7.8 If
Rudin’s Ex. 8
I(x) =
(
0, if x ≤ 0,
1, if x > 0,
P
if {xn } is a sequence of distinct points of (a, b), and if
|cn | converges, prove that
the series
∞
X
f (x) =
cn I(x − xn ),
a≤x≤b
n=1
converges uniformly, and that f is continuous for every x 6= xn .
Proof Put fn (x) = cn I(x − xn ). Since
|fn (x)| = |cn | |In (x − xn )| ≤ |cn |,
and
P
x ∈ [a, b],
|cn | converges, by Theorem 7.10, we know that
∞
X
cn I(x − xn ) converges
n=1
uniformly on [a, b].
m
X
Put gm (x) =
cn I(x − xn ). To prove that f is continuous at every x ∈
/ {xn }, by
n=1
Theorem 7.12, we only need to show that for every m, gm is continuous at such x.
In fact, if we put
δ = min{|x − x1 |, . . . , |x − xm |} > 0,
and if |t − x| < δ, then
|t − xn | ≥ |x − xn | − |x − t| > δ − δ = 0,
n = 1, . . . , m,
so that
I(t − xn ) = 1 = I(x − xn ),
n = 1, . . . m.
Hence, for any > 0, if |t − x| < δ, we have
m
m
X
X
|gm (t) − gm (x)| = cn I(t − xn ) −
cn I(x − xn )
n=1
n=1
m
m
X
X
=
cn −
cn = 0 < .
n=1
n=1
It follows that for each m, gm is continuous at every x ∈
/ {xn }.
7.9 Let {fn } be a sequence of continuous functions which converges uniformly to a
function f on a set E. Prove that
lim fn (xn ) = f (x)
n→∞
for every sequence of points xn ∈ E such that xn → x, and x ∈ E. Is the converse
of this true?
1
Rudin’s Ex. 9
Proof Let > 0 be given. Since {fn } converges to f uniformly on E, there exists
N1 such that n ≥ N1 implies
|fn (t) − f (t)| < /2,
whenever t ∈ E. In particular, if n ≥ N1 , we have
|fn (xn ) − f (xn )| < /2.
From the hypotheses, by Theorem 7.12, f is continuous. Hence, there exists δ > 0
such that |t − x| < δ implies
|f (t) − f (x)| < /2.
For xn → x, there exists N2 such that n ≥ N2 implies
|xn − x| < δ.
Thus, if n ≥ N2 , we have
|f (xn ) − f (x)| < /2.
Put N = max{N1 , N2 }. Then, if n ≥ N , we have
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| < /2 + /2 = .
This implies that lim fn (xn ) = f (x).
n→∞
The converse statement is as follows: Let {fn }, and f be continuous functions on a
set E. If
lim fn (xn ) = f (x)
n→∞
for every sequence of points xn ∈ E such that xn → x, and x ∈ E, prove that {fn }
converges to f uniformly on E.
We give an example to show that the above statement is false. Take E = (0, 1), and
fn (x) =
1
,
nx
n = 1, 2, 3, . . . .
For any sequence xn → x, with x 6= 0, we have
lim fn (xn ) = lim
n→∞
n→∞
1
1
= 0 · = 0.
nxn
x
It is clear that f = 0 is continuous, and for each n, fn is continuous on E. Hence the
sequence {fn } and f satisfy the hypotheses of the statement on E. However, {fn }
does not converge to f on E uniformly. In fact, for n > m, from the inequalities,
1
1 1
1
|fn (x) − fm (x)| = −
≥
,
=
nx mx
(n − m)x
nx
1
we know that the Cauchy criterion of uniform convergence fails, since
can be
nx
unbounded if x is chosen sufficiently close to 0.
2
7.10 Letting (x) denote the fractional part of the real number x, consider the function
f (x) =
∞
X
(nx)
,
n2
n=1
x real.
Find all discontinuities of f , and show that they form a countable dense set. Show
that f is nevertheless Riemann-integrable on every bounded interval.
Proof Put
(nx)
.
n2
fn (x) =
Since
|fn (x)| ≤
1
,
n2
X 1
P
and
converges, by Theorem 7.10,
fn converges uniformly on R. Hence f
2
n
is a well-defined function on R.
If x ∈
/ Q, then nx is not an integer for any n. By Exercise 4.16, we know that fn is
continuous at such x. Hence each fn is continuous on the set of irrational numbers
R − Q. By Theorem 7.12, we know that f is continuous on R − Q.
p
Next we show that f is discontinuous at every point of Q. Let x = ∈ Q, where p
q
and q are co-prime. It is easy to see that
(qt)
(p)
p − [p]
= lim+ 2 = lim+
= 0,
q2
q2
t→p q
t→p
(qt)
(p)
p − [p]
1
lim fq (t) = lim− 2 = lim− 2 = lim−
= 2 > 0.
q
q
q2
q
t→x−
t→x
t→p
t→p
lim fq (t) = lim+
t→x+
t→x
Take an integer N0 such that
∞
X
1
1
< 2,
n2
2q
n=N0
and put N = max{N0 , q}. Since for any n,
lim fn (t) ≤ lim fn (t),
t→x−
t→x+
we have
lim
t→x−
N
X
fn (t) − lim
t→x+
n=1
N
X
fn (t) =
n=1
N X
n=1
lim fn (t) − lim fn (t)
t→x−
t→x+
≥ lim− fq (t) − lim+ fq (t) =
t→x
t→x
fn (t) ≤
∞
X
1
1
< 2,
n2
2q
Since, for any t ∈ R,
0≤
∞
X
n=N +1
fn (t) ≤
∞
X
n=N0
3
n=N0
1
.
q2
Rudin’s Ex. 10
we have
lim [f (x − 1/m) − f (x + 1/m)]
"∞
#
∞
X
X
fn (x − 1/m) −
fn (x + 1/m)
= lim
m→∞
m→∞
"
= lim
m→∞
≥
n=1
n=1
N
X
N
X
fn (x − 1/m) −
n=1
∞
X
fn (x + 1/m) +
n=1
fn (x − 1/m) −
n=N +1
∞
X
#
fn (x + 1/m)
n=N +1
1
1
1
− 2 = 2,
q2
2q
2q
which implies that f is discountinuous at x =
p
.
q
Finally, we show that f is Riemann integrable on every bounded interval. In fact,
since [nt] is Riemann integrable by Theorem 6.9 since it is monotonic. This implies
(nt)
nt − [nt]
that fn (t) =
=
is Riemann integrable on every bounded interval.
n2
n2
k
X
Hence the partial sums
fn is Riemann integrable on every bounded interval. It
n=1
follows from Theorem 7.16 that f is Riemann integrable, since
k
X
fn converges to
n=1
f uniformly.
7.11 Suppose {fn }, {gn } are defined on E, and
P
(a)
fn has uniformly bounded partial sums;
(b) gn → 0 uniformly on E;
(c) g1 (x) ≥ g2 (x) ≥ g3 (x) ≥ · · · for every x ∈ E.
P
Prove that
fn gn converges uniformly on E.
Proof Put
An (x) =
n
X
fn (x).
k=1
By hypothesis (a), there is M such that |An (x)| ≤ M for all n. Let > 0 be given.
4
Rudin’s Ex. 11
by hypothesis (b), there is N such that gN (x) < (/2M ). Thus, if p, q ≥ N , we have
q
q
X
X
[An (x) − An−1 (x)]gn (x)
fn (x)gn (x) = n=p
n=p
q−1
X
An (x)[gn (x) − gn+1 (x)] + Aq (x)gq (x) − Ap−1 (x)gp (x)
=
n=p
≤
q−1
X
|An (x)| |gn (x) − gn+1 (x)| + |Aq (x)| |gq (x)| + |Ap−1 (x)| |gp (x)|
n=p
≤M
q−1
X
!
|gn (x) − gn+1 (x)| + |gq (x)| + |gp (x)|
n=p
= 2M gp (x) ≤ 2M gN (x) < .
P
By the Cauchy Criterion, the series
fn gn converges uniformly on E.
7.14 Suppose f is a real continuous function on R, fn (t) = f (nt) for n = 1, 2, 3, . . . , and
{fn } is equicontinuous on [0, 1]. What conclusion can you draw about f ?
Rudin’s Ex. 15
Proof We shall prove that such an f is constant on [0, ∞). Let > 0 be given.
Since {fn } is equicontinuous on [0, 1], there exists δ > 0 such that for any |s − t| < δ,
s, t ∈ [0, 1], we have
|fn (s) − fn (t)| < ,
n = 1, 2, 3, . . . .
For any two numbers x, y ∈ [0, ∞), we choose a positive integer n such that
|x − y|
< δ,
n
x
≤ 1,
n
y
≤ 1.
n
Then,
x
y |f (x) − f (y)| = fn
− fn
< .
n
n
Since > 0 is arbitrary, we have f (x) = f (y). This means that f is constant on
[0, ∞), since x, y are arbitrary.
7.15 Suppose {fn } is an equicontinuous sequence of functions on a compact set K, and
{fn } converges pointwise on K. Prove that {fn } converges uniformly on K.
Proof Let > 0 be given. Since {fn } is equicontinuous on K, there exists δ > 0
such that for any x, y ∈ K with d(x, y) < δ, we have
|fn (x) − fn (y)| < /3,
n = 1, 2, 3, . . . .
Let Bδ (x) be the open ball of radius δ centered at x ∈ K. The collection {Bδ (x)} is
an open cover of K. Since K is compact, there exists a finite subcover:
K ⊂ Bδ (x1 ) ∪ · · · ∪ Bδ (xn ).
5
Rudin’s Ex. 16
Since {fn } converges pointwise on K to a function, say f , for each i, 1 ≤ i ≤ n,
there is Ni such that n ≥ Ni implies
|fn (xi ) − f (xi )| < /3.
Hence, if we put N = max Ni , then, for n ≥ N , we have
1≤i≤n
|fn (xi ) − f (xi )| < /3,
1 ≤ i ≤ n.
Take any x ∈ K. Assume x ∈ Bδ (xi0 ) for some 1 ≤ i ≤ i0 . By the equicontinuity,
we know that
|fn (x) − fn (xi0 )| < /3,
n = 1, 2, 3, . . . .
Since {fn } converges pointwise to f , we take n → ∞ and have
|f (x) − f (xi0 )| ≤ /3.
Thus, if n ≥ N , we have
|fn (x) − f (x)| ≤ |fn (x) − fn (xi0 )| + |fn (xi0 ) − f (xi0 )| + |f (xi0 ) − f (x)|
< /3 + /3 + /3 = .
This means that {fn } converges to f uniformly on K.
6