SOLUTION OF HW6
MINGFENG ZHAO
October 30, 2011
1. [Page 125, Problem 19] Give an odd function f , defined everywhere, periodic with period 2, and
integrable on every interval. Let g(x) =
Rx
0
f (t) dt.
a. Prove that g(2n) = 0 for all n ∈ Z.
b. Prove that g is even and period with period 2.
Proof. Claim I: g(2) = 0.
In fact, we have
Z
g(2)
2
=
f (t) dt
0
Z
1
=
Z
0
Z
f (t) dt
1
1
=
Z
0
f (t) dt +
Z
f (z + 2) dz
1
=
Z
0
f (t) dt +
f (z) dz
1
=
Z
0
f (−t) − dt Let z = −t
f (t) dt +
0
Z
1
1
Z
0
f (t) dt −
=
0
Z
−f (t) dt
1
Z
0
f (t) dt +
0
Z
f (t) dt
1
1
Z
f (t) dt −
0
Since f (−t) = −f (t)
1
=
=
Since f (z + 2) = f (z)
−1
0
Z
Let t = z + 2
−1
0
=
2
f (t) dt +
1
f (t) dt
0
0.
So g(2) = 0.
Claim II: g is periodic with period 2, that is, g(x + 2) = g(x) for all x ∈ R.
1
2
MINGFENG ZHAO
In fact, for all x ∈ R, we have
x+2
Z
g(x + 2)
=
f (t) dt
0
x
Z
Z
f (t) dt
x
0
Z
=
x+2
f (t) dt +
=
2
g(x) +
x
Z
x
Z
f (z + 2) dz
f (t) dt +
Z
2
= g(x) +
x
Z
f (t) dt +
x
Z
f (z) dz
x
Z
f (t) dt +
x
Z
Since f (z + 2) = f (z)
0
2
g(x) +
Let t = z + 2
0
x
=
f (t) dt
2
2
= g(x) +
=
x+2
Z
f (t) dt +
f (t) dt Let z = t
0
2
g(x) +
f (t) dt
0
=
g(x) + g(2)
=
g(x),
Since g(2) = 0.
So we get that g is periodic with period 2, that is, g(x + 2) = g(x) for all x ∈ R.
Claim III: g(2n) = 0 for all n ∈ Z.
In fact, for all n ∈ Z, since g(x + 2) = g(x) for all x ∈ R, in particular, we get
g(2n) = g(2(n − 1)) = g(2(n − 2)) = · · · = g(2) = 0.
Hence, we get g(2n) = 0 for all n ∈ Z.
2. [Page 138, Problem 21] Show that lim
1−
x→0
√
1 − x2
1
= .
2
x
2
Proof. For all x 6= 0, we have
1−
√
1 − x2
x2
=
1−
√
√
1 − x2 1 + 1 − x2
√
·
x2
1 + 1 − x2
=
1 1 − (1 − x2 )
√
·
x2 1 + 1 − x2
=
1
x2
√
·
x2 1 + 1 − x2
SOLUTION OF HW6
3
1
√
.
1 + 1 − x2
=
So when x → 0, we have
lim
1−
x→0
√
1 − x2
x2
=
lim
x→0
1
√
1 + 1 − x2
1
.
2
=
3. [Page 139, Problem 27] Let f (x) = sin
1
if x 6= 0. Prove that there exists no real number A such
x
that f (x) → A as x → 0.
Proof. If there exists some real number A such that f (x) → A as x → 0, that is, for any 1 > > 0,
there exists some δ > 0 such that
|f (x) − A| < ,
for all |x| < δ.
In particular, for all |x| < δ, and |y| < δ, we have
|f (x) − f (y)| = |f (x) − A − f (y) + A| ≤ |f (x) − A| + |f (y) − A| < 2.
That is, for any > 0, there exists some δ > 0 such that
|f (x) − f (y)| < ,
for all |x|, |y| < δ.
For that 1 > > 0, we take for all n ≥ 1, we define
xn =
1
2nπ +
π
2
,
and yn =
1
2nπ +
Obviously, we can find that
lim xn = lim yn = 0.
n→∞
n→∞
3π
2
.
4
MINGFENG ZHAO
So for the above δ > 0, there exists some N > 0 such that whenever n ≥ N , we have
0 ≤ xn , yn < δ.
On other hand, we know that
|f (xn ) − f (yn )| =
sin 2nπ + π − sin 2nπ + 3π 2
2 =
2
>
2.
Which contradicts with |f (xn ) − f (yn )| < 2. Therefore, there exists no real number A such that
f (x) → A as x → 0.
4. [Page 139, Problem 32] Let f (x) = x sin
1
if x 6= 0. Define f (0) so that f will be continuous at 0.
x
Proof. Define f (0) = 0. Now we show that f is continuous at x = 0. In fact, for all x 6= 0, we have
|f (x)| =
x sin 1 x
=
1 |x| · sin 2
≤
|x| Since | sin t| ≤ 1 for all t ∈ R.
For any > 0, we take δ = , then for all 0 < |x| < δ = , we have
|f (x) − f (0)| = |f (x) − 0|
= |f (x)|
≤
|x|
<
δ
= .
SOLUTION OF HW6
Which implies that f is continuous at x = 0.
5
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu