Homework 2 Solutions
Math 501
Due September 12, 2014
Exercise 1
Let y ∈ R, n ∈ N, and > 0. We claim that there exists δ > 0 such that
|u − y| < δ implies
|un − y n | < .
If n = 1, it is clear that δ = works.
n
o
If n = 2, let δ = min 1, 2|y|+1
and suppose that |u − y| < δ. Note that if
|u − y| < δ ≤ 1, then it is not too hard to see that |u + y| ≤ 2|y| + 1 (draw a
number line). Now observe that
|u2 − y 2 | = |u + y||u − y| < δ|u + y| ≤ δ(2|y| + 1) ≤
(2|y| + 1) = .
2|y| + 1
Here is the proof in general. [The book suggests to use induction, but I
don’t see how that follows from the hint about the identity.] Observe that if
|u − y| < |y|, then it is not hard to see that |u| < 2|y|. It follows then that if
|u − y| < |y| then
|un−1 + un−2 y + · · · + y n−1 | ≤ |u|n−1 + |u|n−2 |y| + · · · + |y|n−1
< 2n−1 |y|n−1 + 2n−2 |y|n−1 + · · · + |y|n−1
n
X
= |y|n−1
2n−k
k=1
= |y|n−1 (2n − 1).
Let
δ = min |y|, n−1 n
|y|
(2 − 1)
1
and suppose |u − y| < δ. Then
|un − y n | = |u − y||un−1 + un−2 y + · · · + y n−1 |
< δ|un−1 + un−2 y + · · · + y n−1 |
< δ|y|n−1 (2n − 1)
< n−1 n
|y|n−1 (2n − 1)
|y|
(2 − 1)
= .
Exercise 2
Let x > 0 and n ∈ N. Define
A = {s > 0 : sn ≤ x},
y = sup A.
It is clear that A is bounded above by max{1, x}. Therefore y is a real number.
We claim that y n = x. To show this, we show that neither y n > x nor y n < x
can hold.
First suppose that y n > x. Let = y n − x. Then there exists 0 < u < y
such that
0 < y n − un < = y n − x.
Therefore,
un = y n − (y n − un ) > y n + x − y n = x,
which implies u is an upper bound for A. However, this together with the fact
that u < y contradicts that y is the least upper bound for A.
Next suppose that y n < x. Let = x − y n . Then there exists u > y such
that
0 < un − y n < = x − y n .
Therefore,
un = (un − y n ) + y n < x − y n + y n = x,
which implies u ∈ A. However, this together with the fact that u > y contradicts
that y is an upper bound for A.
Exercise 3
Let x ∈ R. We suppose x > 0 for simplicity in our arguments and note that
the arguments hold for the case that x < 0 with minor modifications.
Consider the following definitions. Let N be the largest integer less than or
equal to x, let x1 be the largest integer less than or equal to 10(x − N ), and for
2
each n ≥ 1, let xn+1 be the largest integer less than or equal to
x1
xn 10n+1 x − N +
+ ··· + n
10
10 x
xn−1 1
n
− xn .
= 10 10 x − N +
+ · · · + n−1
10
10
(a) It is clear that x − N ∈ [0, 1). Therefore x1 is an integer between 0 and 9.
Observe that since x2 is the largest integer less than or equal to
x1 = 10(10(x − N ) − x1 )
102 x − N +
10
and by construction 10(x − N ) − x1 ∈ [0, 1), which implies 10(10(x − N ) −
x1 ) < 10, it follows that x2 is an integer between 0 and 9. Suppose by way
of induction
x1
xn−1 10n x − N +
+ · · · + n−1
− xn ∈ [0, 1).
10
10
It follows then that xn+1 is an integer between 0 and 9.
(b) Next we show that for each k there exists ` ≥ k such that xk 6= 9. Suppose
by way of contradiction that there exists k such that for all ` ≥ k we have
x` = 9. Let
k−1
X xn
y=N+
10n
n=1
By definition xk = 9, means that
9 ≤ 10k (x − y) < 10,
where the inequality on the right is a consequence of part (a). This implies
9 · 10−k + y ≤ x < 101−k + y.
In general, we note that x` = 9 for each ` > k means by definition that
!
`−1
X
9
`
9 ≤ 10 x − y −
< 10,
10n
n=k
which in turn implies
y+
`
`−1
X
X
9
9
1−`
≤
x
<
10
+
y
+
,
n
10
10n
n=k
n=k
which is equivalent to
y+
`
X
9
≤ x < y + 101−k .
10n
n=k
3
Therefore,
y+
∞
X
9
≤ x < y + 101−k .
10n
(1)
n=k
Note however that
∞
X
9
= 101−k ,
10n
n=k
which shows that (1) is a contradiction since it cannot be the case that
y + 101−k < y + 101−k .
(c) Consider a decimal expansion N.x1 x2 . . . not terminating in an infinite
string of 9’s and consider the set
o
n
x1
x2
x1
+
,... .
S = N, N + , N +
10
10 100
Let xk ≤ 9 for all k ≥ 1, it is clear that S is bounded above by
N+
∞
X
9
=N +1
10k
k=1
Therefore it has a least upper bound. Note that by construction the real
number x with decimal expansion N.x1 x2 . . . is an upper bound by construction. We claim that x is the least upper bound. Let ∈ (0, 1) be a real
number with decimal expansion 0.y1 y2 . . .. Let K = inf{k ≥ 1 : yk > 0}.
Observe that
∞
∞
X
X
xk
9
1
≤
= K
k
k
10
10
10
k=K+1
k=K+1
and
=
∞
X
yk
yK
≥ K.
10k
10
k=1
Next, observe that
x−=N +
∞
X
xk
−
10k
k=1
≤N+
K
∞
X
X
xk
xk
yK
+
− K
k
k
10
10
10
k=1
≤N+
K
X
k=1
≤N+
k=K+1
1 − yK
xk
+
k
10
10K
K
X
xk
,
10k
k=1
which in turn must be smaller than some element of S. Thus for all > 0,
x − is not an upper bound for S, which implies x is the least upper bound.
4
Exercise 4
(a) Let E be a convex set. We claim that for each k ≥ 1, if w1 , . . . , wk ∈ E and
s1 , . . . , sk ∈ [0, 1] such that s1 +· · · sk = 1, then s1 w1 +· · ·+sk wk ∈ E. Note
that the case k = 1 is trivial and the case k = 2 is true by the definition of
convexity.
Suppose the claim holds for k > 2, and let w1 , . . . , wk+1 ∈ E and s1 , . . . , sk+1 ∈
[0, 1] satisfy s1 + · · · + sk+1 = 1. Let
s = s1 + · · · + sk .
and let
sk
s1
w1 + · · · + wk
s
s
Observe that by the induction hypothesis
z=
z=
s1
sk
w1 + · · · + wk ∈ E.
s
s
Moreover, note that.
s1 w1 + · · · + sk wk + sk+1 wk+1 = sz + sk+1 wk+1 .
Further, note that s + sk+1 = 1, which implies by the base case of induction
that sz + sk+1 wk+1 ∈ E, and the claim is proved.
(b) Suppose for any k ≥ 1, we have that if w1 , . . . , wk ∈ E and s1 , . . . , sk ∈ [0, 1]
with s1 + · · · sk = 1, then s1 w1 + · · · + sk wk ∈ E. Note that if k = 2, this
is simply the definition of convexity.
Exercise 5
(a) Let a, b, c > 0 and consider the ellipsoid defined by
y2
z2
x2
E = (x, y, z) : 2 + 2 + 2 ≤ 1 .
a
b
c
Define the following pairing of vectors u, v ∈ R3 :
hu, vi =
u1 v1
u2 v2
u3 v3
+ 2 + 2 .
2
a
b
c
It is clear that hu, vi = hv, ui. Moreover, if u, v, w ∈ R3 and t ∈ R, then
(u1 + tv1 )w1
(u2 + tv2 )w2
(u3 + tv3 )w3
+
+
a2
b2 c2
u1 w1
u2 w2
u3 w3
v1 w 1
v2 w2
v3 w3 =
+
+
+
t
+
+
a2
b2
c2
a2
b2
c2
= hu, wi + thv, wi.
hu + tv, wi =
5
Next note that
u2
u2
u21
+ 22 + 23 ≥ 0
2
a
b
c
and equality holds if and only if u1 = u2 = u3 = 0. Thus we’ve shown that
our pairing is an inner product. Moreover, note that
E = u ∈ R3 : hu, ui ≤ 1 .
hu, ui =
Recall that our argument that showed that the Euclidean closed unit ball is
convex required only that the usual Euclidean dot product satisfy the three
properties of an inner product. Therefore the same argument shows that E
is convex.
Exercise 6
(a) Define
an =
1 + 1/n
−1 − 1/n
if n is odd,
if n is even.
and let
An = {an , an+1 , an+2 , . . .}
and
xn = sup An ,
yn = inf An .
(i) Let n ≥ 1. Observe that if n is even
xn = sup An
1
1
1
1
,1 +
, −1 −
,...
= sup 1 + , −1 −
n
n+1
n+2
n+3
1
=1+ ,
n
and if n is odd
xn = sup An
1
1
1
1
= sup −1 − , 1 +
, −1 −
,1 +
,...
n
n+1
n+2
n+3
1
=1+
.
n+1
Thus the sequence (xn ) is given by
1
1
1
1
1
1
1 + ,1 + ,1 + ,1 + ,1 + ,1 + ,...
2
2
4
4
6
6
which implies
lim sup an = lim xn = lim 1 +
n→∞
n→∞
6
k→∞
1
= 1.
2k
(ii) Let n ≥ 1. Observe that if n is even
yn = inf An
1
1
1
1
= inf 1 + , −1 −
,1 +
, −1 −
,...
n
n+1
n+2
n+3
1
= −1 −
,
n+1
and if n is odd
yn = inf An
1
1
1
1
, −1 −
,1 +
,...
= inf −1 − , 1 +
n
n+1
n+2
n+3
1
= −1 − .
n
Thus the sequence (yn ) is given by
1
1
1
1
1
1
−1, −1 − , −1 − , −1 − , −1 − , −1 − , −1 − , . . .
3
3
5
5
7
7
which implies
lim sup an = lim xn = lim −1 −
n→∞
n→∞
k→∞
1
= −1.
2k
(b) Let (an ) ⊂ R be a bounded sequence.
(i) Let A ⊂ B ⊂ R be subsets. Suppose first B is not bounded above.
Then sup B = ∞ and it is trivially true that sup A ≤ sup B. Suppose
next that A is empty. Then sup A = −∞ and the desired inequality is
again trivially true. Thus, it suffices to assume that A is non-empty
that B is bounded above. It is clear that since A ⊂ B, a bound for B
is also a bound for A. Thus A is bounded above as well and x = sup A
and y = sup B are both real numbers. Note that y is an upper bound
for B, which implies that it is an upper bound for A as well. Since x
is the least upper bound for A, it follows that x ≤ y.
(ii) Observe that An+1 ⊂ An for each n ≥ 1, which implies xn = sup An ≥
sup An+1 = xn+1 by the above result. Thus the sequence (xn ) is
monotone.
(iii) Note further that the sequence (xn ) is bounded because the sequence
(an ) is bounded. Therefore by the Monotone Convergence Theorem
(Exercise 4 of Homework 1), it follows that xn converges.
7