Math 318 HW #2 Solutions
1. Exercise 6.2.16. For each n ∈ N, let fn be a function defined on [0, 1]. If (fn ) is bounded on
[0, 1]—that is, there exists an M > 0 such that |fn (x)| ≤ M for all n ∈ N and x ∈ [0, 1]—and
if the collection of functions (fn ) is equicontinuous (Exercise 6.2.15), follow these steps to
show that (fn ) contains a uniformly convergent subsequence.
(a) Use Exercise 6.2.14 to produce a subsequence (fnk ) that converges at every rational point
in [0, 1]. To simplify the notation, set gk = fnk . It remains to show that (gk ) converges
uniformly on all of [0, 1].
Proof. Since (fn ) is bounded and Q ∩ [0, 1] is countable, Exercise 6.2.14 implies that
there is a subsequence (fnk ) that converges at all points in Q ∩ [0, 1].
(b) Let > 0. By equicontinuity, there exists a δ > 0 such that
gk (x) − gk (y)| <
3
for all |x − y| < δ and k ∈ N. Using this δ, let r1 , r2 , . . . , rm be a finite collection of
rational points with the property that the union of the neighborhoods Vδ (ri ) contains
[0, 1].
Explain why there must exist an N ∈ N such that
|gs (ri ) − gt (ri )| <
3
for all s, t ≥ N and ri in the finite subset of [0, 1] just described. Why does having the
set {r1 , r2 , . . . , rm } be finite matter?
Proof. Let > 0. Since r1 , . . . , rm ∈ Q ∩ [0, 1], we know that each sequence (gk (ri ))
converges and, hence, is Cauchy. Therefore, for each i = 1, . . . , m, there exists Ni ∈ N
such that r, s ≥ Ni implies that
|gs (ri ) − gt (ri )| < .
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Let N = max{N1 , . . . , Nm } (this is where it’s essential that the collection {r1 , . . . , rm }
is finite; if it were infinite, then there might well not be a maximum Ni ). Then, for any
r, s ≥ N and for any i ∈ {1, . . . , m}, we have that s, t ≥ Ni and so
|gs (ri ) − gt (ri )| < ,
3
as desired.
(c) Finish the argument by show that, for an arbitrary x ∈ [0, 1],
|gs (x) − gt (x)| < for all s, t ≥ N .
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Proof. Let > 0. Choose N as in part (b). Then, if s, t ≥ N and x ∈ [0, 1], we have
that x ∈ Vδ (ri ) for some i, and so
|gs (x) − gt (x)| = |(gs (x) − gs (ri )) + (gs (ri ) − gt (ri )) + (gt (ri ) − gt (x))|
≤ |gs (x) − gs (ri )| + |gs (ri ) − gt (ri )| + |gt (ri ) − gt (x)|.
Now, the first and third terms on the right hand side are less than /3 by the equicontinuity part of (b), whereas the second is less than /3 by what we proved in (b).
Therefore, the above is strictly less than /3 + /3 + /3 = , so we conclude that
|gs (x) − gt (x)| < .
Since the choice of x ∈ [0, 1] was arbitrary, we conclude that this holds for all x ∈ [0, 1].
Since the choice of > 0 was arbitrary, we see that, indeed, the sequence (gk ) converges
uniformly on [0, 1].
2. Exercise 6.3.2. Consider the sequence of functions defined by
gn (x) =
xn
.
n
(a) Show that (gn ) converges uniformly on [0, 1] and find g = lim gn . Show that g is differentiable and compute g 0 (x) for all x ∈ [0, 1].
Proof. Let > 0 and choose N > 1/. Then, if n ≥ N and x ∈ [0, 1], we have
n
x
xn
1n
1
1
|gn (x) − 0| = − 0 =
≤
= ≤
< .
n
n
n
n
N
Since the choice of > 0 was arbitrary, we see that (gn ) → 0 uniformly on [0, 1], so
g(x) = 0 for all x.
The function g is obviously differentiable, and g 0 (x) = 0 for all x ∈ [0, 1].
(b) Now, show that (gn0 ) converges on [0, 1]. Is the convergence uniform? Set h = lim gn0 and
compare h and g 0 . Are they the same?
Proof. First, note that
nxn−1
= xn−1 .
n
I claim that lim gn0 (x) = h(x), where h is the function defined by
(
0 if 0 ≤ x < 1
h(x) =
1 if x = 1.
gn0 (x) =
To see this, let > 0 and let x ∈ [0, 1]. If x = 0 or 1, then gn (x) = x for all x, so clearly
lim gn (x) = h(x) when x = 0 or 1.
ln Otherwise h(x) = 0, so we want to show that lim gn0 (x) = 0. Pick N > ln
x + 1, which is
N
−1
equivalent to saying that x
< . Then for any n ≥ N ,
|gn0 (x) − h(x)| = xn−1 − 0 = xn−1 ≤ xN −1 < .
2
Since the choice of > 0 was arbitrary, we see that gn0 (x) → h(x) for all x.
The convergence cannot possibly be uniform, since each gn0 is continuous, yet h is not
continuous, so uniform convergence would violate Theorem 6.2.6.
Finally, notice that, although h and g 0 agree practically everywhere, they are not the
same function, since h(1) = 1 whereas g 0 (1) = 0.
3. Exercise 6.3.5. Use the following advice to supply a proof for Theorem 6.3.2. To get started,
observe that the triangle inequality implies that, for any x ∈ [a, b],
|fn (x) − fm (x)| ≤ |(fn (x) − fm (x)) − (fn (x0 ) − fm (x0 ))| + |fn (x0 ) − fm (x0 )|.
Now, apply the Mean Value Theorem to fn − fm .
Proof. Let > 0. Since (fn (x0 )) converges, it is Cauchy, and so there exists N1 ∈ N such
that m, n ≥ N1 implies
|fn (x0 ) − fm (x0 )| < /2.
Also, since (fn0 ) converges uniformly, the Cauchy criterion for uniform convergence applies,
so there exists N2 ∈ N such that m, n ≥ N2 implies
0
|fn0 (x) − fm
(x)| <
2(b − a)
for any x ∈ [a, b].
Now, let N = max{N1 , N2 } and pick m, n ≥ N . If I define g(x) := fn (x) − fm (x), then g
0 (x). For any x ∈ [a, b], either x = x (in
is differentiable on [a, b] with g 0 (x) = fn0 (x) − fm
0
which case |fn (x) − fm (x)| < /2 < ) or the Mean Value Theorem implies that there exists
c ∈ [a, b] such that x0 < c < x (if x0 < x; modify slightly in the case x0 > x) and
g 0 (c)(x − x0 ) = g(x) − g(x0 ).
Hence,
|fn (x) − fm (x)| = |(fn (x) − fm (x)) − (fn (x0 ) − fm (x0 )) + (fn (x0 ) − fm (x0 ))|
≤ |(fn (x) − fm (x)) − (fn (x0 ) − fm (x0 ))| + |fn (x0 ) − fm (x0 )|
= |g 0 (c)(x − x0 )| + |fn (x0 ) − fm (x0 )|
where the first inequality follows from an application of the triangle inequality and the last
equality follows from the above Mean Value Theorem argument.
The second term on the right in the last line is less than /2 since m, n ≥ N ≥ N1 , while the
first term is bounded by
0
|g 0 (c)||x − x0 | ≤ |g 0 (c)|(b − a) = |fn0 (c) − fm
(c)|(b − a) <
(b − a) = /2.
2(b − a)
Putting this all together, we see that
|fn (x) − fm (x)| < /2 + /2 = .
Since the choice of > 0 was arbitrary, the Cauchy Criterion implies that (fn ) converges
uniformly.
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4. Exercise 6.4.4. In Section 5.4, we postponed the argument that the nowhere-differentiable
function
∞
X
1
g(x) =
h (2n x)
2n
n=0
is continuous on R. Use the Weierstrass M-test to supply the missing proof.
Proof. First, notice that 0 ≤ h(y) ≤ 1 for all y ∈ R. Therefore, for any x ∈ R,
1
h(2n x) = 1 |h(2n x)| ≤ 1 · 1 = 1 .
2n
2n 2n 2n
Now, we know from Example 2.7.5 that
∞
X
1
= 2,
2n
n=0
so the Weierstrass M-test implies that the series
∞
X
1
h(2n x)
2n
n=0
converges uniformly on R.
Notice that
hn (x) =
1
h(2n x) = (qn ◦ h ◦ pn )(x)
2n
where pn (x) = 2n x and qn (x) = 21n x. Since h is continuous and since pn and qn are obviously
continuous (by Theorem 4.3.4), Theorem 4.3.9 implies that hn is continuous.
P
Therefore, since each hn is continuous and since the series g =
hn converges uniformly,
Theorem 6.4.2 implies that g is continuous.
5. Exercise 6.4.7. Let
h(x) =
∞
X
n=1
x2
1
.
+ n2
(a) Show that h is a continuous function defined on all of R.
Proof. Notice that, for any x ∈ R,
1 1
1
x2 + n2 = |x2 + n2 | ≤ n2 .
P 1
P 1
Since the series
converges, the Weierstrass M-test implies that
converges
n2
x2 +n2
uniformly on R.
1
Therefore, since each function hn (x) = x2 +n
2 is continuous on R, Theorem 6.2.6 implies
that the limit function h is also continuous on all of R.
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(b) Is h differentiable? If so, is the derivative function h0 continuous?
Answer. Yes, h is differentiable. To prove this, first note that, if hn (x) =
h0n (x) =
1
,
x2 +n2
then
−2x
.
+ n2 )2
(x2
Second, notice that h0n (0) = 0 and that
−2x
−2
= lim
=0
2
2
x→+∞ 4x(x2 + n2 )
+n )
lim h0n (x) = lim
x→+∞ (x2
x→+∞
by L’Hôpital’s Rule. Since h0n (x) < 0 for all x > 0, the above two facts (along with the
fact that h0n is differentiable) imply that h0n must have a global minimum (at least for
x ≥ 0) at some critical point of h0n . Since
h00n (x) =
6x2 − 2n2
−2(x2 + n2 )2 + 8x2 (x2 + n2 )
=
,
(x2 + n2 )4
(x2 + n2 )3
√
we see that h0n has critical points when 6x2 − 2n2 = 0, meaning when x = ±n/
3. In
√
particular, the above analysis implies that h0n has a global minimum at x = n/ 3.
0
0
Moreover, since
√ hn is an odd function, we know that hn must have a global maximum
at x = −n/ 3. Therefore, for all x ∈ R,
√
√
h0n (n/ 3) ≤ h0n (x) ≤ h0n (−n/ 3).
Equivalently,
√
√
−3 3
3 3
0
≤ hn (x) ≤
,
8n3
8n3
so we see that
|h0n (x)|
Since the series
P 3√ 3
8n3
√
3 3
.
≤
8n3
converges, the Weierstrass M-test implies that the series
∞
X
h0n (x)
n=1
converges uniformly on all of R.
P
In turn, since we already know the series h(x) =
hn (x) converges, Theorem 6.3.1
implies that
∞
∞
X
X
−2x
.
h0 (x) =
h0n (x) =
2
(x + n2 )2
n=1
n=1
Since this series converges uniformly and since each term in continuous, we see that h0
is also continuous.
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