Homework 6 Solutions
Chapter 5, Exercise 9
If FR is defined as given, then one may check that F̂R(ξ) =
(
1− |ξ|
R
)
·1[−R,R](ξ).
This is easily derived from the result of Exercise 2 of the same chapter. Conse-
quently, we see by inversion that
(FR ∗ f)(x) =
∫
R
̂(FR ∗ f)(ξ)e2πiξxdx
=
∫
R
F̂R(ξ)f̂(ξ)e2πiξxdx
=
∫ R
−R
(
1−
|ξ|
R
)
f̂(ξ)e2πiξxdx.
To see that {FR} is a family of good kernels as R→∞, note that∫
R
|FR(x)|dx =
∫
R
FR(x)dx = F̂R(0) = 1,
and using the bound sin2(πtR) ≤ 1, we also have that
R
∫
{|x|>δ}
sin2(πtR)
(πtR)2
dt = 2R
∫ ∞
δ
sin2(πtR)
(πtR)2
dt ≤ 2R
∫ ∞
δ
1
(πtR)2
dt =
2
δπ2
R−1,
and the right side clearly tends to 0 as R→∞.
1
Chapter 5, Exercise 11
The fact that u is continuous on the closure of the upper half plane follows
immediately from Theorem 5.2.1(ii) in the same chapter.
To show that u(t, x)→ 0 as |x|+ t→∞, we apply the hint given to write
u(t, x) =
∫
R
f(x−y)Ht(y)dy =
∫
{|y|>|x|/2}
f(x−y)Ht(y)dy+
∫
{|y|≤|x|/2}
f(x−y)Ht(y)dy.
Let’s call the terms on the right side as A,B, respectively. To bound A, note
that f is bounded (say |f | ≤ C
√
4π), and 1 < 2y/|x| if |y| > |x|/2, so
|A| ≤
∫
{|y|>|x|/2}
|f(x− y)|Ht(y)dy
≤
2C
t1/2
∫ ∞
|x|/2
e−y
2/4tdy
≤
2C
t1/2
∫ ∞
|x|/2
2y
|x|
e−y
2/4tdy
=
8Ct1/2
|x|
e−x
2/16t.
To bound B, note that |x− y| ≥ 1
2
|x| if |y| ≤ 1
2
|x|, thus 1
1+(x−y)2 ≤
1
1+x2/4
. But
since f ∈ S, it follows that |f(x)| ≤ D/(1 + x2) for some D > 0. Thus,
|B| ≤
∫
{|y|≤|x|/2}
|f(x− y)|Ht(y)dy
≤
D
1 + x2/4
∫
R
Ht(x)dx
=
D
1 + x2/4
.
On the other hand, |Ht(y)| ≤ t−1/2, and hence
|u(t, x)| ≤
∫
R
|f(x− y)|Ht(y)dy ≤ t−1/2
∫
R
|f(x)|dx =: Kt−1/2.
Thus, given � > 0, we choose T = T (�) large enough so that KT−1/2 < �,
then we choose M = M(�) large enough so that D
1+x2/4
+ 8CT
1/2
|x| e
−x2/16T < �
whenever |x| > M . Then it is true that |u(t, x)| < � whenever |x|+ t ≥M + T .
2
Chapter 5, Exercise 12
For notational convenience, we write H(t, x) for the heat kernel, and subscripts
will always indicate partial derivatives throughout this problem.
Let u(t, x) = xt−1H(t, x). Then it is clear that u(t, x) = −2Hx(t, x) if t > 0
and x ∈ R, thus we have
ut(t, x) = −2
∂
∂t
∂
∂x
H(t, x) = −2
∂
∂x
∂
∂t
H(t, x) = −2
∂3
∂x3
H(t, x) = uxx(t, x).
If x 6= 0, then limt→0 xt−3/2e−x
2/4t = 0; this follows immediately from the fact
that ex
2/4t ≥ 1
2
(x2/4t)2, which in turn follows from the fact that eu = 1 + u +
1
2
u2 + … ≥ 1
2
u2 for u ≥ 0. If x = 0, then obviously limt→0 xt−3/2e−x
2/4t = 0.
This proves that u(t, x)→ 0 as t→ 0 along lines parallel to the axis {x = 0}.
On the other hand, if we let (t, x) → 0 along the parabola x2 = 4t, then
we get u(t, x(t)) = (4π)−1/2(4t)1/2t−3/2e−1 = ct−1/2 → +∞ as t → 0, hence u
is not continuous at 0.
3
Chapter 5, Exercise 13
Suppose u is harmonic on the strip {(x, y) : −π
2
< y < π
2
}, and u extends
continuously to the boundary, with u(x,−π/2) = u(x, π/2) = 0. Also assume
that u vanishes at infinity. We want to show u = 0. We give two solutions.
Solution 1: Given � > 0, we can choose M large enough so that |x| > M
implies |u(x, y)| ≤ �. Then the boundary condition on u and the maximum
principle imply that |u(x, y)| ≤ � whenever |x| ≤ M (the maximum principle
is an easy corollary of the mean value property, and says that any harmonic
function achieves its maximum on the boundary of a connected domain). Since
we can choose M as large as we’d like, this actually shows that |u(x, y)| < �
everywhere on the strip. Since � is arbitrary, u = 0.
Solution 2: Let’s define H = {(x, y) : x > 0}, and let us define a new func-
tion
v(x, y) = u
(1
2
log(x2 + y2), tan−1(y/x)
)
.
We claim that v is a harmonic function on H, and extends continuously to the
closure of H, where it is zero on the boundary and vanishes at ∞. If we can
prove this then we will be done, because then we would have v = 0 by theorem
5.2.7, from which it follows that u(x, y) = v(ex cos y, ex sin y) = 0.
To prove that v is harmonic, we write v in polar coordinates v(r, θ) = u(log r, θ),
from which we may compute the three quantities
vr =
1
r
ux(log r, θ), vrr =
1
r2
(uxx − ux)(log r, θ), vθθ = uyy(log r, θ).
Consequently we find using the formula for the laplacian in polar coordinates:
∆v = vrr +
1
r
vr +
1
r2
vθθ =
1
r2
(uxx + uyy) = 0,
so that v is harmonic. Showing that v does indeed satisfy the zero boundary
conditions is then an easy consequence of the boundary conditions on u.
Chapter 5, Exercise 14
Note from exercise 9 that F̂R(ξ) =
(
1 − |ξ|
R
)
· 1[−R,R](ξ). Hence by the Pois-
son summation formula, we find that
∞∑
n=−∞
FN (x+ n) =
∞∑
n=−∞
F̂N (n)e2πinx =
N∑
n=−N
(
1−
|n|
N
)
e2πinx.
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