Math 2001: Real analysis
Class test, Oct 25, 2016
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Student ID:
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You have 180 minutes to solve eleven problems.
Show your complete work.
The maximal score for this exam is 50. Scores for individual problems add up to 55,
so you can get a perfect score even if you loose ve points.
• No aids permitted: this is a closed book, closed notes exam.
•
•
•
Problem Your score Max. score
1
5
2
5
3
5
4
5
5
5
6
5
7
5
8
5
9
5
10
5
11
5
Total
50
(1) Let f : R → R be an increasing function, and suppose that a sequence (s ) satises
s
= f (s ) for n = 1, 2, . . . Prove that (s ) is a monotone sequence.
n
n+1
n
n
Solution:
Suppose rst that s ≤ s . If s ≤ s for some n ∈ N, then s = f (s ) ≤
f (s ) = s . Hence, by induction, s ≤ s
for all n ∈ N, that is, the sequence
(s ) is increasing. Similarly, if s ≥ s , then, by the same induction argument,
s ≥s
for all n ∈ N, that is, (s ) is decreasing.
1
n+1
n
n
2
n
n+2
n
1
n+1
n+1
n
n+1
n+1
2
n
(2) Find the boundary bd S, the interior int S and the closure cl S of the set
Remember to show your work.
Solution:
√
S = Q ∪ { 2}.
We rst prove that bd S = R. If x ∈ R and ε > 0, then the interval N (x, ε) =
(x − ε, x + ε) contains a rational number, and hence it has non-empty intersection
with S. Furthermore, it has non-empty intersection with the complement of S: a
simple way to see this is to note that each of the intervals√(x − ε, x) and (x, x + ε)
contains an irrational number, and at most one of them is 2 the other one must
belong to R \ S. By the denition of the boundary, x ∈ bd S.
Since boundary and interior are disjoint, we must have int S = ∅. Similarly, the
closure contains the boundary, and thus cl S = R.
(3) Prove that if (s ) is a bounded increasing sequence, then lim s
n
n→∞
n
.
= sup{s1 , s2 , . . .}
Solution:
Let x = sup{s , s , . . .} (this is well-dened, because the set is non-empty and bounded). Fix ε > 0. By the denition of the supremum, there is an element s of
{s , s , . . .} such that s > x − ε. It follows that if n ≥ N , then s ≥ s > x − ε,
and clearly s ≤ x < x + ε. Therefore, |s − x| < ε for all n ≥ N . Since ε > 0 is
arbitrary, we have lim s = x.
1
2
N
1
2
N
n
n
n
n→∞
n
N
(4) Prove that if lim s
n→∞
n
, then lim (s t ) = ∞.
= lim tn = ∞
n→∞
n→∞
n n
Solution:
By denition, there is N such that s > 1 for all n ≥ N . Furthermore, for any
M there is N such that t > |M | for all n ≥ N . Let N = max{N , N }. Then
s t = t > |M | ≥ M for all n ≥ N (for the rst inequality, note that s ≥ 0 and
t ≥ 0 for n ≥ N ). By denition, this means that lim (s t ) = ∞.
0
1
n n
n
n
n
0
1
0
n
1
n
n→∞
n n
(5) (a) Give the denition of a Cauchy sequence.
(b) Give an example of a sequence which is Cauchy, but not monotone.
(c) Give an example of a sequence which is monotone, but not Cauchy.
Remember to show your work.
Solution:
(a) The sequence (s ) is a Cauchy sequence if for every ε > 0 there is N > 0 such
that |s − s | < ε for all n, m ≥ N .
(b) s = (−1) /n is Cauchy (because it is convergent: |s −0| < ε whenever n > 1/ε),
but it is not monotone: s < s > s .
(c) s = n is monotone (increasing), but not Cauchy (because for ε = 1 we have
|s − s | ≥ ε for all n).
n
n
m
n
n
n
1
n
n
n+1
2
3
(6) Let T be the set of all subsequential limits of a sequence (s ), let S = {s , s , . . .}
and let S be the set of all accumulation points of S. Is T equal to S ? Prove or give
a counterexample.
n
1
0
2
0
Solution:
No, if for example s = 0 (constant sequence), then S = {0} and S = ∅ (for 0 is an
isolated point of S), while T = {0}.
(It is in fact easy to prove that S ⊆ T ⊆ S, and that any countable sets S and T
with this property can be obtained in this way).
0
n
0
(7) Suppose that f is a function dened on some pdeleted neighbourhood of 0, and that
lim f (x) = 1. Prove that the function g(x) = f (x) is well-dened in some deleted
neighbourhood of 0 and that lim g(x) = 1. Use merely the denition of the limit of
a function.
√
x→0
x→0
Hint:
a−1
.
a−1= √
a+1
Solution:
Using the denition of lim f (x) = 1 with ε = 1, we see that there is a δ > p0 such
that |f (x) − 1| < 1 when 0 < |x − 0| < δ ; in particular, f (x) > 0 and g(x) = f (x)
is well-dened when 0 < |x − 0| < δ . This proves the rst statement.
Let ε > 0. There is δ > 0 such that |f (x) − 1| < ε when 0 < |x − 0| < δ . Let
δ = min{δ , δ }. Then, if 0 < |x − 0| < δ , then g(x) is well-dened, and
0
x→0
0
0
1
0
1
p
|f (x) − 1|
|f (x) − 1|
|g(x) − 1| = | f (x) − 1| = p
= |f (x) − 1| < ε.
≤
0+1
f (x) + 1
By denition, lim g(x) = 1.
x→0
1
(8) Let f (x) = x when x is a rational number and f (x) = 0 otherwise. Find all points
at which f is continuous.
Solution:
The function f is continuous at 0: if |x − 0| < ε, then |f (x) − f (0)| < ε (for either
f (x) − f (0) = x or f (x) − f (0) = 0, so the denition is satised with δ = ε).
On the other hand, f is discontinuous at any other point c 6= 0. Indeed, if c > 0
is irrational and ε = c, then every interval (c, c + δ) contains a rational number x,
and |f (x) − f (c)| = |x − 0| = x > c = ε, contrary to the denition of continuity.
Similarly, if c > 0 is rational and ε = c, then every interval (c − δ, c + δ) contains an
irrational number x, and |f (x) − f (c)| = |0 − c| = c = ε, again violating the denition
of continuity. The argument for c < 0 is analogous.
(9) Prove that the polynomial f (x) = x
Solution:
6
− 3x + 1
has at least two real roots.
Since f is continuous, f (0) = 1 > 0, f (1) = 1−3+1 = −1 < 0 and f (2) = 64−6+1 =
59 > 0, by intermediate value theorem, f has a root in (0, 1) and another root in
(1, 2).
(10) Let f : R → R be a dierentiable function. Prove that the following two conditions
are equivalent:
(a) f (−x) = f (x) for all x ∈ R;
(b) f (−x) = −f (x) for all x ∈ R.
0
0
Solution:
Dierentiation of both sides of f (−x) = f (x), we obtain −f (−x) = f (x), that is,
f (−x) = −f (x), as desired. Conversely, suppose that f (−x) = −f (x) and let
g(x) = f (x) + f (−x). Then g (x) = f (x) − f (−x) = 0 for all x ∈ R, and therefore
g is constant. Since g(0) = 0, we must have g(x) = 0 for all x ∈ R, and therefore
f (−x) = f (x) for all x ∈ R.
0
0
0
0
0
0
0
0
0
(11) Let f : [a, b] → R be a continuous function on an interval [a, b], dierentiable in (a, b).
Prove that the following two conditions are equivalent:
(a) |f (x)| ≤ M for all x ∈ (a, b);
(b) |f (x) − f (y)| ≤ M |x − y| for all x, y ∈ [a, b].
0
Solution:
If |f (x)| ≤ M for all x ∈ (a, b), then, by the mean value theorem, for each x, y ∈ [a, b] such
that x < y there is m ∈ (x, y) such that f (y) − f (x) = f (m)(y − x). Thus |f (y) − f (x)| ≤
|f (m)||y − x| ≤ M |y − x|, as desired. The case x > y follows by exchanging the roles of x
and y, while the case x = y is automatic.
Conversely, suppose that |f (x) − f (y)| ≤ M |x − y| for all x, y ∈ [a, b]. Then
0
0
0
|f (y) − f (x)|
≤M
y→x
|y − x|
|f 0 (x)| = lim
for all x ∈ (a, b).
(Extra space for solutions)
(Extra space for solutions)