ACM 105: Applied Real and Functional Analysis.
Solutions to Homework # 1.
Andy Greenberg, Alexei Novikov
Problem 1. Bernoulli sequences.
a. In = {x ∈ l∞ : xn = a}. But by construction of Bernoulli sequences, a can only be ±1. Moreover,
if for In , a = 1, then I˜n , the complement of In , has the same structure with a = −1. Therefore the
properties of In are inherited by I˜n . Note that if we have a sequence xk ∈ In , then limk xk also lies
in In , so In is closed. But by the argument above, I˜n is also closed. However the complement of a
closed set is open, so In is also open. It thus follows that In is both open and closed.
b. We use the fact that for each n, the number of elementary events is finite. Therefore the power set,
i.e., the set of all possible events depending on the outcome of elementary events up till n, is in itself
finite and will be our σ-algebra Fn . Since all sets are included, closeness under all kinds of unions
and intersections is immediate. Note that “usually” (i.e., when it is not finite) this set is too large
to consider as the σ-algebra of measurable events. Now consider F0 = ∪Fi . Since ∀ X1 , X2 ∈ F0
∃ n such that X1 , X2 ∈ Fn (by definition of the union of sets) and each Fn is a σ-algebra, F0 has to
be closed under finite unions and intersections, i.e., F0 is an algebra. However the set
E∞ =
∞
\
Ii = (1, 1, . . . , 1, . . .) ∈
/ F0
i=1
since E∞ does not belong to any Fn for any finite n. Therefore, F0 is not a σ-algebra.
c. Using Carathéodory’s extension theorem we can define the minimal σ-algebra that contains unions
of elementary events and a measure on this σ-algebra. It will contain more elements than F0 , since,
as we have just established, F0 is not closed under infinite intersections, while a σ-algebra always
is.
Problem 2. Probability of long leads. The arc-sine law.
a. Consider the set Gα = {x : gn (x) > α}. Obviously, Gα is a union of a number of elementary
events (namely, of those whose total number of votes is greater than α, counting a vote for Al as
+1 and a vote for Bill as -1). Since each elementary event is measurable by construction and there
is at most countably infinite of them, Gα has to be a measurable set by definition of the σ-algebra
(of measurable sets).
b. Obviously,
f
k−1
[
k
=
Ai (n)
n
i=0
where Ai (n) = “Al is winning in exactly i out of ntossings.” Each of the Ai is a union of(at most,
countable) events of the form {x : fn (x) > 0} ∪ {x : fn (x) = 0} ∩ {x : fn−1 (x) > 0} , each of
which is measurable, because so is fn . Therefore, as a union of a countable number of measurable
events, f (k/n) is measurable.
c, e. See attached plot. We used M = N = 1000 iterations, since for a larger number of runs and
simulations, the convergence is so good that you cannot tell the two curves apart. The solid curve
is the simulation results; the dashed curve is the actual arc-sine.
1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Figure 1: Long leads: simulations vs. limit distributions
d. Here are the approximate numerical values:
k
f
< 0.02n ≈ 11.3%;
n
k
P 0.49n < f
< 0.51n ≈ 1.6%;
n
k
P f
> 0.02n ≈ 88.7%;
n
[ k
k
P
f
> 0.976n
f
< 0.024n
≈ 19.8%;
n
n
[ k
k
P
f
< 0.524n ∩ (win)
f
> 0.476n ∩ (loss) ≈ 3.5%
n
n
P
It turns out that it is much more likely for the winner to win overwhelmingly than by a slim margin.
f, g. See plot on next page.
Problem 3. Measurable Sets and Open/Closed Sets.
We first set off to prove that measurability implies existence of the specified open and close sets (i.e.,
necessity). The converse statement (sufficiency) follows after that.
Necessity. Recall (Royden, p. 53) the definitions of two classes of sets, Gδ , which includes countable
unions of open sets, and Fσ , which includes countable intersections of closed sets. By Carathéordory’s
theorem, both of these types of sets are measurable. We first prove that there exist sets F ⊂ Fσ and
2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
Figure 2: Number of returns: simulations vs. limit distributions
G ⊂ Gδ , such that F ⊂ E ⊂ G and µ(G\F) = 0. Recall the principal definitions. Outer measure of a set
E is
X
µ∗ (E) = inf
µ(Pk )
∪Pk ⊃E
k
where
Pk are disjoint open intervals with P
the natural measure assigned to them. (This implies that
S
Pk = P ∈ Gδ , is measurable and µ(P ) = µ(Pk ), so that µ∗ (E) = inf µ(P ).) Furthermore, a set E is
k
k
measurable if for any set A
µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ Ẽ)
(1)
S
where Ẽ is the complement of E. If we denote X ≡ E Ẽ (i.e., the “whole set”), then by substituting
X into (1) for A we get a simple implication, namely, if E is measurable, then
µ∗ (E) + µ∗ (Ẽ) = µ∗ (X) = µ(X)
(2)
∗
Now proceed to the proof of the intermediate
S claim. Since µ (E) is the infinum over all possible open
coverings of E, there exists a covering On = Pk ⊃ E, On ∈ Gδ , such that
k
µ(On ) ≤ µ∗ (E) +
1
2n
(3)
Applying the same argument to Ẽ we find a covering Ẽ ⊂ On0 ∈ Gδ such that
µ(On0 ) ≤ µ∗ (Ẽ) +
1
2n
If we denote Fn = O˜n0 , then Fn ∈ Fσ , is measurable and Fn ⊂ E ⊂ On . Moreover, the last inequality
implies
1
µ(X) − µ(Fn ) ≤ µ∗ (Ẽ) +
2n
3
Since E is measurable, (2) holds, so we have
µ(X) − µ(Fn ) ≤ µ(X) − µ∗ (E) +
1
2n
or
1
2n
Subtracting (4) from (3) and using the inclusion Fn ⊂ E ⊂ On we obtain
µ(Fn ) ≥ µ∗ (E) −
µ(On ) − µ(Fn ) = µ(On \Fn ) ≤
(4)
1
n
Since the right-hand side tends to 0 as n → ∞, we have found the desired F = lim Fn ∈ Fσ and
G = lim On ∈ Gδ .
It is now easy to find the
S desired open set O and closed set F , following from the measurability of G
and F. Since G ∈ Gδ , G = Pn , where Pn are open sets. Without loss of generality, assume that Pn are
n
P
disjoint. Then, since G is measurable, µ(G) =
µ(Pn ) < ∞, so there exists a positive integer N such
n
that
X
µ(Pn ) <
n>N
We denote
O=
N
[
ε
2
Pn
n=1
Note that as a finite union of open sets, it is itself an open set and O ⊃ E with µ(O) − µ(E) < ε/2.
Taking complements or repeating the argument with F we find the set
F =
N
\
Qn
n=1
where Qn are closed, which is a closed set and E ⊃ F with µ(E) − µ(F ) < ε/2. Together, the two
iequalities give
µ(O) − µ(F ) = µ(O\F ) < ε
as required.
Sufficiency. Let there exist an open set O and a closed set F such that F ⊂ E ⊂ O and µ(O\F ) < ε.
Let us prove that E is measurable, i.e., that (1) holds for any set A. Since both open and closed sets are
measurable, for any set A we have
µ∗ (A) = µ∗ (O ∩ A) + µ∗ (Õ ∩ A)
µ∗ (A) = µ∗ (F ∩ A) + µ∗ (F̃ ∩ A)
(5)
Adding these two up produces, after some rearrangement,
h
i
µ∗ (A) = µ∗ (O ∩ A) + µ∗ (F̃ ∩ A) − µ∗ (A) − µ∗ (Õ ∩ A) − µ∗ (F ∩ A)
The terms outside the square brackets can be bounded from below using the inclusions F ⊂ E ⊂ O and
Õ ⊂ Ẽ ⊂ F̃ and monotonicity of the outer measure:
h
i
µ∗ (A) ≥ µ∗ (E ∩ A) + µ∗ (Ẽ ∩ A) − µ∗ (A) − µ∗ (Õ ∩ A) − µ∗ (F ∩ A)
(6)
We now perform some slick manipulation of the terms inside the square brackets. Add and subtract
µ∗ (O ∩ A):
µ∗ (A) − µ∗ (Õ ∩ A) − µ∗ (F ∩ A) = µ∗ (A) − µ∗ (Õ ∩ A) − µ∗ (O ∩ A) +µ∗ (O ∩ A) − µ∗ (F ∩ A)
|
{z
}
4
The first three terms combine to 0 because of the first of the equalities (5). Note also that since O ⊃ F ,
it follows that O = F ∪ (O\F ), so that the last three terms reduce to
µ∗ (O ∩ A) − µ∗ (F ∩ A) = µ∗ F ∪ (O\F ) ∩ A − µ∗ (F ∩ A) ≤
≤ µ∗ (F ∩ A) + µ∗ ((O\F ) ∩ A) − µ∗ (F ∩ A) = µ∗ ((O\F ) ∩ A) ≤ µ∗ (O\F ) < ε
by subadditivity and monotonicity of outer measure. Thus we bounded the term in the square brackets
in (6) by ε, so it follows that
µ∗ (A) ≥ µ∗ (E ∩ A) + µ∗ (Ẽ ∩ A) − ε
Since ε is arbitrary, this implies µ∗ (A) ≥ µ∗ (E ∩ A) + µ∗ (Ẽ ∩ A) for any A. Subadditivity implies that
always µ∗ (A) ≤ µ∗ (E ∩ A) + µ∗ (Ẽ ∩ A). The two inequalities can hold simultaneously only if
µ∗ (A) = µ∗ (E ∩ A) + µ∗ (Ẽ ∩ A)
i.e., (1) holds, so E is measurable.
Note. The quantity µ(X) − µ∗ (Ẽ) is called the inner measure of set E. A more general theorem
(cf. Royden, Proposition 12.24) says that (2) is not only a necessary, but also a sufficient condition of
measurability: E is measurable (i.e., (1) holds) if and only if µ∗ (E) = µ∗ (E).
It is now easy to establish the results
µ(E) = inf µ(O) = sup µ(F )
E⊂O
E⊃F
Recall that a = inf{x ∈ A} if (a) for any x ∈ A, x ≥ a; and (b) for any ε > 0 ∃ x0 ∈ A such that
x0 < a + ε. Similarly, b = sup{x ∈ B} if (a) for any x ∈ B, x ≤ b; and (b) for any ε > 0 ∃ x0 ∈ A such
that x0 > b − ε.
Since we only consider F ⊂ E ⊂ O, monotonicity of the measure implies µ(F ) ≤ µ(E) ≤
≤ µ(O) for all such sets F and O. Moreover, the property that has just been established ensures
existence of such open set O ⊃ E and closed set F ⊂ E that
µ(F ) < µ(E) + ε
and
µ(O) > µ(E) − ε
Therefore both criteria for the sup (inf) are satisfied.
Problem 4. Egorov’s Theorem.
Theorem (D.F.Egorov). If {fn } is a sequence of measurable functions that converge to a real-valued
function f almost everywhere on a measurable set E with finite measure µ(E) < ∞, then for any ε > 0
there is a subset A ⊂ E with µ(A) < ε, such that fn converges uniformly on E\A.
Remark. Rather than using the hint and citing Proposition 3.24, we choose to prove the Egorov’s Theorem
from the ground up. In fact, you will see that the proof follows the same line of thinking as in Royden’s
text.
Proof. Since fn → f a.e., f is measurable on E. Define the collection of sets
\ 1
Enm =
x : |fi (x) − f (x)| <
m
i>n
For a fixed value of m, Enm is simply the set of all points x such that
|fi (x) − f (x)| <
1
m
holds for all i > n. Obviously, E1m ⊂ E2m ⊂ . . . ⊂ Enm ⊂ . . ., and we can define E m =
∞
S
n=1
Enm . By the
continuity property of measure (which can be shown to follow from countable additivity), lim µ(Enm ) =
µ(E m ), and therefore given any m > 0 and ε > 0, there exists n0 (m) such that
ε
µ(E m \Enm0 (m) ) < m
2
5
n→∞
Now let Eε =
∞
T
m=1
Enm0 (m) . We claim that the desired set A is just E\Eε . To see this we need to prove
that fn → f uniformly on Eε and that µ(E\Eε ) < ε. Since ∀x ∈ Eε we have
|fi (x) − f (x)| <
1
m
for any m ≥ 1 and all i > n0 (m), it follows that sup|fi (x) − f (x)| → 0, and thus uniform convergence on
Eε is established. To estimate the measure of the set E\Eε , we first note that µ(E\E m ) = 0 for every
m. Indeed, if x0 ∈ E\E m , then for any i, no matter how large, the inequality
|fi (x0 ) − f (x0 )| ≥
1
m
holds, so that fn cannot converge to f on E\E m . But since fn → f a.e., the measure of this set has to
be 0. Thus
ε
µ(E\Enm0 (m) ) = µ(E m \Enm0 (m) ) < m
2
Therefore
!
!
∞
∞
∞
∞
\
[
X
X
ε
m
m
µ(E\Eε ) = µ E\
En0 (m) = µ
(E\En0 (m) ) ≤
µ(E\Enm0 (m) ) <
=ε
m
2
m=1
m=1
m=1
m=1
so that µ(E\Eε ) < ε, and we are done.
Problem 5. Luzin’s Theorem.
Theorem (N.N.Luzin). Let f (x) be a measurable real-valued function on [a, b] and let ε > 0 be arbitrary.
Then there is a continuous function g(x) such that the measure of the set {x : f (x) 6= g(x)} is less than
ε.
Proof. We first outline the following steps in the proof.
1. Based on Proposition 3.22 from Royden’s book (proved below), find a sequence of continuous
functions gn such that |gn − f | < εn = ε/(3 · 2n ) on a set [a, b]\En with µ(En ) < εn .
2. Conclude that there exists a sequence of continuous functions gn → f on [a, b]\E with µ(E) < ε/3.
3. Find a subset A = [a, b]\E 0 with µ(E 0 ) < ε/3 such that gn → f uniformly on A.
4. Find a closed set F ⊂ A with µ(A\F ) < ε/3 and conclude that we have built a sequence of
continuous functions gn which converges uniformly to f on a closed set F ⊂ [a, b] with µ([a, b]\F ) <
ε. Since uniform convergence preserves continuity, lim gn = g is continuous on F and g = f on F .
5. Extend g continuously to the whole interval [a, b]. Thus we have a function g which is continuous
on [a, b] and µ({x : g(x) 6= f (x)}) = µ([a, b]\F ) < ε, Q.E.D.
We first prove the following claim, which is part of Proposition 3.22.
Claim. Given a measurable function f on the interval [a, b], for any δ > 0, there exists a set A with
µ([a, b]\A) < δ and a continuous function h such that |f − h| < δ on A.
Proof. Define
m
m
(m + 1)
fn (x) =
if
≤ f (x) <
n
n
n
with m and n > 0 integers. Then the functions fn are simple and fn → f uniformly, since |fn (x)−f (x)| =
= 1/n → 0. Therefore for any δ > 0 there exists an index N such that |fn (x) − f (x)| < δ for all n ≥ N .
Thus we have a simple function (ϕ = fN , say), which is at most δ-close to our measurable function f .
Moreover, since ϕ is simple, there exist a collection of measurable sets Ei such that ϕ ≡ cn = const on
En . By the arguments used in solving Problem 3 of this homework, we can approximate each En with a
K
S
finite union of disjoint open intervals In =
(ak , bk ), so that µ(In \En ) < δ/2n+1 . By defining g(x) = cn
k=1
6
on In , we arrive at a step function g (i.e.,
S a function which assumes constant values on intervals), such
that g(x) = ϕ(x) except on the set E = (En \In ) with
µ(E) <
X
µ(En \In ) =
n
X
n
δ
2n+1
=
δ
2
The desired continuous function h is simply the linear interpolant of g: if, e.g., g = ck1 on the interval
(ak1 , ak2 ) and g = ck2 on the interval (ak2 , ak3 ), then we define

δ
ck1 ,
on ak1 , ak2 − 2n+2





c −c
c +c
δ
δ
h = 2n+1 k2 δ k1 x + k1 2 k2 , on ak2 − 2n+2
, ak2 + 2n+2





δ
ck2 ,
on ak2 + 2n+2
, ak 3
(these are nothing else than formulas for linear interpolation). Then h is different from
P g only on the
union of the small intervals around each endpoint. The measure of this set is less than n 2δ/2n+1 = δ/2.
Thus for any given δ we have the following:
• a simple function ϕ such that |f − ϕ| < δ
• a step function g such that g = ϕ except on a set of measure < δ/2; this implies that |g − f | < δ
except on a set of measure < δ/2
• a continuous function h such that h = g except on a set of measure < δ/2, implying that |h − f | < δ
except on a set of measure < δ/2 + δ/2 = δ
The Claim is thus established.
It is now easy to fill in the details of the proof of Luzin’s theorem. We shall stick with numbers in
the outline above.
1. For each n simply take δ = εn = ε/(3 · 2n ) for a given ε and use the Claim above to come up with
the corresponding gn .
2. Follows immediately, since |gn − f | → 0 as n → ∞.
3. Existence of such a set A is guaranteed by Egorov’s theorem.
4. Existence of such a closed set F ⊂ A is proved in Problem 3 of this homework.
5. Extension is possible due to the result in Problem 2.40 in Royden’s text.
We are done.
Problem 6. Continuity of the Lebesgue Integral.
We need to prove that for any ε > 0 there exists δ > 0 such that
y
Z
|F (x) − F (y)| = f (x) dx < ε
x
if only |x − y| < δ.
This fact is clear if f (x) is bounded. Indeed, in this case,
y
Z
Zy
|F (x) − F (y)| = f (x) dx ≤ |f (x)| dx ≤ sup |f (x)| |x − y| < ε
x
x
7
if we take δ < ε/ sup |f (x)|.
In the general case, consider the following sets:
An = {x : n ≤ |f (x)| < n + 1}
BN =
N
[
An
n=0
CN = R − BN
Thus constructed, An is a disjoint partition of the real line; therefore, by the properties of the Lebesgue
integral,
Z
∞ Z
X
|f (x)| dx
|f (x)| dx =
n=0 A
R
n
Thus there exists N0 such that
∞
X
Z
n=N0 +1 A
Now take 0 < δ <
ε
2(N0 +1) .
|f (x)| dx =
Z
|f (x)| dx <
ε
2
CN0
n
Then
y
Z
Zy
|F (x) − F (y)| = f (x) dx ≤ |f (x)| dx =
x
x
Z
|f (x)| dx +
[x,y]∩BN0
Z
|f (x)| dx ≤
[x,y]∩CN0
≤ (N0 + 1) |x − y| +
Z
|f (x)| dx <
ε ε
+ =ε
2 2
CN0
and we are done.
Problem 7. Convergence Theorems for “Convergence in Measure”.
Theorem (bounded convergence∗ ). Suppose the sequence of functions fn measurable
R on the
R set E with
finite measure is bounded above, i.e., |fn | ≤ M . Then if fn → f in measure, then fn → f .
E
E
Proof. Convergence in measure means that for any ε > 0 there exists a δ > 0 such that the measure
of the set Sn = {x : |fn (x) − f (x)| > δ} is less than ε. (Note that the sets Sn may be very different for
each function fn , depending on the structure of the sequence.) Therefore
Z
Z
Z Z
Z
Z
fn − f = (fn − f ) ≤ |fn − f | = |fn − f | +
|fn − f |
E
E
E
E
Sn
|fn −f |<=δ
By choosing a sufficiently small δ0 , the measure of the set {x : |fn (x) − f (x)| > δ0 } can be made less
than ε/(4M ). Since |fn | < M and fn → f in measure, f is also bounded by M , so that |fn − f | ≤ 2M ,
and thus the first term in the expression above is less than ε/2. As for the second term, it is clearly
bounded above by δµ(E). Choosing δ = min {δ0 , ε/(2µ(E))}, obtain
Z
Z fn − f ≤ ε + ε = ε
2 2
E
E
Since ε is arbitrary, convergence of the integrals is thus proved.
Having established the bounded convergence theorem for convergence in measure, we can now proceed
exactly as in the “convergence a.e.” case, to prove Fatou’s lemma and the monotone and Lebesgue
convergence theorems (cf. Royden’s text).
8
Lemma (Fatou∗ ). If fn is a sequence of nonnegative measurable functions and fn → f in measure on set
E, then
Z
Z
f ≤ lim fn
E
E
Proof. Define a measurable function h supported on a set E 0 ⊂ E with finite measure, with
f, f ≤ M
h=
0, f > M
for some M > 0. Obviously, h ≤ M and h ≤ f on E. Now define the sequence hn (x) =
= min{h(x), fn (x)}. Clearly, hn → h in measure, so by the bounded convergence theorem
Z
Z
Z
Z
h = h = lim hn ≤ lim fn
E0
E
But h ≤ f , so
R
E
h≤
R
E
E0
E
f , and taking the supremum over h we get
Z
Z
f ≤ lim fn
E
E
as required.
Theorem (monotone convergence∗ ). Let fn be an increasing sequence of mesurable functions such that
fn ↑ f in measure. Then
Z
Z
f = lim fn
Proof. By Fatou’s∗ Lemma,
Z
Z
f ≤ lim fn
R
R
But since for each n we have fn ≤ f , it follows that fn ≤ f , which implies
Z
Z
lim fn ≤ f
For both inequalities to hold, we must have
Z
Z
Z
Z
lim fn = lim fn = lim fn = f
as required.
Theorem (Lebesgue’s∗ convergence theorem). Let fn be a sequence of functions measurable on the set
E with finite measure andR suppose
R there exists a measurable function g(x) such that |fn | ≤ g. Then if
fn → f in measure, then fn → f .
E
E
Proof. Obiviously, |f | ≤ g, and therefore f is also measurable. Consider the function g − fn . It is
nonnegative and converges to g − f in measure; therefore, by Fatou’s∗ Lemma,
Z
Z
Z
Z
(g − f ) ≤ lim (g − fn ) ≤ g − lim fn
E
E
E
E
which implies
Z
f ≥ lim
E
Z
fn
(7)
E
Now take the function g + fn , which is also nonnegative and converges in measure to g + f . Applying
Fatou’s∗ Lemma again, we obtain
Z
Z
Z
(g + f ) ≤ g + lim fn
E
E
9
E
implying
Z
f ≤ lim
Z
E
E
Z
Z
fn
(8)
But both (7) and (8) can be true only if
f = lim
E
fn
E
which proves the theorem.
Problem 8. Green’s Function for Poisson’s Equation on a Unit Disk.
a. Since the problem is set in a ball, spherical coordinates are appropriate. Recall that if x1 =
r cos θ cos φ, x2 = r cos θ sin φ, x3 = r sin θ and y1 = ρ cos θ0 cos φ0 , y2 = ρ cos θ0 sin φ0 , y3 = ρ sin θ0 ,
then
s
2
2
0
0
0
|x − y| = r + ρ − 2rρ cos θ cos θ + sin θ sin θ cos(φ − φ )
and the Green’s function is −1/(4π|x − y|). We make use of spherical symmetry to claim that the
value of the integral at any point x lying on a sphere around the origin is the same. Therefore it
suffices to evaluate the integral at one convenient value of θ, say, θ = 0. In this case |x − y| takes a
particularly simple form,
p
|x − y| = r2 + ρ2 − 2rρ cos θ0
and we have
1
u(x) = −
4π
Z
D
dy
1
=−
|x − y|
4π
Z2πZπ Z1
0
0
0
ρ2 sin θ0 dρ dθ0 dφ0
p
r2 + ρ2 − 2rρ cos θ0
There is no dependence on φ0 , so the outer integral reduces to simple multiplication by 2π. Integration dθ0 (e.g., in Maple) yields
Z1
1
u(r) = −
2
ρ
ρ + r − (ρ − r) sgn(ρ − r) dρ
r
0
To integrate the latter expression, we split the interval up in two parts: [0, 1] = [0, r] + [r, 1] and
use linearity of the integral to get
1
u(r) = −
2
Z1
ρ
ρ + r − (ρ − r) sgn(ρ − r) dρ =
r
0
1
=−
2
Zr
0
2ρ2
1
dρ −
r
2
Z1
2ρ dρ = −
1
2
−
r2
r2
1
+1 =
−
3
6
2
r
which is the required expression, since r = |x|.
b. Since
u(x) = −
1
4π
Z
D
f (y)
dy
|x − y|
and the domain of integration (i.e., the unit disk D) is fixed, the only x dependence comes from
the denominator of the integrand. For any h > 0, ∆h is a simple linear operator, so it commutes
with integration. Thus we have
Z
Z
Z
1
f (y)
1
1
∆h u(x) = −∆h
f (y)∆h
dy = −
dy = f (y)∆h G(x − y) dy
4π
|x − y|
4π
|x − y|
D
D
10
D
c. For any fixed y, ∆x G(x − y) = ∆x G(x). We use the fact that G(x) only depends on |x|, so we use
spherical coordinates again:
1 2 1
1
1
2 1
∆G(x) = 2 ρ Gρ ρ + 2 2 Gφφ + 2
sin θGθ )θ = 2 ρ
=0
ρ
ρ sin θ
ρ
4πρ2 ρ
ρ sin θ
This is true for any fixed y; for y = x, this does not hold.
d. In b. above, put h = 1/n. Then we have a sequence of nonnegative measurable
functions gn (x) =
R
∆1/n G(x − y). Since lim gn = g = 0 a.e. (as proved in c.), it follows that g = 0. But we know
n→∞
R
that limn gn = 1 ≥ 0. Therefore we see that Fatou’s lemma still holds. However we can easily
establish that gn is neither bounded nor monotone, so none of the convergence theorems implying
equality of the integrals can be used.
References
[1] Royden H.L., Real Analysis, 3rd ed., Prentice-Hall, 1988.
[2] Kolmogorov A.N., Fomin S.V., Introductory Real Analysis, Dover, 1975.
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