Homework 3 Solutions
1. (Exercise 7.9) Let {fn } be a sequence of continuous functions which converges uniformly to a function f on a set E . Suppose x ∈ E , {xn } ⊆ E , and xn → x.
Claim. limn→∞ fn (xn ) = f (x)
Proof. Let > 0. As fn → f uniformly, f is continuous. Hence, there is a δ > 0 such that whenever
x, y ∈ E and |x − y| < δ , we have |f (x) − f (y)| < . As xn → x, there is an N such that whenever n ≥ N1 ,
we have |xn − x| < δ . Then, if n ≥ N1 , we have |f (xn ) − f (x)| < . Now, nd N2 such that whenever
n ≥ N2 , we have supy∈E |fn (y) − f (y)| < . Then, if N = max{N1 , N2 } and n ≥ N , we have
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| < 2
so fn (xn ) → f (x).
2. (Exercise 7.10)
Let (x) denote the fractional part of the real number x. Let
f (x)
∞
X
(nx)
n2
n=1
=
for x ∈ R.
Claim. f
is dened on R.
If fn (x) = (nx)
n2 , we have |fn (x)| ≤
M -test (theorem 7.10).
Proof.
Claim.
1
n2 ,
so this series converges uniformly on R by the Weierstrass
Every point of Q is a discontinuity for f .
Suppose z = pq ∈ Q be in lowest terms (so q > 0 and gcd(p, q) = 1). Let gn = nk=1 fn , so {gn }
are the partial sums of our series. Consider gq . Note there is a δ > 0 such that I = (z − δ, z + δ) contains only
z with denominator less than or equal to q . Hence, we see gq is continuous
on I\{z} and gq has a jump disconP
η
1
tinuity at z . Set η = gq (z−) − gq (z+) > 0. Let N be such that ∞
n=N n2 < 4 and such that N ≥ q . We note
gN has similar properties to gq at z ; namely that there is a J = (z − δ2 , z + δ2 ) such that gN is continuous on
J\{z} and such that gN has a jump discontinuity at z . Indeed, we see gN (z−) − gN (z+) ≥ gq (z−) − gN (z+).
Hence, as supx∈R |gN (x)−f (x)| < η4 , we havem if f (z−) and f (z+) exist, f (z−)−f (z+) ≥ η −2( η4 ) = η2 > 0.
If either of f (z−) or f (z+) do not exist, then f is not continuous at z , so in either case f is discontinuous at z .
Proof.
Claim.
P
Every point of R\Q is a point of continuity for f .
1
Let z ∈ R\Q. Let > 0 and let N be such that ∞
/ Q, there is a δ1 > 0
n=N n2 < . As z ∈
such that for I = (z − δ1 , z + δ1 ), gN is continuous on I . Hence, we may nd a δ > 0 such that for x with
|x − z| < δ , |gN (x) − gN (z)| < . Then, we have for |x − z| < δ ,
Proof.
P
|f (x) − f (z)| ≤ |f (x) − gN (x)| + |gN (x) − gN (z)| + |gN (z) − f (z)| < 3
so f is continuous at z .
Claim. f ∈ R
on every bounded interval
Proof. For a given bounded interval and given n, gn has nitely many discontinuities, so gn is Riemann
integrable on this interval. Hence, f is integrable on this interval as well as it is a uniform limit of the gn .
1
3. (Exercise 7.12) Let {fn } dened on (0, ∞) be Riemann integrable on [t, T ] whenever 0 < t < T < ∞.
|fn | ≤ g , fn → f uniformly on every compact subset of (0, ∞), and
ˆ ∞
g(x)dx < ∞
0
Claim.
ˆ
n→∞
ˆ
∞
∞
f (x)dx
fn (x)dx =
lim
0
0
´∞
´T
We note the hypotheses give g ≥ 0. Hence, 0 g(x)dx makes sense as either limT →∞ limt→0 t g(x)dx
´T
or limt→0 limT →∞ t g(x)dx. Indeed, we rst show these two limits exist (as possibly ∞) and are equal (that
is, equal if nite or, if one is innite, then so is the other).
Proof.
´T
First, for xed T , we see t g(x)dx increases as t → 0. Hence, we may evaluate this limit in [0, ∞], say
as LT . We note that if t < T < T 0 , then
ˆ
ˆ
T0
ˆ
T
g(x)dx =
T0
g(x)dx +
t
g(x)dx
t
T
´ T0
where T g(x)dx is constant in t (and nite), so either LT is nite for every T , or LT is innite for every
T . Moreover, this shows that if T < T 0 , then LT ≤ LT 0 . Thus, we may evaluate the limit limT →∞ LT
´T
in [0, ∞], as this is an increasing sequence, say as L. Similarly, we set limT →∞ t g(x)dx = Mt and
M = limt→0 Mt , where the justication for taking this limit is similar. We claim M = L, where M, L ∈ [0, ∞].
First suppose L = ∞. By the above, we either have
L = ∞ for all T or LT nite for all T with
´T T
limT →∞ LT = ∞ . Suppose LT = ∞ for all T , so limt→0 t g(x)dx = ∞. We see
ˆ
ˆ
T
T →∞
and so if limt→0
T
g(x)dx ≥
Mt = lim
t
g(x)dx
t
´T
g(x)dx = ∞, we have limt→∞ Mt = ∞, so M = ∞. Now, suppose LT < ∞ for all T but
t
limT →∞ LT = ∞. For N ∈ R, nd S such that whenever T > S , we LT > N . Hence, we have for T > S ,
ˆ
ˆ
U
t→0 U →∞
t→0
T
g(x)dx ≥ lim
M = lim Mt = lim lim
g(x)dx = LT > N
t→0
t
t
so M = ∞.
The analysis for M = ∞ =⇒ L = ∞ is similar and will be omitted here.
Thus, M is innite if and only if L is innte. Suppose both are nite. We now show in this case that
M = L. Let > 0. Find S such that if T > S , we have |LT − L| < . Likewise, nd s such that if t < s, we
have |Mt − M | < . We have, for such t, T ,
|L − M | ≤ |L − LT | + |LT − Mt | + |Mt − M | < 2 + |LT − Mt |
´T
´R
´T
Now, LT − Mt = limr→0 r g(x)dx − limR→∞ t g(x)dx. We look at LT − r g(x)dx. By denition, there
´T
is a δT such that when r < δT we have |LT − r g(x)dx| < . We note for R0 > R we have
ˆ
ˆ
R0
r
ˆ
R0
g(x)dx =
g(x)dx +
R
g(x)dx
r
2
R
and so LR0 = LR +
´ R0
R
g(x)dx. Hence, if |LR −
ˆ
ˆ
R0
|LR0 −
r
g(x)dx| < , then
ˆ
R0
g(x)dx| = ||L +
r
´R
ˆ
R
g(x)dx −
R0
g(x)dx − +
R
r
g(x)dx| < R
That is. we may take δT independent of T for T large (ie nd δS , and this will work for all T > S ). All
´T
together this gives a δ such that if r < δ and T > S , then |LT − r g(x)dx| < .
Similarly, we may nd an N such that if R > N and t < s, then |Mt −
t < min(s, δ) and T > max(S, N ), we have
´R
t
g(x)dx| < . Hence, if
|L − M | ≤ 2 + |LT − Mt |
and
ˆ
|LT − Mt | ≤ |LT −
ˆ
T
T
g(x)dx| + |
t
g(x)dx − Mt | < 2
t
so |L − M | < 4. This shows L = M .
´∞
´∞
Now, we claim that 0 fn (x)dx and 0 f (x)dx make sense as well. Indeed, note |f | ≤ g and, for such
´T
´∞
a function, given > 0´ we may
´ nd t,´T such that | t f (x)dx − t0 f (x)dx| < (use the bounds above for
g and the fact that | f | ≤ |f | ≤ g for various intervals. For such a t, T we know fn converges to f
uniformly on [t, T ], and so there is an N such that if n ≥ N , we have
ˆ
ˆ
T
|
T
fn (x)dx −
f (x)dx| < t
t
Hence, if n ≥ N , we have
ˆ
|
ˆ
∞
fn (x)dx −
0
ˆ
∞
f (x)dx|
≤
ˆ
T
fn (x)dx −
0
0
<
ˆ
∞
|
t
ˆ
T
fn (x)dx| + |
fn (x)dx −
t
ˆ
T
f (x)dx| + |
t
ˆ
T
f (x)dx −
t
3
which proves the claim.
4. (Exercise 7.14) Let f be a continuous real function on R such that 0 ≤ f ≤ 1, f (t + 2) = f (t) for
P∞
t ∈ R, P
f = 0 on [0, 31 ], and f = 1 on [ 32 , 1]. Let Φ(t) = (x(t), y(t)) where x(t) = n=1 2−n f (32n−1 t) and
∞
y(t) = n=1 2−n f (32n t).
Claim. Φ
is continuous and maps [0, 1] onto [0, 1]2 .
Proof. As both x, y are uniformly convergent sequences of continuous functions (eg by the Weierstrass
P∞
P∞
M -test), Φ is continuous. To show onto, let x0 = n=1 2−n a2n−1 and y0 = n=1 2−n a2n where ak ∈ {0, 1}
for each k, so we've written a base-2 expansion of an arbitrary
(x0 , y0 ) in [0, 1]2 (that every point has
P∞ point
−n−1
(2an ). We claim Φ(t0 ) = (x0 , y0 ). To
such an expansion was covered in discussion). Set t0 = n=1 3
see this, we show f (3k t0 ) = ak . If we can show this, then x(t0 ) is
∞
X
2−n f (32n−1 t0 ) =
n=1
∞
X
2−n a2n−1 = x0
n=1
and likewise y(t0 ) = y0 .
Write
3 k t0 =
k−1
X
3k−n−1 (2an ) + 3−1 (2ak ) +
n=0
∞
X
n=k+1
3
3k−n−1 (2an )
∞
f (x)dx|
0
Now,
P
k−n−1
(2an ) is an integer multiple of 2 as n ≤ k − 1 in all summands,
n=0 3
∞
k−n−1
3
(2a
)) from the property that f (t + 2) = f (t) for all t. We note
n
n=k+1
Pk−1
|
∞
X
3k−n−1 (2an )| ≤
n=k+1
∞
X
3k−n−1 2 = 2
∞
X
3−n−2 =
n=0
n=k+1
and 3−1 (2ak ) is either 0 or 32 . Hence, 3−1 (2ak ) +
on ak . Indeed, we see f (3k t0 ) = ak , as claimed.
P∞
n=k+1
2 1
91−
1
3
so f (3k t0 ) = f (3−1 (2ak ) +
=
23
1
=
92
3
3k−n−1 (2an ) lies in either [0, 31 ] or [ 23 , 1], depending
5. Let {fn } be a sequence of monotonically increasing function from [0, 1] to [0, 1]. Suppose fn converge
pointwise on [0, 1] to f and that f is continuous. Prove fn → f uniformly.
Proof. Let > 0. As [0, 1] is compact, f is uniformly continuous, so there is a δ > 0 such that if
k
for 0 ≤ k ≤ M . As
|x − y| < δ , then |f (x) − f (y)| < . Let M > 1δ and consider the points xk = M
fn → f , we can nd an N such that if n ≥ N , then |f (xk ) − fn (xk )| < for every k ∈ {0, . . . , M }.
Now, let x ∈ [0, 1] be arbitrary and nd k such that xk ≤ x ≤ xk+1 . Suppose n ≥ N . Then, as
fn is increasing, we have fn (xk ) ≤ fn (x) ≤ fn (xk+1 ). Now, |fn (xk ) − f (xk )| < and likewise for
xk+1 , so f (xk ) − < fn (x) < f (xk+1 ) + . As each fn is increasing, we see f is also increasing (for
x < y we have fn (y) − fn (x) ≥ 0 for all n so f (y) − f (x) ≥ 0), so f (xk ) ≤ f (x) ≤ f (xk+1 ). Thus,
|fn (x) − f (x)| < f (xk+1 ) + − (f (xk ) − ) = f (xk+1 ) − f (xk ) + 2. From our choice of xk with uniform
continuity, we have f (xk+1 ) − f (xk ) < , so |fn (x) − f (x)| < 3. As this holds independent of x, we have
uniform convergence.
6.
Let f : R → R be continuous and set fn (x) =
Claim. fn
1
n
Pn−1
k=0
f (x + nk ).
converges uniformly on every bounded interval.
´ x+1
Let g(x) = x f (t)dt. Let I = [a, b] with a < b. We show fn converges uniformly to g on I .
Let > 0. As f is uniformly continuous on [a, b + 1], we may nd a δ > 0 such that for x, y ∈ [a, b + 1], if
|x − y| < δ , then |f (x) − f (y)| < . Let N > 1δ . Suppose n ≥ N , so n1 < δ . Then, for x ∈ [a, b], we have
Proof.
ˆ x+1
n−1
k
1X
f (x + ) −
f (t)dt|
n
n
x
k=0
n−1
X ˆ x+ k+1
n
k
(f (x + ) − f (t))dt|
= |
k
n
k=0 x+ n
k+1
ˆ
n−1
X x+ n
k
≤
|f (x + ) − f (t)|dt
k
n
k=0 x+ n
n−1
X ˆ x+ k+1
n
<
dt
|fn (x) − g(x)| = |
k=0
= n
= k
x+ n
n
so fn converges uniformly to g on [a, b + 1].
4