Math 318 Exam #1 Solutions
1. (a) Suppose (fn ) and (gn ) are two sequences of functions that converge uniformly on a subset A ⊂ R. Is it true that the sequence
(fn gn ) converges uniformly on A? Prove or give a counterexample.
Answer. No. For each n, let
fn (x) =
1
x
and gn (x) =
1
n
on the interval (0, 1). Then certainly fn → x1 uniformly (since
it’s a constant sequence) and gn → 0 uniformly (since the value
of gn is independent of x). However, I claim that the sequence
1
((fn gn )(x)) =
nx
does not converge uniformly. Notice that, for any given > 0 and
1
any fixed x ∈ (0, 1), we can pick N > x
so that n ≥ N implies
1
= 1 ≤ 1 < ,
−
0
nx
nx
Nx
so the sequence (1/(nx)) converges to the zero function pointwise.
However, for any fixed N ∈ N, we can choose x such that 0 <
x < 1/N so that
1
1
N x − 0 = N x > 1,
so the convergence cannot be uniform.
(b) If your answer to (a) was “yes”, does the result still hold if one
of the sequences does not converge uniformly? If your answer to
(a) was “no”, what additional assumptions on (fn ) and (gn ) will
yield uniform convergence of (fn gn )?
Answer. Notice that the limiting function 1/x above was not
bounded (though the limit of the gn obviously is). Here’s the
modified claim: Suppose fn → f uniformly and gn → g uniformly
on A ⊆ R and that both f and g are bounded. Then (fn gn )
converges uniformly on A.
Proof. Let > 0. Suppose M1 > 0 is an upper bound for f
(meaning |f (x)| ≤ M1 for all x ∈ A) and that M2 is an upper
bound for g. Since fn → f uniformly, there exists N1 ∈ N such
that n ≥ N1 implies
|fn (x) − f (x)| < 1
for all x ∈ A. Equivalently, |fn (x)| < |f (x)| + 1 ≤ M1 + 1 for all
x ∈ A.
In turn, since gn → g uniformly, there exists N2 ∈ N such that
n ≥ N2 implies
|gn (x) − g(x)| <
.
2(M1 + 1)
Finally, since fn → f uniformly, there exists N3 ∈ N such that
n ≥ N3 implies
|fn (x) − f (x)| <
.
2M2
Now, if N = max{N1 , N2 , N3 } and n ≥ N , we have
|fn (x)gn (x) − f (x)g(x)| = |fn (x)gn (x) − fn (x)g(x) + fn (x)g(x) − f (x)g(x)|
≤ |fn (x)gn (x) − fn (x)g(x)| + |fn (x)g(x) − f (x)g(x)|
= |fn (x)||gn (x) − g(x)| + |g(x)||fn (x) − f (x)|
≤ (M1 + 1)|gn (x) − g(x)| + M2 |fn (x) − f (x)|
< (M1 + 1)
+ M2
2(M + 1)
2M2
=
for all x ∈ A, so we see that, indeed, (fn gn ) → (f g) uniformly on
A.
2. (a) Let (fn ) be a sequence of continuous functions. Suppose fn → f
uniformly on A ⊆ R. Prove that
lim fn (xn ) = f (x)
n→∞
for all x ∈ A and all sequences (xn ) in A converging to x.
Proof. Fix x ∈ A and let (xn ) be a sequence converging to x. Let
> 0. The fact that fn → f uniformly on A implies that there
exists N1 ∈ N such that n ≥ N1 implies
|fn (y) − f (y)| < /2
for all y ∈ A.
In turn, since fn → f uniformly and each fn is continuous, the
limit function f is also continuous. Hence, by Theorem 4.3.2(iv),
limn→∞ f (xn ) = f (x), so there exists N2 ∈ N such that n ≥ N2
implies
|f (xn ) − f (x)| < /2.
Therefore, if N = max{N1 , N2 } and n ≥ N , then
|fn (xn ) − f (x)| = |fn (xn ) − f (xn ) + f (xn ) − f (x)|
≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)|
< /2 + /2
= .
Since the choice of > 0 was arbitrary, we see that limn→∞ fn (xn ) =
f (x). In turn, since the choice of x ∈ A was arbitrary, we see that
this is true for all x ∈ A, as desired.
(b) Is the converse true?
Answer. No. Consider the functions fn (x) = nx on all of R. Then
(fn (x)) → 0, but not uniformly. However, the conclusion of part
(a) will hold. To see this, let x ∈ R and (xn ) → x. Let > 0.
Since (xn ) → x, there exists N1 ∈ N such that n ≥ N1 implies
|xn − x| < 1, meaning that |xn | < |x| + 1. Also, we can certainly
pick N2 such that N2 > |x|+1
.
Therefore, if N = max{N1 , N2 } and n ≥ N , we see that
|fn (xn )−f (x)| = |fn (xn )−0| =
|xn |
|x| + 1
|x| + 1
|x| + 1
≤
≤
≤
< .
n
n
N
N2
Since the choice of > 0 was arbitrary, we see that, indeed,
limn→∞ fn (xn ) = f (x). Since the choice of sequence (xn ) → x
was arbitrary, the same holds for any sequence converging to x.
Finally, since the choice of x was arbitrary, this is true for any
x ∈ R.
3. Let fn be a monotone increasing,
continuous function on [0, 1] for each
P
n ∈ N. Suppose f (x) = ∞
f
n=0 n (x) converges for every x ∈ [0, 1].
Show that the function f is continuous on [0, 1].
Proof. Let > 0. The goal is to show that the series satisfies the
hypotheses
of the Cauchy Criterion. To that end, note that, since
P
fn (0) converges, there exists N0 ∈ N such that n > m ≥ N0 implies
|fm+1 (0) + . . . + fn (0)| < /2.
P
Likewise, since
fn (1) converges, there exists N1 ∈ N such that n >
m ≥ N1 implies
|fm+1 (1) + . . . + fn (1)| < /2.
Let N = max{N0 , N1 } and let n > m ≥ N . Since each fi is increasing,
the function fm+1 + . . . + fn is also increasing, so Lemma 3.1, stated
below, implies that
|fm+1 (x) + · · · + fn (x)| ≤ |fm+1 (0) + . . . + fn (0)| + |fm+1 (1) + . . . + fn (1)|
< /2 + /2
=
P
for any x ∈ [0, 1]. Therefore, the series
fn satisfies the hypotheses
P
of the Cauchy Criterion for Uniform Convergence, so we see that fn
converges uniformly.
Since each fn is continuous, Theorem 6.4.2 implies that f is also continuous.
Lemma 3.1. Suppose g : [a, b] → R is monotone increasing. Then
|g(x)| ≤ |g(a)| + |g(b)|
for any x ∈ [a, b].
Proof. The fact that g is increasing implies that
g(a) ≤ g(x) ≤ g(b)
for any x ∈ [a, b]. Therefore,
|g(x)| ≤ max{|g(a)|, |g(b)|} ≤ |g(a)| + |g(b)|,
as desired.
(1)
P∞
P
Let Sn = nk=1 ak be the nth partial sum
Pn
Sk
σn = k=1 .
n
P∞
We say that the series n=1 an is Cesaro summable to L if lim σn = L.
A
P consequence of HW 4 Problem #5 from last semester is that if
an = L, then the series is Cesaro summable to L.
P∞
n+1 is Cesaro summable to 1 .
(a) Show that the series
n=1 (−1)
2
(You showed on HW 3 Problem #5 that this is also the Abel
sum of the series.)
4. Consider a series
and define
n=1 an .
Proof. Clearly,
S1 = 1
S2 = 1 − 1 = 0
S3 = 1 − 1 + 1 = 1
S4 = 1 − 1 + 1 − 1 = 0
..
.
Sk =
1 (−1)k+1
+
2
2
so
Pn
Pn
σn =
k=1 Sk
n
=
k=1
1
2
+
n
(−1)k+1
2
=
n
2
+ S2n
1 Sn
= +
.
n
2 2n
Since Sn is always either 0 or 1, the term S2nn certainly goes to
zero as n → ∞, so
1 Sn
1
lim σn = lim
+
= .
n→∞
n→∞ 2
2n
2
Hence, the series is indeed Cesaro summable to 12 .
P
(b) Does there exist a series ∞
n=1 an that is Abel summable but not
Cesaro summable?
Answer. Yes. Consider an = (−1)n n. Then on [0, 1) we have
that
!
∞
∞
∞
X
X
X
d
d
1
−x
n
n
n
n−1
n n
(−1) nx = x
(−1) nx
=x
(−1) x
=x
=
.
dx
dx 1 + x
(1 + x)2
n=1
n=1
n=0
Therefore, since the function on the right is continuous at 1, we
see that
lim
x→1−
∞
X
(−1)n nxn = lim
x→1−
n=1
−x
1
=−
(1 + x)2
4
is the Abel sum of the series.
However, the partial sums look like
S1 = −1
S2 = −1 + 2 = 1
S3 = −1 + 2 − 3 = −2
S4 = −1 + 2 − 3 + 4 = 2
..
.
S2k−1 = −k
S2k = k
Therefore,
(
0
σn =
− n+1
2n
if n is even
if n is odd.
The even terms go to zero, but the odd terms go to −1/2, so
lim σn does not exist, so the series is not Cesaro summable.
P∞
5. Let n=0 an xn be a power series with each an ≥ 0. Suppose that the
radius of
is 1, so that the power series defines a function
Pconvergence
n at least on (−1, 1).
a
x
f (x) = ∞
n=0 n
Prove that the power series converges at x = 1 (meaning f (1) makes
sense) if and only if f is bounded on [0, 1).
Proof. (⇒) If the power series converges at x = 1, then Abel’s Theorem implies that f is continuous on the compact set [0, 1] and, therefore, bounded on that set. Hence, f is bounded on [0, 1).
(⇐) On the other hand, suppose f is bounded onP
[0, 1). My goal is to
show that the sequence of partial sums sk (1) = kn=0 an is bounded
above; since each an ≥ 0, the sequence (sk (1)) is monotone increasing,
so boundedness plus the Monotone Convergence Theorem will imply
convergence.
To see that
P the sk (1) are bounded, let M > 0 be an upper bound for
f (x) = an xn on [0, 1), meaning that |f (x)| ≤ M for all x ∈ [0, 1).
Now, I claim that M is an upper bound for sk (1) for all k ∈ N. Suppose
not. Then there exists K ∈ N such that sK (1) > M . But then, since
sK (x) ≤ f (x) ≤ M for all x ∈ [0, 1), we see that
|sK (1) − sK (x)| > sK (1) − M
for any x ∈ [0, 1).
But then this means that sK cannot possibly be continuous at x = 1,
since there is no δ that will work for = sK (1) − M > 0. On the other
hand, sK is clearly continuous since it is just a polynomial. From
this contradiction, then, we can conclude that M is indeed an upper
bound for sk (1) for every k and, therefore, that the sequence (sk (1))
converges, which is exactly what it means to say that the power series
converges at x = 1.