Math 131A, Lecture 1
Real Analysis
Sample Final Exam
Instructions: You have three hours to complete the exam. There are ten problems, worth a
total of one hundred points. Write your solutions in the space below the questions. If a question
is in multiple parts, clearly label each part. If you need more space use the back of the page.
Questions are not in order of difficulty. Do not forget to write your name in the space below.
Name:
Question Points Score
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
10
10
10
Total:
100
Problem 1.
(a) [5pts.] Define the expression limx→a+ f (x) = L.
Solution: We say that limx→a+ f (x) = L if there is an interval (a, b) such that
if (xn ) is a sequence in (a, b) converging to a, then (f (xn )) converges to L.
(b) [5pts.] Prove that if limx→a+ f (x) = limx→a− = L, then limx→a f (x) = L.
Solution: We know there is some interval (c, a) such that any sequence (xn ) in
(c, a) converging to a has f (xn ) → L, and also some interval (a, b) such that any
sequence (xn ) in (a, b) converging to a has f (xn ) → L. Let (tn ) be any sequence
in (c, a) ∪ (a, b), and divide tn into two subsequences (tnk ) consisting of those
tn such that tnk < a and (tm` ) consisting of those tm` such that tm` > a. Then
for > 0, because the left-hand limit is L, there is some N1 such that nk > N1
implies |f (tnk ) − L| < , and because the right-hand limit is L, there is some N2
such that m` > N2 implies |f (tm` ) − L| < . So for n > max{N1 , N2 }, we must
have |f (tn ) − L| < . Since was arbitrary, f (tn ) → L; since (tn ) was arbitrary,
limx→a f (x) = L.
Problem 2.
(a) [5pts.] Let f be a bounded function on [a, b]. Assuming that the upper Darboux
sums U (f, P ) have already been defined, define the upper Darboux integral U (f ).
Solution: U (f ) = inf{U (f, P ) : P is a partition of [a, b]}.
(b) [5pts.] Prove that if f (x) and g(x) are integrable on [a, b] and f (x) ≤ g(x) for all
Rb
Rb
x ∈ [a, b], then a f (x) dx ≤ a g(x) dx.
Solution: Since f and g are integrable, h = f − g is integrable. Moreover,
h(x) ≥ 0 for all x, so for any partition P of [a, b], we see that L(h, P ) ≥ 0. Ergo
Rb
Rb
Rb
Rb
h
=
L(h)
≥
0.
But
h
=
f
−
g. The result follows
a
a
a
a
Problem 3.
(a) [5pts.] Prove that ||a| − |b|| ≤ |a − b|.
Solution: By the triangle inequality, |a| ≤ |a−b|+|b|. Therefore |a|−|b| ≤ |a−
b|. Similarly, |b|−|a| ≤ |a−b|, so −|a−b| ≤ |a|−|b|. Therefore ||a|−|b|| ≤ |a−b|.
(b) [5pts.] Prove that if f (x) is continuous at x0 then |f |(x) = |f (x)| is continuous at
x0 .
Solution: Let > 0. Then since f is cts at x0 , there exists δ > 0 such that
|x−x0 | < δ implies |f (x)−f (x0 )| < . But then if |x−x0 | < δ, ||f |(x)−|f |(x0 )| <
|f (x) − f (x0 )| < . Ergo |f | is continuous at x0 .
Problem 4.
(a) [5pts.] State the Mean Value Theorem.
Solution: Let f be a continuous function on [a, b] which is differentiable on
(a, b). Then there is some x ∈ (a, b) such that
f 0 (x) =
f (b) − f (a)
.
b−a
(b) [5pts.] Suppose f is differentiable on R and f 0 (x) ≤ 4 for all x ∈ R. Prove that
there is at most one a > 2 such that f (a) = a2 .
Solution: Suppose there are two numbers b > a such that f (a) = a2 and
f (b) = b2 . Then since f is differentiable, hence continuous, on R, by MVT
there is some x ∈ (a, b) such that
f (b) − f (a)
b−a
2
b − a2
=
b−a
=b+a
f 0 (x) =
But f 0 (x) ≤ 4 for all x ∈ R, so b + a ≤ 4, implying that at most one of a and b
is greater than 2.
Problem 5.
(a) [5pts.] Give a definition of a Cauchy sequence.
Solution: We say a sequence (sn ) is Cauchy if for any > 0, there is some N
such that n, m > M implies that |sn − sm | < .
(b) [5pts.] Prove that if f is a uniformly continuous function on (a, b) and (sn ) is a
Cauchy sequence in (a, b), then (f (sn )) is also a Cauchy sequence. [Note: Do not
try to use the fact that Cauchy sequences converge in R.]
Solution: Let > 0. Since f is uniformly continuous on (a, b), there is some
δ > 0 such that if x, y ∈ (a, b) and |x − y| < δ, then |f (x) − f (y)| < . Choose
N such that n, m > N implies that |sn − sm | < δ, then n, m > N implies that
|f (sn ) − f (sm )| < . Since was arbitrary, (f (sn )) is Cauchy.
Problem 6.
(a) [4pts.] Define the radius of convergence of a power series
P∞
n=0
an (x − c)n .
1
1
Solution: Let β = lim sup |an | n . Then R = β1 , with the convention that “ ∞
is 0.
00
P
P
(b) [4pts.] Suppose that
an xn has finite radius of convergence R > 0 and
an ≥ 0
for all n. Show that if the series converges at R, it converges at −R.
P
Solution:
We
know
an Rn converges. But |an (−R)n | = an Rn , so by comparP
ison
an Rn also converges.
(c) [2pts.] Give an example of a power series that converges on exactly (−1, 1].
Solution:
P∞
n=1
(−x)n
n
is an example.
Problem 7.
(a) [5pts.] Define lim inf sn .
Solution: We say lim inf sn = limN →∞ inf{sn : n > N }.
(b) [5pts.] Prove that if lim inf sn = lim sup sn = s then (sn ) converges to s.
Solution: Since s = lim sup sn , there exists a positive number N1 such that
|s − sup{sn : n > N1 }| < . In particular, sup{sn : n > N1 } < s + , so for
n > N1 , sn < s + . Similarly, since lim sup sn = s, there exists N2 such that
n > N2 implies that s− < sn . Therefore if n ≥ max{N1 , N2 }, s− < sn < s+ ,
so |s − sn | < . Ergo sn → s.
Problem 8.
(a) [5pts.] State the Bolzano-Weierstrauss Theorem.
Solution: Every bounded sequence has a convergent subsequence.
(b) [5pts.] Prove that a continuous function f on a closed interval [a, b] is bounded.
Solution: Suppose not. Wlog f is unbounded above. Therefore for all n ∈ N,
there is some xn ∈ [a, b] such that f (xn ) > n. Now (xn ) is a bounded sequence
(since it is contained in [a, b]), so it has a convergent subsequence (xnk ), with
limit some x0 ∈ [a, b], since [a, b] is closed. So by continuity, f (xnk ) → f (x0 ).
But this is nonsense, because f (xnk ) diverges to infinity. So f must be bounded
on [a, b].
Problem 9.
(a) [3pts.] Let f be a function defined on an open interval I. What does it mean for f
to be differentiable at a point a ∈ I?
Solution: We say that f is differentiable at a if the limit
f 0 (a) = lim
x→a
f (x) − f (a)
x−a
exists, in which case we say f 0 (a) is the derivative of f at a.
(b) [4pts.] Let u(x) and v(x) be defined on an interval I and both differentiable at a
a ∈ I. Prove that (uv)0 (a) = u0 (a)v(a) + u(a)v 0 (a).
Solution: We compute
lim
x→a
uv(x) − uv(a)
u(x)v(x) − u(a)v(x) + u(a)v(x) − u(a)v(a)
= lim
x→a
x−a
x−a
v(x) − v(a)
u(x) − u(a)
= lim
· v(x) + u(a) ·
x→a
x−a
x−a
0
0
= u (a)v(a) + u(a)v (a)
(c) [3pts.] Give an example of two nonzero functions u(x) and v(x) and a point a such
that one of u and v is not differentiable at a but uv(x) is differentiable at a. (If
you’re feeling ambitious, try making both u and v not differentiable at a.)
1
2
Solution: There are lots of examples, but e.g. u(x) = x 3 and v(x) = x 3 are
each not differentiable at a = 0 and their product uv(x) = x is.
Problem 10.
(a) [5pts.] Let f (x) be bounded on [a, b]. Assuming that the upper and lower Darboux
integrals have been defined, what does it mean to say that f (x) is integrable on
[a, b]?
Solution: We say f is integrable if the upper Darboux integral U (f ) and lower
Rb
Darboux integral L(f ) are equal, and we say this number is a f .
(b) [5pts.] Prove that if f (x) is continuous on [a, b] and f (x) ≥ 0 for all x ∈ [a, b] then
Z
b
f (x) dx = 0
a
implies that f (x) = 0 for all x ∈ [a, b]. (Note: You may assume that a function
that is bounded on [a, b] and continuous at all but finitely many points of [a, b] is
integrable on [a, b].)
Solution: Suppose not. Then there is some x0 ∈ [a, b] such that f (x0 ) = α > 0,
Rb
which implies that f (x) > α2 on some interval (c, d) containing x. Then a f ≥
Rd
f ≥ α2 (d − c) > 0. Contradiction.
c