SOLUTION OF HW7
MINGFENG ZHAO
October 30, 2011
1. [Page 124, Problem 18] Let f (x) = x − [x] −
1
2
if x ∈ Rn \Z, and f (x) = 0 if x ∈ Z. Define a new
function P as follows:
x
Z
f (t) dy,
P (x) =
for all x ∈ R.
0
a. Draw the graph of f over the interval [−3, 3], and prove that f is periodic with period 1, i.e,
f (x + 1) = f (x) for all x ∈ R.
b. Prove that P (x) = 12 (x2 − x) if 0 ≤ x ≤ 1, and that P is periodic with period 1.
c. Express P (x) in terms of [x].
d. Determine a constant c such that
R1
0
(P (t) + c) dt = 0.
e. For the constant c of the part d, let Q(x) =
Rx
0
(P (t) + c) dt. Prove that Q is periodic with
period 1 and that
Q(x) =
x2
x
x3
−
+ ,
6
4
12
Proof. By the definition of f , we know that





t + 3 − 12 = t +









t + 2 − 12 = t +







1


 t+1− 2 =t+



f (t) =
t − 0 − 12 = t −







 t−1− 1 =t−

2








t − 2 − 12 = t −








 0,
for all 0 ≤ x ≤ 1.
5
2,
if −3 < t < −2
3
2,
if −2 < t < −1
1
2,
if −1 < t < 0
1
2,
if 0 < t < 1
3
2,
if 1 < t < 2
5
2,
if 2 < t < 3
if t = −3, −2, −1, 0, 1, 2, 3.
1
2
MINGFENG ZHAO
So the graph of f over [−3, 3] should be:
Claim I: f (x + 1) = f (x) for all x ∈ R.
For any x ∈ R, by the result of part a of the problem 4, in Page 64, we know that [x + 1] = [x] + 1,
which implies that
f (x + 1)
x + 1 − [x + 1] −
1
2
= x + 1 − [x] − 1 −
1
2
=
= x − [x] −
1
2
= f (x).
Hence, we get that f (x + 1) = f (x) for all x ∈ R.
b. Since for all 0 ≤ t < 1, we have
f (t) = t − 0 −
1
1
=t− .
2
2
Then we have
Z
P (x)
x
=
f (t) dt
0
Z
=
0
x
1
t−
2
dt
SOLUTION OF HW7
=
x
t2
t
− 2
2 0
=
1 2
(x − x).
2
3
Claim II: P (x + 1) = P (x) for all x ∈ R.
In fact, we have
Z
P (x + 1)
x+1
=
f (t) dt
0
Z
=
x
Z
0
f (t) dt
x
Z
=
x+1
f (t) dt +
1
P (x) +
f (t) dt
x
Z
=
1
1
P (x) +
Z
f (z + 1) dz
1
P (x) +
Z
1
Z
Since f (z + 1) = f (z)
f (t) dt
Let z = t
x
Z
f (t) dt +
P (x) +
f (z) dz
0
x
=
x
Z
f (t) dt +
P (x) +
Let t = z + 1
0
x
=
x
Z
f (t) dt +
x
=
x+1
Z
f (t) dt +
0
1
f (t) dt
0
=
P (x) + P (1)
=
P (x),
Since P (1) =
1 2
(1 − 1) = 0.
2
Hence, we know that P (x + 1) = P (x) for all x ∈ R.
c. Since P (x + 1) = P (x) for all x ∈ R, then
P (x)
= P (x − [x])
=
1
(x − [x])2 − x + [x]
2
=
1 2
(x − 2x[x] + [x]2 − x + [x]).
2
Since 0 ≤ x − [x] < 1
4
MINGFENG ZHAO
d. If
R1
0
(P (t) + c) dt = 0, then we must have
Z
1
c = −
P (t) dt
0
Z
1
1 2
(t − 2t[t] + [t]2 − t + [t]) dt
2
0
Z 1
1
(t2 − t) dt
= − ·
2 0
1
1 t3
t2 = − ·
−
2
3
2 0
1 1 1
= − ·
−
2 3 2
= −
=
1
.
12
e. Claim III: Q(x + 1) = Q(x) for all x ∈ R.
In fact, we have
Z
Q(x + 1)
x+1
=
(P (t) + c) dt
0
Z
=
x
Z
0
(P (t) + c) dt
x
Z
=
x+1
(P (t) + c) dt +
1
Q(x) +
Z
(P (t) + c) dt
x
Z
=
1
1
Q(x) +
Z
Z
(P (z + 1) + c) dz
1
Q(x) +
Z
1
Q(x) +
Z
(P (t) + c) dt +
Z
Q(x) +
0
1
(P (t) + c) dt
0
=
Q(x).
Hence, we have Q(x + 1) = Q(x) for all x ∈ R.
Claim IV: Q(x) =
(P (z) + c) dz
Since P (z + 1) = P (z)
(P (t) + c) dt
Let z = t
0
x
=
x
(P (t) + c) dt +
Z
Let t = z + 1
0
x
=
x
(P (t) + c) dt +
x
=
x+1
(P (t) + c) dt +
x2
x
x3
−
+
for all 0 ≤ x ≤ 1.
6
4
12
x
SOLUTION OF HW7
5
In fact, for all 0 ≤ x ≤ 1, we have
Z
Q(x)
x
(P (t) + c) dt
=
0
=
=
=
x
1 2
1
(t − 2t[t] + [t]2 − t + [t]) +
dt
2
12
0
Z x 2
t
t
1
− +
dt Since 0 ≤ t < x ≤ 1, then [t] = 0
2
2 12
0
x
t3
t2
t − + 6
4
12 0
Z
=
x3
x2
x
−
+ .
6
4
12
2. [Page 125, Problem 20] Give an even function f , defined every where, periodic with period 2, and
Z x
integrable on every interval. Let g(x) =
f (t) dt, and A = g(1).
0
a. Prove that g is odd and that g(x + 2) − g(x) = g(2) for all x ∈ R.
b. Compute g(2) and g(5) in terms of A.
c. For what value of A will g be periodic with period 2?
Proof. a. Claim I: g is odd, that is, g(x) = −g(−x) for all x ∈ R.
In fact, for all x ∈ R, we have
−x
Z
g(−x)
=
f (t) dt
0
x
Z
f (−z) − dz
=
Let t = −z
0
Z
x
= −
f (−z) dz
0
Z
= −
x
f (z) dz
Since f (z) = f (−z)
0
= −g(x).
Hence, we get that g is odd, that is, g(x) = −g(−x) for all x ∈ R.
Claim II: g(x + 2) − g(x) = g(2) for all x ∈ R.
6
MINGFENG ZHAO
In fact, we have
x+2
Z
g(x + 2)
=
f (t) dt
0
x
Z
=
Z
0
f (t) dt
x
Z
=
x+2
f (t) dt +
2
Z
f (t) dt
2
x
Z
2
= g(x) +
Z
f (z + 2) dz
Z
2
Z
x
f (z) dz
f (t) dt +
2
g(x) +
Z
f (t) dt +
x
f (t) dt Let z = t
x
Z
Since f (z + 2) = f (z)
0
x
Z
Let t = z + 2
0
= g(x) +
=
x
f (t) dt +
x
=
x+2
f (t) dt +
g(x) +
0
2
g(x) +
f (t) dt
0
=
g(x) + g(2).
Hence, we get that g(x + 2) − g(x) = g(2) for all x ∈ R.
b. When x = 2, we have
Z
g(2)
2
=
f (t) dt
0
Z
1
=
Z
2
f (t) dt +
0
f (t) dt
1
Z
0
= A+
f (z + 2) dz
Let t = z + 2
−1
Z
0
= A+
f (z) dz
Since f (z + 2) = f (z)
−1
Z
0
f (−t) − dt
= A+
Let z = −t
1
Z
0
= A−
f (−t) dt
1
Z
1
= A+
f (−t) dt
0
Z
= A+
f (t) dt
0
=
2A.
1
Since f (t) = f (−t)
SOLUTION OF HW7
7
Hence g(2) = 2A. On the other hand, when x = 5, we have
g(5)
=
g(3) + g(2)
By the result of part a
=
g(1) + g(2) + g(2)
=
g(1) + 2g(2)
=
A + 2(2A)
=
5A.
By the result of part a
Hence we get g(5) = 5A.
c. If g is periodic with period 2, in particular, we get
g(0)
= g(2)
=
But we know that g(0) =
R0
0
2A.
f (t) dt = 0, hence we get A = 0. Now for all x ∈ R, by the result of
part a, we have
g(x + 2) − g(x) = g(2) = 0.
That is, if A = 0, g is periodic with period 2.
3. Prove that the limit f (x) as x goes to p exists if and only if the right-hand limit and left-hand
limit of f (x) as x goes to p exist and are the same.
Proof. (⇒) Assume limx→p f (x) exists, then there exists some A ∈ R such that for any > 0, there
exists some δ > 0 such that
|f (x) − A| < ,
for all 0 < |x − p| < δ.
8
MINGFENG ZHAO
In particular, for all 0 < x − p < δ, we have
|f (x) − A| < .
Which implies that
lim f (x) = A.
x→p+
And for all 0 < p − x < δ, we have
|f (x) − A| < .
Which implies that
lim f (x) = A.
x→p−
Therefore, we conclude that
lim f (x) = lim f (x) = lim f (x) = A.
x→p+
x→p−
x→p
(⇐) Assume there exists some A ∈ R such that
lim f (x) = lim f (x) = A.
x→p+
x→p−
Since lim f (x) = A, then for any > 0, there exists some δ1 > 0 such that
x→p+
|f (x) − A| < ,
for all 0 < x − p < δ1 .
Since lim f (x) = A, then for the above > 0, there exists some δ2 > 0 such that
x→p−
|f (x) − A| < ,
for all 0 < p − x < δ2 .
Taking δ = min {δ1 , δ2 }, then for all 0 < |x − p| < δ, we have −δ < x − p < δ, hence
|f (x) − A| < .
Which implies that
lim f (x) = A.
x→p
SOLUTION OF HW7
9
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu