Sequences of Functions
Uniform convergence
9.1 Assume that fn → f uniformly on S and that each fn is bounded on S. Prove that {fn} is uniformly bounded on S.
Proof : Since fn → f uniformly on S, then given ε = 1, there exists a positive integer n0 such that as n ≥ n0, we have
|fn(x) − f (x)| ≤ 1 for all x ∈ S. (*) Hence, f (x) is bounded on S by the following
|f (x)| ≤ |fn0(x)| + 1 ≤ M (n0) + 1 for all x ∈ S. (**) where |fn0(x)| ≤ M (n0) for all x ∈ S.
Let |f1(x)| ≤ M (1) , ..., |fn0−1(x)| ≤ M (n0− 1) for all x ∈ S, then by (*) and (**),
|fn(x)| ≤ 1 + |f (x)| ≤ M (n0) + 2 for all n ≥ n0. So,
|fn(x)| ≤ M for all x ∈ S and for all n where M = max (M (1) , ..., M (n0− 1) , M (n0) + 2) .
Remark: (1) In the proof, we also shows that the limit function f is bounded on S.
(2) There is another proof. We give it as a reference.
Proof : Since Since fn→ f uniformly on S, then given ε = 1, there exists a positive integer n0 such that as n ≥ n0, we have
|fn(x) − fn+k(x)| ≤ 1 for all x ∈ S and k = 1, 2, ...
So, for all x ∈ S, and k = 1, 2, ...
|fn0+k(x)| ≤ 1 + |fn0(x)| ≤ M (n0) + 1 (*) where |fn0(x)| ≤ M (n0) for all x ∈ S.
Let |f1(x)| ≤ M (1) , ..., |fn0−1(x)| ≤ M (n0− 1) for all x ∈ S, then by (*),
|fn(x)| ≤ M for all x ∈ S and for all n
where M = max (M (1) , ..., M (n0− 1) , M (n0) + 1) .
9.2
Define two sequences {fn} and {gn} as follows:fn(x) = x
1 + 1
n
if x ∈ R, n = 1, 2, ...,
gn(x) =
1
n if x = 0 or if x is irrational,
b + n1 if x is rational, say x = ab, b > 0.
Let hn(x) = fn(x) gn(x) .
(a) Prove that both {fn} and {gn} converges uniformly on every bounded interval.
Proof : Note that it is clear that
n→∞lim fn(x) = f (x) = x, for all x ∈ R and
n→∞lim gn(x) = g (x) =
0 if x = 0 or if x is irrational, b if x is ratonal, say x = ab, b > 0.
In addition, in order to show that {fn} and {gn} converges uniformly on every bounded interval, it suffices to consider the case of any compact interval [−M, M ] , M > 0.
Given ε > 0, there exists a positive integer N such that as n ≥ N, we
have M
n < ε and 1 n < ε.
Hence, for this ε, we have as n ≥ N
|fn(x) − f (x)| = x n ≤ M
n < ε for all x ∈ [−M, M ] and
|gn(x) − g (x)| ≤ 1
n < ε for all x ∈ [−M, M ] .
That is, we have proved that {fn} and {gn} converges uniformly on every bounded interval.
Remark: In the proof, we use the easy result directly from definition of uniform convergence as follows. If fn → f uniformly on S, then fn → f uniformly on T for every subset T of S.
(b) Prove that hn(x) does not converges uniformly on any bounded in- terval.
Proof : Write hn(x) =
x
n 1 + 1n
if x = 0 or x is irrational a + na 1 + 1b + bn1
if x is rational, say x = ab . Then
n→∞lim hn(x) = h (x) =
0 if x = 0 or x is irrational a if x is rational, say x = ab .
Hence, if hn(x) converges uniformly on any bounded interval I, then hn(x) converges uniformly on [c, d] ⊆ I. So, given ε = max (|c| , |d|) > 0, there is a positive integer N such that as n ≥ N, we have
max (|c| , |d|) > |hn(x) − h (x)|
=
nx 1 + n1 = |x|n
1 + 1n
if x ∈ Qc∩ [c, d] or x = 0
an 1 + 1b +bn1
if x ∈ Q ∩ [c, d] , x = ab which implies that (x ∈ [c, d] ∩ Qc or x = 0)
max (|c| , |d|) > |x|
n
1 + 1 n
≥ |x|
n ≥ max (|c| , |d|) n
which is absurb. So, hn(x) does not converges uniformly on any bounded interval.
9.3
Assume that fn → f uniformly on S, gn→ f uniformly on S.(a) Prove that fn+ gn→ f + g uniformly on S.
Proof : Since fn → f uniformly on S, and gn → f uniformly on S, then given ε > 0, there is a positive integer N such that as n ≥ N, we have
|fn(x) − f (x)| < ε
2 for all x ∈ S and
|gn(x) − g (x)| < ε
2 for all x ∈ S.
Hence, for this ε, we have as n ≥ N,
|fn(x) + gn(x) − f (x) − g (x)| ≤ |fn(x) − f (x)| + |gn(x) − g (x)|
< ε for all x ∈ S.
That is, fn+ gn→ f + g uniformly on S.
Remark: There is a similar result. We write it as follows. If fn → f uniformly on S, then cfn→ cf uniformly on S for any real c. Since the proof is easy, we omit the proof.
(b) Let hn(x) = fn(x) gn(x) , h (x) = f (x) g (x) , if x ∈ S. Exercise 9.2 shows that the assertion hn → h uniformly on S is, in general, incorrect.
Prove that it is correct if each fn and each gn is bounded on S.
Proof : Since fn → f uniformly on S and each fn is bounded on S, then f is bounded on S by Remark (1) in the Exercise 9.1. In addition, since gn → g uniformly on S and each gn is bounded on S, then gn is uniformly bounded on S by Exercise 9.1.
Say |f (x)| ≤ M1 for all x ∈ S, and |gn(x)| ≤ M2 for all x and all n. Then given ε > 0, there exists a positive integer N such that as n ≥ N, we have
|fn(x) − f (x)| < ε
2 (M2+ 1) for all x ∈ S and
|gn(x) − g (x)| < ε
2 (M1+ 1) for all x ∈ S which implies that as n ≥ N, we have
|hn(x) − h (x)| = |fn(x) gn(x) − f (x) g (x)|
= |[fn(x) − f (x)] [gn(x)] + [f (x)] [gn(x) − g (x)]|
≤ |fn(x) − f (x)| |gn(x)| + |f (x)| |gn(x) − g (x)|
< ε
2 (M2+ 1)M2+ M1 ε 2 (M1 + 1)
< ε 2+ ε
2
= ε
for all x ∈ S. So, hn → h uniformly on S.
9.4 Assume that fn → f uniformly on S and suppose there is a constant M > 0 such that |fn(x)| ≤ M for all x in S and all n. Let g be continuous on the closure of the disk B (0; M ) and define hn(x) = g [fn(x)] , h (x) = g [f (x)] , if x ∈ S. Prove that hn → h uniformly on S.
Proof : Since g is continuous on a compact disk B (0; M ) , g is uniformly continuous on B (0; M ) . Given ε > 0, there exists a δ > 0 such that as
|x − y| < δ, where x, y ∈ S, we have
|g (x) − g (y)| < ε. (*)
For this δ > 0, since fn → f uniformly on S, then there exists a positive integer N such that as n ≥ N, we have
|fn(x) − f (x)| < δ for all x ∈ S. (**) Hence, by (*) and (**), we conclude that given ε > 0, there exists a positive integer N such that as n ≥ N, we have
|g (fn(x)) − g (f (x))| < ε for all x ∈ S.
Hence, hn → h uniformly on S.
9.5 (a) Let fn(x) = 1/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that {fn} converges pointwise but not uniformly on (0, 1) .
Proof : First, it is clear that limn→∞fn(x) = 0 for all x ∈ (0, 1) . Supppos that {fn} converges uniformly on (0, 1) . Then given ε = 1/2, there exists a positive integer N such that as n ≥ N, we have
|fn(x) − f (x)| =
1 1 + nx
< 1/2 for all x ∈ (0, 1) .
So, the inequality holds for all x ∈ (0, 1) . It leads us to get a contradiction since
1
1 + N x < 1
2 for all x ∈ (0, 1) ⇒ lim
x→0+
1
1 + N x = 1 < 1/2.
That is, {fn} converges NOT uniformly on (0, 1) .
(b) Let gn(x) = x/ (nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that gn → 0 uniformly on (0, 1) .
Proof : First, it is clear that limn→∞gn(x) = 0 for all x ∈ (0, 1) . Given ε > 0, there exists a positive integer N such that as n ≥ N, we have
1/n < ε
which implies that
|gn(x) − g| =
x 1 + nx
=
1
1 x + n
< 1 n < ε.
So, gn→ 0 uniformly on (0, 1) .
9.6 Let fn(x) = xn. The sequence {fn(x)} converges pointwise but not uniformly on [0, 1] . Let g be continuous on [0, 1] with g (1) = 0. Prove that the sequence {g (x) xn} converges uniformly on [0, 1] .
Proof : It is clear that fn(x) = xn converges NOT uniformly on [0, 1]
since each term of {fn(x)} is continuous on [0, 1] and its limit function f = 0 if x ∈ [0, 1)
1 if x = 1.
is not a continuous function on [0, 1] by Theorem 9.2.
In order to show {g (x) xn} converges uniformly on [0, 1] , it suffices to shows that {g (x) xn} converges uniformly on [0, 1). Note that
n→∞lim g (x) xn = 0 for all x ∈ [0, 1).
We partition the interval [0, 1) into two subintervals: [0, 1 − δ] and (1 − δ, 1) . As x ∈ [0, 1 − δ] : Let M = maxx∈[0,1]|g (x)| , then given ε > 0, there is a positive integer N such that as n ≥ N, we have
M (1 − δ)n< ε which implies that for all x ∈ [0, 1 − δ] ,
|g (x) xn− 0| ≤ M |xn| ≤ M (1 − δ)n < ε.
Hence, {g (x) xn} converges uniformly on [0, 1 − δ] .
As x ∈ (1 − δ, 1) : Since g is continuous at 1, given ε > 0, there exists a δ > 0 such that as |x − 1| < δ, where x ∈ [0, 1] , we have
|g (x) − g (1)| = |g (x) − 0| = |g (x)| < ε which implies that for all x ∈ (1 − δ, 1) ,
|g (x) xn− 0| ≤ |g (x)| < ε.
Hence, {g (x) xn} converges uniformly on (1 − δ, 1) .
So, from above sayings, we have proved that the sequence of functions {g (x) xn} converges uniformly on [0, 1] .
Remark: It is easy to show the followings by definition. So, we omit the proof.
(1) Suppose that for all x ∈ S, the limit function f exists. If fn → f uniformly on S1(⊆ S) , then fn → f uniformly on S, where # (S − S1) <
+∞.
(2) Suppose that fn → f uniformly on S and on T. Then fn → f uni- formly on S ∪ T.
9.7 Assume that fn→ f uniformly on S and each fn is continuous on S.
If x ∈ S, let {xn} be a sequence of points in S such that xn→ x. Prove that fn(xn) → f (x) .
Proof : Since fn → f uniformly on S and each fn is continuous on S, by Theorem 9.2, the limit function f is also continuous on S. So, given ε > 0, there is a δ > 0 such that as |y − x| < δ, where y ∈ S, we have
|f (y) − f (x)| < ε 2.
For this δ > 0, there exists a positive integer N1 such that as n ≥ N1, we have
|xn− x| < δ.
Hence, as n ≥ N1, we have
|f (xn) − f (x)| < ε
2. (*)
In addition, since fn → f uniformly on S, given ε > 0, there exists a positive integer N ≥ N1 such that as n ≥ N, we have
|fn(x) − f (x)| < ε
2 for all x ∈ S which implies that
|fn(xn) − f (xn)| < ε
2. (**)
By (*) and (**), we obtain that given ε > 0, there exists a positie integer N such that as n ≥ N, we have
|fn(xn) − f (x)| = |fn(xn) − f (xn)| + |f (xn) − f (x)|
< ε 2+ ε
2
= ε.
That is, we have proved that fn(xn) → f (x) .
9.8
Let {fn} be a seuqnece of continuous functions defined on a compact set S and assume that {fn} converges pointwise on S to a limit function f.Prove that fn→ f uniformly on S if, and only if, the following two conditions hold.:
(i) The limit function f is continuous on S.
(ii) For every ε > 0, there exists an m > 0 and a δ > 0, such that n > m and |fk(x) − f (x)| < δ implies |fk+n(x) − f (x)| < ε for all x in S and all k = 1, 2, ...
Hint. To prove the sufficiency of (i) and (ii), show that for each x0 in S there is a neighborhood of B (x0) and an integer k (depending on x0) such that
|fk(x) − f (x)| < δ if x ∈ B (x0) .
By compactness, a finite set of integers, say A = {k1, ..., kr} , has the property that, for each x in S, some k in A satisfies |fk(x) − f (x)| < δ. Uniform convergence is an easy consequences of this fact.
Proof : (⇒) Suppose that fn → f uniformly on S, then by Theorem 9.2, the limit function f is continuous on S. In addition, given ε > 0, there exists a positive integer N such that as n ≥ N, we have
|fn(x) − f (x)| < ε for all x ∈ S Let m = N, and δ = ε, then (ii) holds.
(⇐) Suppose that (i) and (ii) holds. We prove fk→ f uniformly on S as follows. By (ii), given ε > 0, there exists an m > 0 and a δ > 0, such that n > m and |fk(x) − f (x)| < δ implies |fk+n(x) − f (x)| < ε for all x in S and all k = 1, 2, ...
Consider
fk(x0)(x0) − f (x0)
< δ, then there exists a B (x0) such that as x ∈ B (x0) ∩ S, we have
fk(x0)(x) − f (x) < δ
by continuity of fk(x0)(x) − f (x) . Hence, by (ii) as n > m fk(x0)+n(x) − f (x)
< ε if x ∈ B (x0) ∩ S. (*) Note that S is compact and S = ∪x∈S(B (x) ∩ S) , then S = ∪pk=1(B (xk) ∩ S) . So, let N = maxpi=1(k (xp) + m) , as n > N, we have
|fn(x) − f (x)| < ε for all x ∈ S with help of (*). That is, fn→ f uniformly on S.
9.9
(a) Use Exercise 9.8 to prove the following theorem of Dini: If {fn} is a sequence of real-valued continuous functions converginf pointwise to a continuous limit function f on a compact set S, and if fn(x) ≥ fn+1(x) for each x in S and every n = 1, 2, ..., then fn→ f uniformly on S.Proof : By Exercise 9.8, in order to show that fn→ f uniformly on S, it suffices to show that (ii) holds. Since fn(x) → f (x) and fn+1(x) ≤ fn(x) on S, then fixed x ∈ S, and given ε > 0, there exists a positive integer N (x) = N such that as n ≥ N, we have
0 ≤ fn(x) − f (x) < ε.
Choose m = 1 and δ = ε, then by fn+1(x) ≤ fn(x) , then (ii) holds. We complete it.
Remark: (1) Dini’s Theorem is important in Analysis; we suggest the reader to keep it in mind.
(2) There is another proof by using Cantor Intersection Theorem.
We give it as follows.
Proof : Let gn = fn− f, then gnis continuous on S, gn→ 0 pointwise on S, and gn(x) ≥ gn+1(x) on S. If we can show gn → 0 uniformly on S, then we have proved that fn → f uniformly on S.
Given ε > 0, and consider Sn := {x : gn(x) ≥ ε} . Since each gn(x) is continuous on a compact set S, we obtain that Sn is compact. In addition, Sn+1 ⊆ Sn since gn(x) ≥ gn+1(x) on S. Then
∩Sn6= φ (*)
if each Sn is non-empty by Cantor Intersection Theorem. However (*) contradicts to gn → 0 pointwise on S. Hence, we know that there exists a positive integer N such that as n ≥ N,
Sn = φ.
That is, given ε > 0, there exists a positive integer N such that as n ≥ N, we have
|gn(x) − 0| < ε.
So, gn→ 0 uniformly on S.
(b) Use he sequence in Exercise 9.5(a) to show that compactness of S is essential in Dini’s Theorem.
Proof : Let fn(x) = 1+nx1 , where x ∈ (0, 1) . Then it is clear that each fn(x) is continuous on (0, 1) , the limit function f (x) = 0 is continuous on (0, 1) , and fn+1(x) ≤ fn(x) for all x ∈ (0, 1) . However, fn→ f not uniformly on (0, 1) by Exercise 9.5 (a). Hence, compactness of S is essential in Dini’s Theorem.
9.10
Let fn(x) = ncx (1 − x2)n for x real and n ≥ 1. Prove that {fn} converges pointwsie on [0, 1] for every real c. Determine those c for which the convergence is uniform on [0, 1] and those for which term-by-term integration on [0, 1] leads to a correct result.Proof : It is clear that fn(0) → 0 and fn(1) → 0. Consider x ∈ (0, 1) , then |1 − x2| := r < 1, then
n→∞lim fn(x) = lim
n→∞ncrnx = 0 for any real c.
Hence, fn→ 0 pointwise on [0, 1] . Consider
fn0 (x) = nc 1 − x2n−1
(2n − 1)
1
2n − 1 − x2
, then each fn has the absolute maximum at xn= √ 1
2n−1. As c < 1/2, we obtain that
|fn(x)| ≤ |fn(xn)|
= nc
√2n − 1
1 − 1 2n − 1
n
= nc−12
r n
2n − 1
1 − 1 2n − 1
n
→ 0 as n → ∞. (*)
In addition, as c ≥ 1/2, if fn→ 0 uniformly on [0, 1] , then given ε > 0, there exists a positive integer N such that as n ≥ N, we have
|fn(x)| < ε for all x ∈ [0, 1]
which implies that as n ≥ N,
|fn(xn)| < ε which contradicts to
n→∞lim fn(xn) =
√1
2e if c = 1/2
∞ if c > 1/2 . (**)
From (*) and (**), we conclude that only as c < 1/2, the seqences of functions converges uniformly on [0, 1] .
In order to determine those c for which term-by-term integration on [0, 1] , we consider
Z 1 0
fn(x) dx = nc 2 (n + 1) and
Z 1 0
f (x) dx = Z 1
0
0dx = 0.
Hence, only as c < 1, we can integrate it term-by-term.
9.11 Prove that P xn(1 − x) converges pointwise but not uniformly on [0, 1] , whereas P (−1)nxn(1 − x) converges uniformly on [0, 1] . This illus- trates that uniform convergence of P fn(x) along with pointwise con- vergence of P |fn(x)| does not necessarily imply uniform conver- gence of P |fn(x)| .
Proof : Let sn(x) =Pn
k=0xk(1 − x) = 1 − xn+1, then sn(x) → 1 if x ∈ [0, 1)
0 if x = 1 .
Hence, P xn(1 − x) converges pointwise but not uniformly on [0, 1] by The- orem 9.2 since each sn is continuous on [0, 1] .
Let gn(x) = xn(1 − x) , then it is clear that gn(x) ≥ gn+1(x) for all x ∈ [0, 1] , and gn(x) → 0 uniformly on [0, 1] by Exercise 9.6. Hence, by Dirich- let’s Test for uniform convergence, we have proved thatP (−1)nxn(1 − x) converges uniformly on [0, 1] .
9.12 Assume that gn+1(x) ≤ gn(x) for each x in T and each n = 1, 2, ..., and suppose that gn → 0 uniformly on T. Prove thatP (−1)n+1gn(x) con- verges uniformly on T.
Proof : It is clear by Dirichlet’s Test for uniform convergence.
9.13
Prove Abel’s test for uniform convergence: Let {gn} be a sequence of real-valued functions such that gn+1(x) ≤ gn(x) for each x in T and for every n = 1, 2, ... If {gn} is uniformly bounded on T and ifP fn(x) converges uniformly on T, then P fn(x) gn(x) also converges uniformly on T.Proof : Let Fn(x) = Pn
k=1fk(x) . Then sn(x) =
n
X
k=1
fk(x) gk(x) = Fng1(x)+
n
X
k=1
(Fn(x) − Fk(x)) (gk+1(x) − gk(x))
and hence if n > m, we can write
sn(x)−sm(x) = (Fn(x) − Fm(x)) gm+1(x)+
n
X
k=m+1
(Fn(x) − Fk(x)) (gk+1(x) − gk(x))
Hence, if M is an uniform bound for {gn} , we have
|sn(x) − sm(x)| ≤ M |Fn(x) − Fm(x)| + 2M
n
X
k=m+1
|Fn(x) − Fk(x)| . (*)
Since P fn(x) converges uniformly on T, given ε > 0, there exists a positive integer N such that as n > m ≥ N, we have
|Fn(x) − Fm(x)| < ε
M + 1 for all x ∈ T (**)
By (*) and (**), we have proved that as n > m ≥ N,
|sn(x) − sm(x)| < ε for all x ∈ T.
Hence, P fn(x) gn(x) also converges uniformly on T.
Remark: In the proof, we establish the lemma as follows. We write it as a reference.
(Lemma) If {an} and {bn} are two sequences of complex numbers, define
An=
n
X
k=1
ak.
Then we have the identity
n
X
k=1
akbk= Anbn+1−
n
X
k=1
Ak(bk+1− bk) (i)
= Anb1+
n
X
k=1
(An− Ak) (bk+1− bk) . (ii)
Proof : The identity (i) comes from Theorem 8.27. In order to show (ii), it suffices to consider
bn+1 = b1+
n
X
k=1
bk+1− bk.
9.14 Let fn(x) = x/ (1 + nx2) if x ∈ R, n = 1, 2, ... Find the limit function f of the sequence {fn} and the limit function g of the sequence {fn0} .
(a) Prove that f0(x) exists for every x but that f0(0) 6= g (0) . For what values of x is f0(x) = g (x)?
Proof : It is easy to show that the limit function f = 0, and by fn0 (x) =
1−nx2
(1+nx2)2, we have
n→∞lim fn0 (x) = g (x) = 1 if x = 0 0 if x 6= 0 .
Hence, f0(x) exists for every x and f0(0) = 0 6= g (0) = 1. In addition, it is clear that as x 6= 0, we have f0(x) = g (x) .
(b) In what subintervals of R does fn → f uniformly?
Proof : Note that
1 + nx2
2 ≥√
n |x|
by A.P. ≥ G.P. for all real x. Hence,
x 1 + nx2
≤ 1
2√ n which implies that fn→ f uniformly on R.
(c) In what subintervals of R does fn0 → g uniformly?
Proof : Since each fn0 = 1−nx2
(1+nx2)2 is continuous on R, and the limit function g is continuous on R − {0} , then by Theorem 9.2, the interval I that we consider does not contains 0. Claim that fn0 → g uniformly on such interval I = [a, b] which does not contain 0 as follows.
Consider
1 − nx2 (1 + nx2)2
≤ 1
1 + nx2 ≤ 1 na2,
so we know that fn0 → g uniformly on such interval I = [a, b] which does not contain 0.
9.15 Let fn(x) = (1/n) e−n2x2 if x ∈ R, n = 1, 2, ... Prove that fn → 0 uniformly on R, that fn0 → 0 pointwise on R, but that the convergence of {fn0} is not uniform on any interval containing the origin.
Proof : It is clear that fn → 0 uniformly on R, that fn0 → 0 pointwise on R. Assume that fn0 → 0 uniformly on [a, b] that contains 0. We will prove that it is impossible as follows.
We may assume that 0 ∈ (a, b) since other cases are similar. Given ε = 1e, then there exists a positive integer N0 such that as n ≥ max N0,1b := N (⇒ N1 ≤ b), we have
|fn0 (x) − 0| < 1
e for all x ∈ [a, b]
which implies that
2 N x e(N x)2
< 1
e for all x ∈ [a, b]
which implies that, let x = N1, 2 e < 1
e
which is absurb. So, the convergence of {fn0} is not uniform on any interval containing the origin.
9.16 Let {fn} be a sequence of real-valued continuous functions defined on [0, 1] and assume that fn → f uniformly on [0, 1] . Prove or disprove
n→∞lim
Z 1−1/n 0
fn(x) dx = Z 1
0
f (x) dx.
Proof : By Theorem 9.8, we have
n→∞lim Z 1
0
fn(x) dx = Z 1
0
f (x) dx. (*)
Note that {fn} is uniform bound, say |fn(x)| ≤ M for all x ∈ [0, 1] and all n by Exercise 9.1. Hence,
Z 1 1−1/n
fn(x) dx
≤ M
n → 0. (**)
Hence, by (*) and (**), we have
n→∞lim
Z 1−1/n 0
fn(x) dx = Z 1
0
f (x) dx.
9.17 Mathematicinas from Slobbovia decided that the Riemann integral was too complicated so that they replaced it by Slobbovian integral, de- fined as follows: If f is a function defined on the set Q of rational numbers in [0, 1] , the Slobbovian integral of f, denoted by S (f ) , is defined to be the limit
S (f ) = lim
n→∞
1 n
n
X
k=1
f k n
,
whenever the limit exists. Let {fn} be a sequence of functions such that S (fn) exists for each n and such that fn → f uniformly on Q. Prove that {S (fn)} converges, that S (f ) exists, and S (fn) → S (f ) as n → ∞.
Proof : fn→ f uniformly on Q, then given ε > 0, there exists a positive integer N such that as n > m ≥ N, we have
|fn(x) − f (x)| < ε/3 (1)
and
|fn(x) − fm(x)| < ε/2. (2) So, if n > m ≥ N,
|S (fn) − S (fm)| =
k→∞lim 1 k
k
X
j=1
fn
j k
− fm
j k
= lim
k→∞
1 k
k
X
j=1
fn j
k
− fm j k
≤ lim
k→∞
1 k
k
X
j=1
ε/2 by (2)
= ε/2
< ε
which implies that {S (fn)} converges since it is a Cauchy sequence. Say its limit S.
Consider, by (1) as n ≥ N, 1
k
k
X
j=1
fn j
k
− ε/3
≤ 1 k
k
X
j=1
f j k
≤ 1 k
k
X
j=1
fn j
k
+ ε/3
which implies that
"
1 k
k
X
j=1
fn j k
#
− ε/3 ≤ 1 k
k
X
j=1
f j k
≤
"
1 k
k
X
j=1
fn j k
# + ε/3
which implies that, let k → ∞
S (fn) − ε/3 ≤ lim
k→∞sup1 k
k
X
j=1
f j k
≤ S (fn) + ε/3 (3)
and
S (fn) − ε/3 ≤ lim
k→∞inf 1 k
k
X
j=1
f j k
≤ S (fn) + ε/3 (4)
which implies that
k→∞lim sup 1 k
k
X
j=1
f j k
− lim
k→∞inf 1 k
k
X
j=1
f j k
≤
k→∞lim sup 1 k
k
X
j=1
f j k
− S (fn)
+
k→∞lim inf 1 k
k
X
j=1
f j k
− S (fn)
≤ 2ε
3 by (3) and (4)
< ε. (5)
Note that (3)-(5) imply that the existence of S (f ) . Also, (3) or (4) implies that S (f ) = S. So, we complete the proof.
9.18 Let fn(x) = 1/ (1 + n2x2) if 0 ≤ x ≤ 1, n = 1, 2, ... Prove that {fn} converges pointwise but not uniformly on [0, 1] . Is term-by term integration permissible?
Proof : It is clear that
n→∞lim fn(x) = 0
for all x ∈ [0, 1] . If {fn} converges uniformly on [0, 1] , then given ε = 1/3, there exists a positive integer N such that as n ≥ N, we have
|fn(x)| < 1/3 for all x ∈ [0, 1]
which implies that
fN 1 N
= 1 2 < 1
3
which is impossible. So, {fn} converges pointwise but not uniformly on [0, 1] . Since {fn(x)} is clearly uniformly bounded on [0, 1] , i.e., |fn(x)| ≤ 1 for all x ∈ [0, 1] and n. Hence, by Arzela’s Theorem, we know that the sequence of functions can be integrated term by term.
9.19 Prove that P∞
n=1x/nα(1 + nx2) converges uniformly on every finite interval in R if α > 1/2. Is the convergence uniform on R?
Proof : By A.P. ≥ G.P., we have
x nα(1 + nx2)
≤ 1
2nα+12 for all x.
So, by Weierstrass M-test, we have proved thatP∞
n=1x/nα(1 + nx2) con- verges uniformly on R if α > 1/2. Hence, P∞
n=1x/nα(1 + nx2) converges uniformly on every finite interval in R if α > 1/2.
9.20 Prove that the series P∞
n=1((−1)n/√
n) sin (1 + (x/n)) converges uniformly on every compact subset of R.
Proof : It suffices to show that the seriesP∞
n=1((−1)n/√
n) sin (1 + (x/n)) converges uniformly on [0, a] . Choose n large enough so that a/n ≤ 1/2, and therefore sin 1 + n+1x ≤ sin 1 +xn for all x ∈ [0, a] . So, if we let fn(x) = (−1)n/√
n and gn(x) = sin 1 + xn , then by Abel’s test for uniform con- vergence, we have proved that the series P∞
n=1((−1)n/√
n) sin (1 + (x/n)) converges uniformly on [0, a] .
Remark: In the proof, we metion something to make the reader get more. (1) since a compact set K is a bounded set, say K ⊆ [−a, a] , if we can show the series converges uniformly on [−a, a] , then we have proved it. (2) The interval that we consider is [0, a] since [−a, 0] is similar. (3) Abel’s test for uniform convergence holds for n ≥ N, where N is a fixed positive integer.
9.21 Prove that the series P∞
n=0(x2n+1/ (2n + 1) − xn+1/ (2n + 2)) con- verges pointwise but not uniformly on [0, 1] .
Proof : We show that the series converges pointwise on [0, 1] by con- sidering two cases: (1) x ∈ [0, 1) and (2) x = 1. Hence, it is trivial. De- fine f (x) = P∞
n=0(x2n+1/ (2n + 1) − xn+1/ (2n + 2)) , if the series converges uniformly on [0, 1] , then by Theorem 9.2, f (x) is continuous on [0, 1] . However,
f (x) =
1
2 log (1 + x) if x ∈ [0, 1) log 2 if x = 1 . Hence, the series converges not uniformly on [0, 1] .
Remark: The function f (x) is found by the following. Given x ∈ [0, 1), then both
∞
X
n=0
t2n = 1
1 − t2 and 1 2
∞
X
n=0
tn= 1 2 (1 − t)
converges uniformly on [0, x] by Theorem 9.14. So, by Theorem 9.8, we
have Z x
0
∞
X
n=0
t2n− 1 2
∞
X
n=0
tn = Z x
0
1
1 − t2 − 1 2 (1 − t)dt
= Z x
0
1 2
1
1 − t+ 1 1 + t
−1 2
1 1 − t
dt
= 1
2log (1 + x) . And as x = 1,
∞
X
n=0
x2n+1/ (2n + 1) − xn+1/ (2n + 2)
=
∞
X
n=0
1
2n + 1− 1 2n
=
∞
X
n=0
(−1)n+1
n + 1 by Theorem8.14.
= log 2 by Abel’s Limit Theorem.
9.22 Prove thatP ansin nx andP bncos nx are uniformly convergent on R if P |an| converges.
Proof : It is trivial by Weierstrass M-test.
9.23
Let {an} be a decreasing sequence of positive terms. Prove that the series P ansin nx converges uniformly on R if, and only if, nan → 0 as n → ∞.Proof : (⇒) Suppose that the seriesP ansin nx converges uniformly on R, then given ε > 0, there exists a positive integer N such that as n ≥ N, we have
2n−1
X
k=n
aksin kx
< ε. (*)
Choose x = 2n1 , then sin 12 ≤ sin kx ≤ sin 1. Hence, as n ≥ N, we always
have, by (*) (ε >)
2n−1
X
k=n
aksin kx
=
2n−1
X
k=n
aksin kx
≥
2n−1
X
k=n
a2nsin1
2 since ak> 0 and ak &
= 1 2sin1
2
(2na2n) .
That is, we have proved that 2na2n → 0 as n → ∞. Similarly, we also have (2n − 1) a2n−1 → 0 as n → ∞. So, we have proved that nan→ 0 as n → ∞.
(⇐) Suppose that nan → 0 as n → ∞, then given ε > 0, there exists a positive integer n0 such that as n ≥ n0, we have
|nan| = nan< ε
2 (π + 1). (*)
In order to show the uniform convergence of P∞
n=1ansin nx on R, it suffices to show the uniform convergence of P∞
n=1ansin nx on [0, π] . So, if we can show that as n ≥ n0
n+p
X
k=n+1
aksin kx
< ε for all x ∈ [0, π] , and all p ∈ N then we complete it. We consider two cases as follows. (n ≥ n0)
As x ∈h
0,n+pπ i , then
n+p
X
k=n+1
aksin kx
=
n+p
X
k=n+1
aksin kx
≤
n+p
X
k=n+1
akkx by sin kx ≤ kx if x ≥ 0
=
n+p
X
k=n+1
(kak) x
≤ ε
2 (π + 1) pπ
n + p by (*)
< ε.
And as x ∈h
π n+p, πi
, then
n+p
X
k=n+1
aksin kx
≤
m
X
k=n+1
aksin kx +
n+p
X
k=m+1
aksin kx
, where m = hπ x i
≤
m
X
k=n+1
akkx + 2am+1
sinx2 by Summation by parts
≤ ε
2 (π + 1)(m − n) x + 2am+1 sinx2
≤ ε
2 (π + 1)mx + 2am+1
π
x by 2x
π ≤ sin x if x ∈ h 0,π
2 i
≤ ε
2 (π + 1)π + 2am+1(m + 1)
< ε
2+ 2 ε 2 (π + 1)
< ε.
Hence, P∞
n=1ansin nx converges uniformly on R.
Remark: (1) In the proof (⇐), if we can make sure that nan& 0, then we can use the supplement on the convergnce of series in Ch8, (C)- (6) to show the uniform convergence ofP∞
n=1ansin nx =P∞
n=1(nan) sin nxn by Dirichlet’s test for uniform convergence.
(2)There are similar results; we write it as references.
(a) Suppose an & 0, then for each α ∈ 0,π2 , P∞n=1ancos nx and P∞
n=1ansin nx converges uniformly on [α, 2π − α] .
Proof: The proof follows from (12) and (13) in Theorem 8.30 and Dirichlet’s test for uniform convergence. So, we omit it. The reader can see the textbook, example in pp 231.
(b) Let {an} be a decreasing sequence of positive terms. P∞
n=1ancos nx uniformly converges on R if and only if P∞
n=1an converges.
Proof: (⇒) Suppose thatP∞
n=1ancos nx uniformly converges on R, then let x = 0, then we have P∞
n=1an converges.
(⇐) Suppose thatP∞
n=1an converges, then by Weierstrass M-test, we have proved that P∞
n=1ancos nx uniformly converges on R.
9.24 Given a convergent series P∞
n=1an. Prove that the Dirichlet series P∞
n=1ann−s converges uniformly on the half-infinite interval 0 ≤ s < +∞.
Use this to prove that lims→0+P∞
n=1ann−s =P∞ n=1an. Proof : Let fn(s) =Pn
k=1ak and gn(s) = n−s, then by Abel’s test for uniform convergence, we have proved that the Dirichlet seriesP∞
n=1ann−s converges uniformly on the half-infinite interval 0 ≤ s < +∞. Then by Theorem 9.2, we know that lims→0+P∞
n=1ann−s=P∞ n=1an. 9.25 Prove that the series ζ (s) =P∞
n=1n−s converges uniformly on every half-infinite interval 1 + h ≤ s < +∞, where h > 0. Show that the equation
ζ0(s) = −
∞
X
n=1
log n ns
is valid for each s > 1 and obtain a similar formula for the kth derivative ζ(k)(s) .
Proof : Since n−s ≤ n−(1+h) for all s ∈ [1 + h, ∞), we know that ζ (s) = P∞
n=1n−s converges uniformly on every half-infinite interval 1 + h ≤ s < +∞
by Weierstrass M-test. Define Tn(s) =Pn
k=1k−s, then it is clear that 1. For each n, Tn(s) is differentiable on [1 + h, ∞),
2. lim
n→∞Tn(2) = π2 6 . And
3. Tn0 (s) = −
n
X
k=1
log k
ks converges uniformly on [1 + h, ∞) by Weierstrass M-test. Hence, we have proved that
ζ0(s) = −
∞
X
n=1
log n ns
by Theorem 9.13. By Mathematical Induction, we know that ζ(k)(s) = (−1)k
∞
X
n=1
(log n)k ns .
0.1 Supplement on some results on Weierstrass M- test.
1. In the textbook, pp 224-223, there is a surprising result called Space- filling curve. In addition, note the proof is related with Cantor set in exercise 7. 32 in the textbook.
2. There exists a continuous function defined on R which is nowhere differentiable. The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 154.
Remark: The first example comes from Bolzano in 1834, however, he did NOT give a proof. In fact, he only found the function f : D → R that he constructed is not differentiable on D0(⊆ D) where D0 is countable and dense in D. Although the function f is the example, but he did not find the fact.
In 1861, Riemann gave g (x) =
∞
X
n=1
sin (n2πx) n2
as an example. However, Reimann did NOT give a proof in his life until 1916, the proof is given by G. Hardy.
In 1860, Weierstrass gave
h (x) =
∞
X
n=1
ancos (bnπx) , b is odd, 0 < a < 1, and ab > 1 + 3π 2 , until 1875, he gave the proof. The fact surprises the world of Math, and produces many examples. There are many researches related with it until now 2003.
Mean Convergence
9.26 Let fn(x) = n3/2xe−n2x2. Prove that {fn} converges pointwise to 0 on [−1, 1] but that l.i.m.n→∞fn6= 0 on [−1, 1] .
Proof : It is clear that {fn} converges pointwise to 0 on [−1, 1] , so it
remains to show that l.i.m.n→∞fn6= 0 on [−1, 1] . Consider Z 1
−1
fn2(x) dx = 2 Z 1
0
n3x2e−2n2x2dx since fn2(x) is an even function on [−1, 1]
= 1
√2 Z
√ 2n 0
y2e−y2dy by Change of Variable, let y =√ 2nx
= 1
−2√ 2
Z
√2n
0
yd e−y2
= 1
−2√ 2
"
ye−y2
√2n
0 −
Z
√2n
0
e−y2dy
#
→
√π 4√
2 since Z ∞
0
e−x2dx =
√π
2 by Exercise 7. 19.
So, l.i.m.n→∞fn6= 0 on [−1, 1] .
9.27
Assume that {fn} converges pointwise to f on [a, b] and that l.i.m.n→∞fn = g on [a, b] . Prove that f = g if both f and g are continuous on [a, b] .Proof : Since l.i.m.n→∞fn = g on [a, b] , given εk = 21k, there exists a nk such that
Z b a
|fnk(x) − g (x)|pdx ≤ 1
2k, where p > 0 Define
hm(x) =
m
X
k=1
Z x a
|fnk(t) − g (t)|pdt, then
a. hm(x) % as x % b. hm(x) ≤ hm+1(x)
c. hm(x) ≤ 1 for all m and all x.
So, we obtain hm(x) → h (x) as m → ∞, h (x) % as x %, and
h (x) − hm(x) =
∞
X
k=m+1
Z x a
|fnk(t) − g (t)|pdt % as x %
which implies that
h (x + t) − h (x)
t ≥ hm(x + t) − hm(x)
t for all m. (*)
Since h and hm are increasing, we have h0 and h0m exists a.e. on [a, b] . Hence, by (*)
h0m(x) =
m
X
k=1
|fnk(t) − g (t)|p ≤ h0(x) a.e. on [a, b]
which implies that
∞
X
k=1
|fnk(t) − g (t)|p exists a.e. on [a, b] .
So, fnk(t) → g (t) a.e. on [a, b] . In addition, fn → f on [a, b] . Then we conclude that f = g a.e. on [a, b] . Since f and g are continuous on [a, b] , we have
Z b a
|f − g| dx = 0
which implies that f = g on [a, b] . In particular, as p = 2, we have f = g.
Remark: (1) A property is said to hold almost everywhere on a set S (written: a.e. on S) if it holds everywhere on S except for a set of measurer zero. Also, see the textbook, pp 254.
(2) In this proof, we use the theorem which states: A monotonic function h defined on [a, b] , then h is differentiable a.e. on [a, b] . The reader can see the book, The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 113.
(3) There is another proof by using Fatou’s lemma: Let {fk} be a measruable function defined on a measure set E. If fk ≥ φ a.e. on E and φ ∈ L (E) , then
Z
E
lim
k→∞inf fk≤ lim
k→∞inf Z
E
fk.
Proof : It suffices to show that fnk(t) → g (t) a.e. on [a, b] . Since l.i.m.n→∞fn = g on [a, b] , and given ε > 0, there exists a nk such that
Z b a
|fnk− g|2dx < 1 2k
which implies that
Z b a
m
X
k=1
|fnk− g|2dx <
m
X
k=1
1 2k which implies that, by Fatou’s lemma,
Z b a
m→∞lim inf
m
X
k=1
|fnk − g|2dx ≤ lim
m→∞inf Z b
a m
X
k=1
|fnk− g|2dx
=
∞
X
k=1
Z b a
|fnk− g|2dx < 1.
That is,
Z b a
∞
X
k=1
|fnk− g|2dx < 1 which implies that
∞
X
k=1
|fnk− g|2 < ∞ a.e. on [a, b]
which implies that fnk → g a.e. on [a, b] .
Note: The reader can see the book, Measure and Integral (An In- troduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 75.
(4) There is another proof by using Egorov’s Theorem: Let {fk} be a measurable functions defined on a finite measurable set E with finite limit function f. Then given ε > 0, there exists a closed set F (⊆ E) , where
|E − F | < ε such that
fk → f uniformly on F.
Proof : If f 6= g on [a, b] , then h := |f − g| 6= 0 on [a, b] . By continuity of h, there exists a compact subinterval [c, d] such that |f − g| 6= 0. So, there exists m > 0 such that h = |f − g| ≥ m > 0 on [c, d] . Since
Z b a
|fn− g|2dx → 0 as n → ∞,
we have
Z d c
|fn− g|2dx → 0 as n → ∞.
then by Egorov’s Theorem, given ε > 0, there exists a closed susbet F of [c, d] , where |[c, d] − F | < ε such that
fn→ f uniformly on F which implies that
0 = lim
n→∞
Z
F
|fn− g|2dx
= Z
F
n→∞lim |fn− g|2dx
= Z
F
|f − g|2dx ≥ m2|F |
which implies that |F | = 0. If we choose ε < d−c, then we get a contradiction.
Therefore, f = g on [a, b] .
Note: The reader can see the book, Measure and Integral (An In- troduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 57.
9.28 Let fn(x) = cosnx if 0 ≤ x ≤ π.
(a) Prove that l.i.m.n→∞fn = 0 on [0, π] but that {fn(π)} does not converge.
Proof : It is clear that {fn(π)} does not converge since fn(π) = (−1)n. It remains to show that l.i.m.n→∞fn= 0 on [0, π] . Consider cos2nx := gn(x) on [0, π] , then it is clear that {gn(x)} is boundedly convergent with limit function
g =
0 if x ∈ (0, π) 1 if x = 0 or π . Hence, by Arzela’s Theorem,
n→∞lim Z π
0
cos2nxdx = Z π
0
g (x) dx = 0.
So, l.i.m.n→∞fn= 0 on [0, π] .
(b) Prove that {fn} converges pointwise but not uniformly on [0, π/2] . Proof : Note that each fn(x) is continuous on [0, π/2] , and the limit function
f = 0 if x ∈ (0, π/2]
1 if x = 0 .
Hence, by Theorem9.2, we know that {fn} converges pointwise but not uniformly on [0, π/2] .
9.29 Let fn(x) = 0 if 0 ≤ x ≤ 1/n or 2/n ≤ x ≤ 1, and let fn(x) = n if 1/n < x < 2/n. Prove that {fn} converges pointwise to 0 on [0, 1] but that l.i.m.n→∞fn 6= 0 on [0, 1] .
Proof : It is clear that {fn} converges pointwise to 0 on [0, 1] . In order to show that l.i.m.n→∞fn 6= 0 on [0, 1] , it suffices to note that
Z 1 0
fn(x) dx = 1 for all n.
Hence, l.i.m.n→∞fn 6= 0 on [0, 1] .
Power series
9.30 If r is the radius of convergence ifP an(z − z0)n, where each an 6= 0, show that
n→∞lim inf
an an+1
≤ r ≤ lim
n→∞sup
an an+1
.
Proof : By Exercise 8.4, we have 1
limn→∞sup
an+1
an
≤ r = 1
limn→∞sup |an|n1 ≤ 1 limn→∞inf
an+1
an
.
Since
1 limn→∞sup
an+1
an
= lim
n→∞inf
an an+1
and
1 limn→∞inf
an+1
an
= lim
n→∞sup
an an+1
,
we complete it.
9.31 Given that two power series P anzn has radius of convergence 2.
Find the radius convergence of each of the following series: In (a) and (b), k is a fixed positive integer.
(a) P∞ n=0aknzn Proof : Since
2 = 1
limn→∞sup |an|1/n, (*)
we know that the radius of P∞
n=0aknzn is 1
limn→∞sup |akn|1/n = 1
limn→∞sup |an|1/nk = 2k.
(b) P∞
n=0anzkn Proof : Consider
n→∞lim sup anzkn
1/n = lim
n→∞sup |an|1/n|z|k < 1 which implies that
|z| < 1
limn→∞sup |an|1/n
!1/k
= 21/k by (*).
So, the radius of P∞
n=0anzkn is 21/k. (c) P∞
n=0anzn2 Proof : Consider
lim sup anzn2
1/n
= lim
n→∞sup |an|1/n|z|n and claim that the radius of P∞
n=0anzn2 is 1 as follows.
If |z| < 1, it is clearly seen that the series converges. However, if |z| > 1,
n→∞lim sup |an|1/n lim
n→∞inf |z|n ≤ lim
n→∞sup |an|1/n|z|n
