
γR
19
2
e2πξ (e−2πaξ − e2πaξ )
sin πa
(3)
For the integrals of f on the vertical sides, one observes it tends to zero as R → ∞. Indeed, if
z = R + iy with 0 ≤ y ≤ 2, then
|e−2πizξ | ≤ e4π|ξ|
and
1
| cosh πz + cos πa| ≥ | (eπR − e−πR ) − 1| → ∞ as R → ∞,
2
which shows that the integral over the vertical segment on the right goes to 0 as R → ∞. A
similar argument shows that the integral of f over the vertical segment on the left also goes to
0 as R → ∞. So
2
e2πξ (e−2πaξ − e2πaξ ) = lim
R→∞
sin πa
Hence I =
2
(e−2πaξ
sin πa
2πξ
e
− e2πaξ ) 1−e
4πξ =
I
f (z) = (1 − e4πξ )I
γR
2 sinh 2πaξ
.
sin πa sinh 2πξ
(e) We omit the proof for harmonicity of u in the strip. (For the interchange of the order of
differentiation and integration, see [12, page 154]).
To show the unifrom convergence to the boundary conditions, we use the idea of approximate
to the identity (good kernels).
Given > 0. First we note that the kernel ϕa studied in (c) is positive and
R
R
ϕa = ϕ
ca (0) =
Φa (0) = a. Second, by the uniform continuity of f1 (due to decay at infinity), there is η > 0
such that |f1 (x − t) − f1 (x)| < for all x ∈ R and |t| < η. Therefore we obtain an upper bound
estimate for ku(·, y) − f1 (·)k∞ as follows:
Z
Z
u(x, y)−f1 (x) =
ϕ1−y (t)f0 (x − t) dt + ϕy (t)[f1 (x − t) − f1 (x)] dt − (1 − y)f1 (x)
R
R
Z
Z
≤ (1 − y)kf0 k∞ +
ϕy (t)2kf1 k∞ dt +
ϕy (t)|f1 (x − t) − f1 (x)| dt + (1 − y)kf1 k∞
|t|≥η
|t|<η
Z
≤ (1 − y)(kf0 k∞ + kf1 k∞ ) + 2kf1 k∞
ϕy (t) dt + y.
|t|≥η
The first and last are obviously less than whenever y is close enough to 1. For the second term,
R ∞ πt
sin πy
one note that the integrand is bounded by eπt +e
(e + e−πt −
−πt −2 . Although one can find
η
R∞
πt
2)−1 dt = η (eπte −1)2 dt can be calculated explicitly, we mention that for the situation without
antiderivative, one can also observe prove this integral is finite by observing the exponential
decay at infinity and a strict positive lower bound (due to |t| ≥ η) for the denominator. Hence
the target term is less than provided y is close enough to 1.
Finally, we prove that u(x, y) → 0 as |x| → ∞ uniform in y.
20
Z
∞
Z
ϕ1−y (t)f0 (x − t) dt +
u(x, y) =
−∞
For |t| >
|x|
2
∞
ϕy (t)f1 (x − t) dt =
Z
Z
−∞
|t|>
|x|
2
+
|t|≤
|x|
2
···
part, we use the decreasing property of ϕa for each a to see the integral is dominated
by
ϕ1−y (
For |t| ≤
|x|
2
part, one can dominate the integral by
|f0 (z)| kϕ1−y kL1 (R) +
sup
|z−x|≤
kf0 kL1 (R) + kf1 kL1 (R)
|x|
|x|
)kf0 kL1 (R) + ϕy ( )kf1 kL1 (R) ≤
.
|x|
|x|
2
2
eπ 2 + e−π 2 − 2
|x|
2
|f1 (z)| kϕy kL1 (R) ≤
sup
|z−x|≤
|x|
2
|f0 (z)| +
sup
|z−x|≤
|x|
2
|f1 (z)|.
sup
|z−x|≤
|x|
2
Note that all these upper bounds are independent of y and tends to zero as |x| → ∞. So we
complete the proof for (e).
4. (Tychonoff ’s counterexample) For a > 1, consider the function defined by

0
if t ≤ 0
g(t) =
e−t−a if t > 0
(4)
(a) Show that there exists 0 < θ < 1 depending on a so that for t > 0,
|g (k) (t)| ≤
k! − 1 t−a
e 2 .
(θt)k
Moreover, show that g is in the Gevrey class 1 +
1
a
on R, that is, there are M, R
only depending on a, such that for all k ∈ Z≥0
1
|g (k) (x)| ≤ (k!)1+ a M Rk .
Note that real analytic functions are of Gevrey class 1.
(b) Apply (a) to show that for each x ∈ R and t > 0, the series
u(x, t) :=
∞
X
g (n) (t)
n=0
x2n
(2n)!
converges. Moreover, u solves the heat equation , vanishes for t = 0, continuous up
to the initial t = 0 and satisfies the estimate
|u(x, t)| ≤ c1 ec2 |x|
for some constants c1 , c2 > 0.
21
2a/(a−1)
2
This example shows that the growth condition |u(x, t)| ≤ Aea|x| we meet in the class
is almost optimal, since for each > 0, we can find a > 0 such that the growth
exponent 2a/(a − 1) = 2 + 2/(a − 1) = 2 + . (Also see Exercise 12 and Problem 6.)
Remark 14. The Gevrey part in (a) is taken from [11, Page 73 problem 3].
Proof. (a) The function z 7→ g(z) is holomorphic in C \ {0}. We identify the t-axis as the real
axis of the complex plane. If t > 0 is fixed, the circle ∂Bθt (t) does not meet the origin provided
0 < θ < 1 which will be determined later, and by the Cauchy integral formula (see Book II),
we have for all k ∈ Z≥0
dk g
k!
(t) =
k
dt
2πi
Z
g(z)
dz
(z − t)k+1
∂θt (t)
and hence
dk g
k! 1 k
( )
| k (t)| ≤
dt
2π θt
Z
2π
e−Re(z
−a )
dφ
0
Note that for z ∈ ∂θt (t), z = t+θteiφ = rφ eiϕφ and hence z −a = rφ−a e−iaϕφ . Since (1−θ)t ≤ rφ ≤
k
1
(1+θ)t for all φ, Re(z −a ) ≥ ( (1+θ)t
)a > 12 t−a if we pick θ small enough. So | ddtkg (t)| ≤
Note that this upper bound has maximum at t−a =
|
2k
,
a
1 −a
k!
e− 2 t .
(θt)k
that is,
dk g
kk 1 2 k
k! − 1 t−a
2k k − k
1+ a1
2
a e a = (k!)
(
(t)|
≤
e
≤
k!(
)
)a ( a )a
dtk
(θt)k
aθa
ek k!
aθ
1 k
a
1
f a1 21
,
≤ (k!)1+ a M
aa θ
f is independent of a.
where the last inequality is a consequence of Stirling formula and M
(b) First, we use (a) to check the uniform convergence of u(x, t).
∞
X
∞
∞
2n
X n!
X x2 n n!
x2n
− 21 t−a x
− 12 t−a
|g (t)|
|u(x, t)| ≤
≤
e
=
e
(2n)! n=0 (θt)n
(2n)!
θt (2n)!
n=0
n=0
∞ 2
X
1 −a x2
x n1
− 12 t−a
= e− 2 t e θt
≤e
θt n!
n=0
(n)
(5)
which implies that u(x, t) converges uniformly on x ∈ [−M, M ] and t > t0 > 0 for each
1 −a
x2
M, t0 > 0 and u(x0 , 0) = 0 since a > 1 and lim(x,t)→(x0 ,0) e− 2 t e θt = 0. Similar for ∂t u, ∂x u
and ∂xx u with
∞
∂u X (n+1)
x2n
=
and
g
(t)
∂t
(2n)!
n=0
∞
∞
X
∂ 2 u X (j)
x2j−2
x2n
(n+1)
=
=
.
g
(t)
g
(t)
∂x2
(2j
−
2)!
(2n)!
n=0
j=1
So we see that u ∈ C 2 (R × (0, T )) ∩ C(R × [0, T )) solves the heat equation with zero initial
condition.
22
Finally, for the optimal space-growth rate, we fix x and see the upper bound function obtained
1 −a
x2
1
aθ a−1
with
in (5) fx (t) := e− 2 t e θt attains maximum at t = ( 2x
2)
2 1
a
1
2a
1
aθ a−1
1 2x2 a−1
x2 2x2 a−1
1
a−1
a−1
fx (( 2 ) ) = exp(− (
)
+ (
) ) = exp ( a ) (1 − )|x|
.
2x
2 aθ
θ aθ
aθ
a
1
So we have c1 = 1 and c2 = ( aθ2a ) a−1 (1 − a1 ) here.
4 21 Prove that the function
Z ∞h
i
4
4
exy cos(xy + 2ty 2 ) + e−xy cos(xy − 2ty 2 ) ye−y 3 cos y 3 dy
u(x, t) =
0
is another non-trivial solution of the Cauchy Problem in R × R+ with vanishing
initial data. This was observed by Rosenbloom and Widder [16]. I saw it in [4,
page 180].
4
√ 4
Proof. Set g(x, t) = ex cos(x + 2t) + e−x cos(x − 2t) and a(y) = e−y 3 y cos( 3y 3 ). Hence
Z ∞
a(y)g(xy, ty 2 ) dy.
u(x, t) =
0
First, we check u(x, 0) =
R∞
0
a(y)(exy + e−xy ) cos(xy) dy ≡ 0.
Note that
|e±xy±ixy | = |
∞
∞
X
X
(xy)n
2n |xy|n
(±1 ± i)n
|≤
n!
n!
0
0
and hence
|(e
xy
+e
−xy
) cos(xy)| = |
∞
X
an (xy)n
n!
0
|≤
∞
X
2n+1 |xy|n
0
n!
= 2e2|xy|
where 2an = (1 + i)n + (1 − i)n + (−1 + i)n + (−1 − i)n = 2m 4(−1)m if n = 4m for some m ∈ Z≥0
and 2an = 0 if n is not a multiple of 4.
4
R∞
Since 0 e−y 3 ye2|xy| dy < ∞ for each x ∈ R, we have (from LDCT)
Z
u(x, 0) =
0
∞
Z ∞
∞
∞
√
X
X
4
a4m (xy)4m
a4m x4m
dy =
Re
e−y 3 (1−i 3) y 4m+1 dy
a(y)
(4m)!
(4m)!
0
m=0
m=0
The last integral vanishes for each m ∈ Z≥0 since for all Re z > 0
Z ∞
4
3
3
3
e−y 3 z y 4m+1 dy = Γ(3n + )z −3n− 2
4
2
0
√
and the right hand side is purely imaginary when z = 1 − i 3. Hence we proved u(·, 0) ≡ 0.
(This formula can be proved by residue theorem as problem 3(c) or just check it’s true for
z ∈ R+ and then use theory of analytical continuation to extend this formula to all Re z > 0.)
23
Second, we note that u(0, t) = 2
R∞
0
a(y) cos(2ty 2 ) dy is not identically zero by the uniqueness
theorem for Fourier-cosine transforms of functions a(y) ∈ L1 (0, ∞).
Finally, we examine u is C ∞ and solves the heat equation. It’s clear that g(xy, ty 2 ) satisfies the
heat equation for each fixed y. It remains to show we can interchange the order of integration
in y and differentiations in x and in t.
There is a standard way by using the LDCT and mean-value theorem to achieve this (see e.g.
[12, page 154]). Here we use that corollary and point out what the dominating function for
derivatives is:
∂ 2n g
∂ ng
(x,
t)
=
(x, t) ≤ 2n+1 cosh x,
2n
n
∂x
∂t
and then one consider |x| ≤ R for each R > 0 (since differentiation is a local property) so that
the further dominating integral is finite, that is,
Z ∞
4
n+1
2
e−y 3 y 2n+1 cosh(Ry) dy < ∞.
0
5. (Weak Maximum Principle for Heat Equation)
Theorem 15. Suppose that u(x, t) is a real-valued solution of the heat equation in the upper
half-plane, which is continuous on its closure. Let R denote the rectangle
R = {(x, y) ∈ R2 : a ≤ x ≤ b, 0 ≤ t ≤ c}
and ∂ 0 R be the part of the boundary of R which consists of the two vertical sides and its base
on the line t = 0. Then
min u(x, t) = min u(x, t) and
(x,t)∈∂ 0 R
(x,t)∈R
max u(x, t) = max u(x, t).
(x,t)∈∂ 0 R
(x,t)∈R
The steps leading to a proof of this result are outlined below.
(a) Show that it suffices to prove that if u ≥ 0 on ∂ 0 R, then u ≥ 0 in R.
(b) For > 0, let v(x, t) = u(x, t) + t. Then, v has a minimum on R, say at (x1 , t1 ).
Show that x1 = a or b, or else t1 = 0. To do so, suppose on the contrary that
a < x1 < b and 0 < t1 ≤ c, and prove that vxx (x1 , t1 ) − vt (x1 , t1 ) ≤ −. However, show
also that the left-hand side must be non-negative.
(c) Deduce from (b) that there is some t1 ∈ [0, c] such that u(x, t) ≥ (t1 − t) for any
(x, t) ∈ R and let → 0.
24
Remark 16. How about the boundary condition is changed to zero Neumann condition. A
more precise statement is in my solution file to Evans’ PDE Chapter 7.
Proof. (a) Due to this linear PDE has no zero-order term, we can consider v := u−min(x,t)∈∂ 0 R u(x, t)
and V = max(x,t)∈∂ 0 R u(x, t) − u separately. Note that those extreme values are finite by the
compactness of ∂ 0 R and continuity of u up to the closure. This is why we need the space
domain to be bounded.
(b) Just note at the minimum point, vt ≤ 0 and vxx ≥ 0.
(c) From (b), there is (x1 , t1 ) ∈ ∂ 0 R such that
u(x, t) + t = v(x, t) ≥ v(x1 , t1 ) = u(x1 , t1 ) + t1 ≥ t1 .
Remark 17. As explained in (a), one can replace the space domain R by any bounded domain
Ω in any dimension Rd . The proof is almost identical.
Remark 18. One can check [17, Chapter 9.B] and [14, Chapter 2] for strong maximum principle, that is, to exclude the possibility of existence of interior extrema.
6. The examples in Problem 4 are optimal in the sense of the following uniqueness
theorem due to Tychonoff.
Theorem 19. Suppose u(x, t) satisfies the following conditions:
(i) it is continuous on Rd × [0, T ] and solves the heat equation in Rd × (0, T ) with zero initial
2
condition. (ii) |u(x, t)| ≤ M ea|x| for some M, a, and all x ∈ Rd , 0 ≤ t < T .
Then u ≡ 0.
Remark 20. This growth condition (ii) can be relaxed to |u(x, t)| ≤ C exp ( at )α + a|x|2
for some C > 0, a > 0 and 0 < α < 1. It is proved in [2]. It’s also shown the condition
0 < α < 1 is optimal, one can’t relax to α = 1 by constructing a non-trivial function u with
|u(x, t)| ≤ M exp( at ) for some constants M, a > 0 (note that in contrast to Problem 4 and 4 12 ,
this u is bounded in x for fixed t).
Proof. One can find some discussions and proof in [4, Section 5.4], [5, Section 2.3] or [11,
Section 7.1(b)].
6 12 . There is another well-known uniqueness theorem for heat equation in Rd × R+ due
to David Widder:
25
Theorem 21. Let T > 0 and ST := Rd × (0, T ]. Suppose u ∈ C 2,1 (ST ) := {ut , uxi ,xj ∈ C(ST )}
is a nonnegative solution to heat equation in ST and u(·, t) → 0 in L1loc (Rd ) as t → 0+ . Then
u ≡ 0 in ST .
Remark 22. This theorem is very intuitive for physicists and useful in thermodynamics. I
learned the proof from [4, Section 5.14] which is suitable for arbitrary dimension in contrast to
Widder’s original proof [11, Section 7.1(d)] only suitable for one dimensional case. Also note
that Tychonoff’s function defined in Problem 4 is sign-changing.
Proof. First, there is a more general result which will be proved in the end:
Theorem 23. Let u be a solution for heat equation and satisfies all the regularity (smoothness)
hypotheses and initial data describe above. If u satisfies
Z
sup
|u(y, s)|Ht−s (x − y) dy := Mx,t < ∞ ∀(x, t) ∈ ST ,
s∈[0,t)
Rd
then u ≡ 0 in ST .
Using Theorem 23, Widder’s theorem is proved provided that for each (x0 , t0 ) ∈ ST
Z
Z
|u(y, s)|Ht0 −s (x0 − y) dy =
u(y, s)Ht0 −s (x0 − y) dy ≤ u(x0 , t0 ) ∀s ∈ [0, t0 ).
Rd
(6)
Rd
Note that we may assume T = 1, (x0 , t0 ) = (0, 1) and s = 0 in (6) by introducing the change
of variables
τ=
y − x0
η=√
t0 − s
t−s
,
t0 − s
and the function
U (η, τ ) = u(x0 +
√
t0 − sη, s + (t0 − s)τ ).
To prove (6), one fix ρ > 0 and consider the Cauchy problem that v ∈ H(S1 ) solves heat
equation in S1 and satisfies the initial condition v(x, 0) = ζ(x)u(x, 0)χB2ρ (0) (x), where χE is
the characteristic function on E and x 7→ ζ(x) ∈ C0∞ (B2ρ ) is nonnegative and equals to 1 on
Bρ (it’s called cut-off function or bump function, see Exercise 4). Since the initial datum is
compactly supported in B2ρ , the unique bounded solution v is given by
Z
v(x, t) =
ζ(y)u(y, 0)Ht (x − y) dy.
|y|<2ρ
(Uniqueness comes from energy method or weak maximum principle.)
Lemma 24. u ≥ v in S1 .
26
By Lemma 24, we have
Z
u(0, 1) ≥ v(0, 1) =
Z
ζ(y)u(y, 0)H1 (−y) dy ≥
|y|<2ρ
u(y, 0)H1 (−y)
|y|<ρ
Then we prove (6) by letting ρ → ∞ and hence complete the proof of Widder’s theorem.
Proof of Lemma 24. Let n0 be a positive integer larger than 2ρ, and for n ≥ n0 , consider the
sequence of homogeneous Dirichlet problems (see Remark 26 for the existence.)



∂t vn − ∆vn = 0 in Bn × (0, 1)



vn = 0 on ∂Bn × (0, 1)




vn (x, 0) = ζ(x)u(x, 0)χB (0) (x).
2ρ
(7)
We regard the functions vn as defined in the whole of S1 by zero extension. By weak maximum
principle (see Remark 17 with Ω = Bn ) and nonnegativity of u
0 ≤ vn ≤ vn+1 ≤ ku(·, 0)kL∞ (B2ρ ) and vn ≤ u
(8)
for all n ≥ n0 . By the second of these, the proof of the lemma reduces to showing that the
increasing sequence vn converges to v. We decompose this into two steps:
Step 1: Use Arezla-Ascoli theorem to conclude, on every compact subset of S1 , vn → w in C ∞ .
Hence w solves the heat equation in S1 .
Step 2: Show w(·, t) → ζ(·)u(·, 0) in L1loc (Rd ) as t → 0.
Assume these steps are accomplished for the moment. Since w is bounded by ku(·, 0)kL∞ (B2ρ ) ,
w = v by the uniqueness theorem of bounded solutions of the Cauchy problem (Problem 6).
Hence we complete the proof for Lemma 24.
We first assume Step 1 is true and prove Step 2 by testing the weak formulation of (7) with
Heaviside function which is 1 on R+ and -1 on R− . However, this Heaviside function is not
continuous, we need to approximate it by continuous functions so that one can use integration
by parts.
Given BR ⊂ Rd with R ≥ 2ρ. First, we rewrite (7) as


fn (x, t) = vn (x, t) − ζ(x)u(x, 0)






∂t fn − ∆fn = ∆[ζ(·)u(·, 0)] in S1


fn = 0 on ∂Bn × (0, 1)





f (x, 0) = 0.
n
27
(9)
For δ > 0, let


[lef t] 1, if u > δ




u
, if |u| ≤ δ
hδ (u) =

δ




− 1, if u < −δ.
(10)
(Think this as the approximation to the Heaviside function (as δ → 0) and note that although
there is a corner at ±δ we realize h0δ (s) = 1δ χ|s|≤δ as the weak derivative of hδ ).
Multiply (9) by hδ (fn ) and integrate over Bn × (0, t) for t ∈ (0, 1) with a well-known chain rule
for weak derivative (see, e.g. [8, Theorem 7.8]) to obtain
Z tZ
Z
Z Z
Z fn
1 t
2
|Dfn | χ{|fn |<δ} (y) dyds =
∆[ζ(y)u(y, 0)]hδ (fn ) dyds.
hδ (ξ) dξ dy +
δ 0 Bn
0
B2ρ
Bn ×{t}
0
Discard the second term on the left-hand side, which is nonnegative, and let δ → 0 to get
(denote the Lebesgue measure of Ω ⊂ Rd by |Ω|) for n ≥ R
Z
Z
|vn (y, t) − ζ(y)u(y, 0)| dy ≤ t|B2ρ |k∆[ζu(·, 0)]kL∞ (B2ρ ) .
|vn (y, t) − ζ(y)u(y, 0)| dy ≤
Bn
BR
Letting n → ∞, we have
kw(·, t) − ζ(·)u(·, 0)kL1 (BR ) ≤ t|B2ρ |k∆[ζu(·, 0)]kL∞ (B2ρ ) → 0 as t → 0+ .
Since BR is given, we complete the proof for Step 2.
To prove Step 1, we need the following interior gradient estimate whose proof can be found in
[4, Section 5.12 and 5.12c].
Proposition 25. Let u be a solution of the heat equation that satisfies all the regularity hypotheses as Widder’s theorem. There exist constants γ and C depending only on space dimension d
n
o
2
such that for every parabolic box (x0 , t0 )+Q4ρ := (x, t) : |x−x0 | < 4ρ, t ∈ [t0 −(4ρ) , t0 ] ⊂ ST
Z
C |α| |α|!
sup
≤γ
−
|u| dy ds
(11)
ρ|α| (x0 ,t0 )+Q4ρ
(x0 ,t0 )+Qρ
R
R
1
for all multi-indices α, where −Ω |u| dy ds = |Ω|
|u| dy ds and |Ω| is the Lebesgue measure of
Ω
|Dxα u|
Ω ⊂ Rd+1 . Moreover
Z
∂ku
C 2k (2k)!
sup | k | ≤ γ
−
|u| dy ds
ρ2k
(x0 ,t0 )+Qρ ∂t
(x0 ,t0 )+Q4ρ
for all positive integers k.
Consider compact subsets of the type K = BR × [, 1 − ] for ∈ (0, 12 ) and R ≥ 2ρ. By
the uniform upper bound in (8) and the above interior gradient estimates (applied to vn ), we
28
know for every multi-index α and every k ∈ N, there exists a constant M = M (d, , R, |α|, k)
independent of n such that
kDα vn kL∞ (K) + k
∂k
vn kL∞ (K) ≤ M ku(·, 0)kL∞ (B2ρ ) ∀n ≥ 2R.
∂tk
By using the Ascoli-Arezla theorem with a diagonal process we have, up to a subsequence,
vn → w in C ∞ (K). Using a further diagonal process (indexed by the domains) we have, up
to a further subsequence, vn → w in C ∞ on every compact subsets of S1 . So we complete the
proof for Step 1.
Remark 26. The existence of (7) is a consequence of the existence of eigenpairs {(λnk , wkn )}∞
k=0
for −∆ on each ball Bn , that is, −∆wkn = λnk wkn in Bn and wkn = 0 on ∂Bn . The essential tools
for the proof are Fredholm alternatives and Sobolev compact embedding. Check [4, Chapter 4
and Section 5.10], [5, Section 6.2, 6.5 and 7.4] or [13, Section III.5] for the existence of eigenpairs,
P
n −λn
k t w n (x) where
orthogonality of {wkn }k , and convergence of the solution vn (x, t) = ∞
k
k=0 ck e
P∞ n n
n
ck are chosen so that vn (x, 0) = k=0 ck wk (x). Another method based on maximum principle
can be found in [14, Section 3.3-3.4] (Perron’s method).
Proof of Theorem 23. Let (x, t) ∈ ST be arbitrary but fixed. For ρ > 2|x|, we let y 7→ ζ(t) ∈
C02 (B2ρ ) be a non-negative cut-off (bump) function satisfying ζ = 1 in Bρ , |Dζ| ≤
1
ρ
and
|D2 ζ| ≤ ( ρ2 )2 .
For δ > 0, let hδ (·) be the approximation to the Heaviside function introduced in (10). Multiply
the heat equation by hδ (u(y))ζ(y)Ht−s (x−y), and integrate by parts in dy ds over the cylindrical
domain B2ρ × (τ, t − ) for 0 < τ < t − and 0 < < t. This gives
Z
Z
Z
Z u
1 t−
hδ (ξ) dξ H (x − y)ζ(y) dy +
|Du|2 Ht−s (x − y)χ{|u|<δ} (y)ζ(y) dy ds
δ τ
B2ρ ×{t−}
0
B2ρ
Z
Z u
=
hδ (ξ) dξ Ht−τ (x − y)ζ(y) dy
Z
+
B2ρ ×{τ }
0
t− Z u
hδ (ξ) dξ
τ
Ht−s (x − ·)∆ζ − 2(DHt−s )(x − ·)Dζ (y) dy ds.
0
As τ → 0, the first integral on the right-hand side tends to 0 since it’s majorized by
Z
const.
|u(y, τ )| dy → 0 as τ → 0 by the hypothesis.
B2ρ
29
Discard the second term on the left-hand side since it’s nonnegative and let first δ → 0 and
then τ → 0 to obtain
Z
Z
|u(y, t − )|H (x − y) dy ≤
Bρ
|u(y, t − )|H (x − y)ζ(y) dy
B2ρ
2
≤ 2
ρ
Z
0
t−
Z
2
|u(y, s)|Ht−s (x − y) dy ds +
ρ
B2ρ
Z
t−
Z
|u(y, s)||Dy Ht−s (x − y)| dy ds.
0
B2ρ \Bρ
By the hypothesis, it’s obvious that the first term of right-hand side of this inequality tends to
0 as ρ → ∞. For the second term, one notes that it’s majorized by (up to a constant multiple)
Z
Z
Z
Z
|x − y|
4 t−
1 t−
|u(y, s)|Ht−s (x − y)
dy ds ≤
|u(y, s)|Ht−s (x − y) dy ds,
ρ 0
t−s
0
ρ<|y|<2ρ
ρ<|y|<2ρ
which tends to 0 as ρ → ∞ by the hypothesis.
Hence for all r > 2|x| and ∈ (0, t)
Z
|u(y, t − )|H (x − y) dy = 0.
Br
Finally, letting → 0, we expect the left-hand side tends to |u(x, t)| and hence the proof is
completed. This interpretation can be proved as follows:
Fix r > 2|x|. Write
Z
Z
|x−y|2
− d2 − 4
|u(y, t − )|H (x − y) =
|u(y,
t
−
)|(4π)
e
dy
1
Br
|y|<r,|y−x|> 4
Z
|x−y|2
− d2 − 4
+
|u(y,
t
−
)|(4π)
e
dy
1
|y−x|≤ 4
=: I1 + I2 .
Since the continuous function u is uniformly bounded on Br × [ 2t , t], we have
d
1
2
I1 ≤ rd M (4π)− 2 e− 4 → 0 as → 0.
We also note that
Z
Z
|x−y|2
|x−y|2
− d2 − 4
− d2 − 4
I2 =
|u(y,
t
−
)|
−
|u(x,
t)|
(4π)
e
dy
+
|u(x,
t)|
(4π)
e
dy
1
1
|y−x|≤ 4
|y−x|≤ 4
Z
h
i Z
|z|2
|z|2
− d2 − 4
− d2 − 4
= o() + |u(x, t)|
(4π) e
dz −
(4π)
e
dz
1
Rd
|z|> 4
Z
h
i
− d2 −|w|2
= o() + |u(x, t)| 1 −
e
dw
1 π
|z|> 2√4
where o() → 0 as → 0 (due to the continuity of u). Note the last integral tends to 0 as → 0
too. So we are done.
30
7. (Combined with Problem 4.11 in Book III)
The Hermite function hk (x) are defined by the generating identity
∞
X
hk (x)
k=0
tk
2
2
= e−(x /2−2xt+t )
k!
(a) Show that an alternate definition is given by the Rodrigues’ formula
d k 2
2
hk (x) = (−1)k ex /2
e−x .
dx
They also satisfy the ”creation” and ”annihilation” identities (x −
and (x +
d
)hk (x)
dx
d
)hk (x)
dx
= hk+1
= 2khk−1 (x) for k ≥ 0 where h−1 = 0.
Conclude from the above expression that each hk is of the form Pk (x)e−x
2 /2
, where
Pk is a polynomial of degree k. In particular, the Hermite functions belong to the
Schwartz space and h0 (x) = e−x
2 /2
, h1 (x) = 2xe−x
2 /2
. By (d), one can also show the
linear span of P0 , · · · , Pm is the set of all polynomials of degree ≤ m. (Folland [7,
Exercise 8.23(f )]).
2
2
(b) Prove that the family {hk }∞
k=0 is complete in L , that is if f ∈ L (R), and for all
k ≥ 0,
Z
f (x)hk (x) = 0,
(f, hk ) =
R
then f ≡ 0. (An equivalent formulation of this fact is that the set of Hermite
2
2
−x
)).
polynomials {Pk }∞
0 forms an orthogonal basis for L (R, e
(c) Define h∗k (x) = hk ((2π)1/2 x). Then
hb∗k (ξ) = (−i)k h∗k (x).
Therefore, each h∗k is an eigenfunction for the Fourier transform.
(d) Show that hk is an eigenfunction for the operator L = −d2 /dx2 + x2 , and in fact,
prove that
Lhk = (2k + 1)hk
In particular, we conclude that the functions hk are mutually orthogonal in L2 (R).
R
(e) Show that R [hk (x)]2 dx = π 1/2 2k k!.
(f ) Denote the normalized hk by Hk , that is kHk kL2 = 1. Suppose that K(x, y) =
R
P∞ Hk (x)Hk (y)
, and also F (x) = T (f )(x) = R K(x, y)f (y) dy. Then T is a symmetric
k=0
λk
P
Hilbert-Schmidt operator, and if f ∼ ∞
k=0 ak Hk , then
F ∼
∞
X
ak
k=0
31
λk
Hk .
(g) One can show on the basis of (a) and (b) that whenever f ∈ L2 (R), not only
is F ∈ L2 (R), but also x2 F (x) ∈ L2 (R). Moreover, F can be corrected on a set of
measure zero, so is C 1 , F 0 is absolutely continuous, and F 00 ∈ L2 (R). Finally, the
operator T is the inverse of L in the sense that for every f ∈ L2 (R),
LT (f ) = LF = −F 00 + x2 F = f.
Proof. (a) Note that e−(x
2 /2−2xt+t2 )
[
= ex
2 /2
2
e−(t−x) and one can prove by induction that
d k 2
2
e−z ](x) = Pk (x)e−x
dz
where pk is a polynomial of degree k which is even if k is even and is odd if k is odd.
From the definition, we know for each x ∈ R,
x2 /2
d k
2
e−(t−x) |t=0 = ex
dt
d k 2
k x2 /2
= (−1) e [
e−z ](x).
dz
hk (x) = e
2 /2
2
Pk (−x)e−x = (−1)k ex
2 /2
Pk (x)e−x
2
The ”creation” identity can be proved by this formula easily, we omit it. For the ”annihilation”
k
2
d
x2 /2
k x2
e−z ](x) and apply the binomial theorem
identity, one differentiate e hk (x) = (−1) e [ dz
to the right-hand side.
To show P0 , · · · , Pm spans the set of polynomials of degree ≤ m, it suffices to check P0 , · · · , Pm
are linearly independent. This can be proved by the orthogonality of hk easily, we omit it.
(Another easy proof is by noting that each Pk is of exact degree k, cf. Folland [6, Lemma 6.1].)
(b) There are two equivalent ways (Folland and Stein) to prove this result, their principal idea
both rely on the Fourier transform. (We present both of them for completeness).
PK (2xt)k
(Method I) Given t ∈ R, note that
converges pointwisely to e2xt as K → ∞,
k=0 k!
P
PK |2xt|k
2
(2xt)k
| K
≤ e|2xt| for all K ∈ N and f (x)e−x /2 e|2xt| ∈ L1x (R). Hence by
k=0 k! | ≤
k=0 k!
LDCT and (a)
Z
−x2 /2+2xt
f (x)e
R
Z
dx =
f (x)e
−x2 /2
= lim
K→∞
K Z
X
k=0
lim
K→∞
R
K
X
(2xt)k
k!
k=0
−x2 /2
f (x)e
Z
dx = lim
K→∞
f (x)e
R
ck (t)Pk (x) dx = lim
K→∞
R
The desired result follows from Exercise 8.
32
K
X
(2xt)k
−x2 /2
k=0
K
X
k=0
k!
dx
Z
ck (t)
f (x)hk (x) dx = 0
R
(Method II) Follow the hint of Folland [7, Exercise 8.23(g)], it is enough to show gb ≡ 0, where
P
2
(−2πixξ)k
g(x) = f (x)e−x /2 . Given ξ ∈ R, note that K
converges pointwisely to e−2πixξ as
k=0
k!
P
PK |−2πixξ|k
2
(−2πixξ)k
K → ∞, | K
|
≤
≤ e2π|xξ| for all K ∈ N and f (x)e−x /2 e2π|xξ| ∈ L1x (R).
k=0
k=0
k!
k!
Hence by LDCT and (a),
Z
gb(ξ) =
Z
−2πixξ
g(x)e
dx =
R
= lim
= lim
K→∞
f (x)e
−x2 /2
f (x)e
R
K
X
k=0
K
X
(−2πixξ)k
k=0
lim
K→∞
R
Z
K→∞
−x2 /2
k!
K
X
(−2πixξ)k
k=0
dx = lim
K→∞
k!
K
X
dx
Z
ck (ξ)
k=0
f (x)e−x
2 /2
Pk (x) dx
R
Z
f (x)hk (x) dx = 0.
ck (ξ)
R
(c) Using the induction argument and taking Fourier transform for both sides of creation (or
d
k−1
b
annihilation) identity, we have (−2πi)−1 dhdξ
− 2πiξ hd
k−1 = hk (ξ). Note that the induction
√
ξ
k−1
hypothesis is equivalent to √12π hd
hk−1 ( 2πξ). Combining these, we have
k−1 ( √2π ) = (−i)
√
1
ξ
1 1 dhd
ξ
ξ k−1
√
hb∗k (ξ) = √ hbk ( √ ) = √
( √ ) − 2πiξ hd
(
)
k−1
2π
2π
2π −2πi dξ
2π
2π
√
√
1
dhk−1 √
=
(−i)k−1 2π
( 2πξ) − i 2πξ(−i)k−1 hk−1 ( 2πξ)
−2πi
dξ
h dh
i
√
√
√
k−1
= (−i)k −
( 2πξ) + 2πξhk−1 ( 2πξ)
dξ
p
= (−i)k hk ( 2πξ) = (−i)k h∗k (ξ).
Based on this, it remains to check the desired result holds for k = 0, which is well known result
for Gaussian.
(d) This is proved by the creation and annihilation identities.
(e) Using induction argument and (d), we have
Z
Z
Z
h
i
1
d
d
1
2
[hk (x)] dx =
hk (x)(L − I)hk (x) =
hk (x) (x − )(x + )hk (x) dx
2k R
2k R
dx
dx
R
Z h
i
2
1
d
1
dhk
lim hk (x)(xhk (x) +
(x))
=
(x + )hk (x) +
2k R
dx
2k |x|→±∞
dx
Z
(2k)2
hk−1 (x)2 dx + lim hk (x)hk−1 (x)
=
|x|→±∞
2k R
= 2kπ 1/2 2k−1 (k − 1)! + 0 = π 1/2 2k k!,
where the boundary terms vanish since they are of products of polynomials and e−x
(f)
(g)
33
2 /2
.
8. To refine the results in Chapter 4, and to prove that
fα (x) =
∞
X
2−nα e2πi2
nx
n=0
is nowhere differentiable even in the case α = 1, we need to consider a variant of
the delayed means ∆N , which in turn will be analyzed by the Poisson summation
formula. (A simplified proof can be found in Jones [12, Section 16.H]).
(a) Fix an indefinitely differentiable function Φ satisfying Φ ≡ 1 on B1 (0) and vanishes outside B2 (0). By the Fourier inversion formula, there exists ϕ ∈ S so that
ϕ(ξ)
b = Φ(ξ). Let ϕN (x) = N ϕ(N x) so that ϕc
N (ξ) = Φ(ξ/N ). Finally, set
e N (x) =
∆
∞
X
ϕN (x + n).
n=−∞
e N (x) = P∞ Φ(n/N )e2πinx , thus
Observe by the Poisson summation formula that ∆
−∞
e N is a trigonometric polynomial of degree ≤ 2N , with terms whose coefficients
∆
are 1 when |n| ≤ N . Let
e N (f ) = f ∗ ∆
eN
∆
Note that
e N 0 (fα )
SN (fα ) = ∆
where N 0 is the largest integer of the form 2k with N 0 ≤ N .
e N (x) = ϕN (x) + EN (x) where
(b) If we set ∆
EN (x) =
X
ϕN (x + n),
|n|≥1
then one sees that:
0
e 0 (x)| ≤ cN 2 .
(i) sup|x|≤1/2 |EN
(x)| → 0 as N → ∞. (ii) |∆
N
e 0 (x)| ≤ c/(N |x|3 ) for |x| ≤ 1/2.
(iii) |∆
N
R
R
e 0 (x) dx = 0 and −
e 0 (x) dx → 1 as N → ∞.
Moreover, |x|≤1/2 ∆
x∆
N
N
|x|≤1/2
(c) The above estimates imply that if f 0 (x0 ) exists, then
e 0 )(x0 + hN ) → f 0 (x0 ) as N → ∞,
(f ∗ ∆
N
whenever |hN | ≤ C/N . Then, conclude that both the real and imaginary parts of f1
are nowhere differentiable, as in the proof given in Section 3, Chapter 4.
P
n
n
(d) Let a > 1, |b| < 1. Also prove the Weierstrauss function W (x) = ∞
n=1 b cos(a x)
is nowhere differentiable if a|b| ≥ 1.
34
Remark 27. One can also check another characterization in [12, Section 16.H] that is different
from the methods of delay means.
Proof. (a)
(b)
(c)
(d)
Remark 28. For the differentiablity of Riemann function
P∞
n=1
sin πn2 x
,
n2
see Jarnicki and Pflug
[10, Chapter 13].
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36