140B HOMEWORK 5 SOLUTIONS
Problem 1
Fix ε > 0 and let fn (x) = f (xn ). Since f and xn are continuous functions, fn (x) is continuous,
hence integrable.
Moreover, since xn → 0 as n → ∞ for all x ∈ [0, 1/2], we have that fn (x) → f (0) pointwise for all
x ∈ [0, 1/2].
We claim more, namely uniform convergence fn ⇉ f (0) over [0, 1/2]. Indeed, by continuity of f at
0 there exists δ > 0 such that
x < δ Ô⇒ ∣f (x) − f (0)∣ < ε.
1
Choose N such that 2N < δ. Then for all n > N and x ∈ [0, 1/2] we have
∣fn (x) − f (0)∣ = ∣f (xn ) − f (0)∣ < ε
since xn − 0 <
1
2N
< δ. Thus fn converges uniformly to f (0) on [0, 1/2].
Finally, by Theorem 7.16, we can integrate once we established uniform convergence, yielding
1/2
lim ∫
n→∞
0
f (xn )dx = ∫
1/2
0
f (0) dx =
f (0)
.
2
Problem 2
(i) We must check that d defines a metric:
– First, if f ≠ g there exists x such that f (x) ≠ g(x), so
d(f, g) ≥ sup ∣f (x) − g(x)∣ > 0,
and it is clear that d(f, f ) = 0.
– Second,
d(f, g) = sup ∣f (x) − g(x)∣ + sup ∣f ′ (x) − g ′ (x)∣ = sup ∣g(x) − f (x)∣ + sup ∣g ′ (x) − f ′ (x)∣ = d(g, f ).
– Finally, for any C 1 function h on [a, b] we have
d(f, g) = sup ∣f (x) − g(x)∣ + sup ∣f ′ (x) − g ′ (x)∣
= sup ∣f (x) − h(x) + h(x) − g(x)∣ + sup ∣f ′ (x) − h′ (x) + h′ (x) − g ′ (x)∣
≤ sup ∣f (x) − h(x)∣ + sup ∣h(x) − g(x)∣ + sup ∣f ′ (x) − h′ (x)∣ + sup ∣h′ (x) − g ′ (x)∣
= d(f, h) + d(h, g).
In the above, we have made use of the inequality
sup ∣u(x) + v(x)∣ ≤ sup ∣u(x)∣ + sup ∣v(x)∣
which was shown in class.
Therefore d defines a metric.
(ii) To show C 1 ([a, b]) is complete, we need to show that every Cauchy sequence converges to a
function in C 1 ([a, b]). Note: it is not sufficient to show that the sequence of functions converges to
some function. You need to show that the limit function lives in C 1 ([a, b]).
1
2
140B HOMEWORK 5 SOLUTIONS
Let {fn } be a Cauchy sequence in C 1 ([a, b]). For all ε > 0 we can choose N such that n, m > N
implies
′
ε > ∣∣fm − fn ∣∣ = sup ∣fm (x) − fn (x)∣ + sup ∣fm
(x) − fn′ (x)∣.
In particular
′
sup ∣fm (x) − fn (x)∣ < , and sup ∣fm
(x) − fn′ (x)∣ < .
From the first inequality on the previous line, fn satisfies the Cauchy criterion, so fn converges
uniformly to a limit function f . From the second inequality, fn′ also satisfies the Cauchy criterion,
hence converges uniformly on [a, b] to some function g. Since each fn′ is continuous and the convergence is uniform, g is also continuous. Furthermore, by Theorem 7.17, f ′ (x) = g(x). Therefore
the limit f ∈ C 1 ([a, b]), showing that the metric space is complete.
Problem 3
First we’ll show that ∣ sin(α)∣ ≤ α for α > 0. By the Mean Value Theorem there exists c ∈ (0, α)
such that
sin(α) − sin(0)
∣ = ∣ sin′ (c)∣ ≤ 1
∣
α−0
so we have ∣ sin(α)∣ ≤ ∣α∣ when α > 0. On (1, ∞), this implies
∣2n sin(
1
2n
2n
)∣
≤
≤
.
3n x
3n x 3n
n
Since ∑ ( 32 ) is a geometric series it converges. By the Weierstrass M-test, ∑n≥1 2n sin( 3n1x ) converges uniformly.
Let fn (x) = 2n sin( 3n1x ). Then
∣fn′ (x)∣ = ∣
1
−2n
2 n
cos(
)∣
≤
(
) .
3n x2
3n x
3
n
Since ∑n≥1 ( 23 ) converges, the Weierstrass M-test implies that ∑n≥1 fn′ (x) converges uniformly on
n
1
(1, ∞). By Theorem 7.17, f is differentiable and f ′ (x) = ∑n≥1 3−2
n x2 cos( 3n x ).
Problem 4
(i) Fix a > 0. Then we have
for all x ∈ [a, ∞). Since ∑n≥1
formly on [a, ∞).
1
n3 a
n
n
1
≤ 4 = 3
4
1+n x n x n a
converges, the Weierstrass M-test implies f (x) converges uni-
(ii) Suppose that f (x) converges uniformly over (0, ∞). By Cauchy criterion with = 31 , there
exists N such that m, n > N implies
m
k
1
< .
4
3
k=n+1 1 + k x
∑
Make m = n + 1, so that the above sum has only one term, namely
m
1
< .
4
1+m x 3
This inequality fails if we set for instance x = m14 since the left hand side is m/2 > 1/3. Therefore
the series does not converge uniformly over (0, ∞).
140B HOMEWORK 5 SOLUTIONS
3
(iii) Fix a > 0. We work over the interval [a, ∞) first. Consider the partial sums
n
k
−k 5
′
Ô⇒
s
(x)
=
.
∑
n
4
4 2
k=1 1 + k x
k=1 (1 + k x)
n
sn (x) = ∑
Since
k5
1
1
≤ 3 2 ≤ 3 2,
4
2
(1 + k x)
k x
k a
and ∑k≥1 k31a2 converges, the Weierstrass M-test implies that {s′n } converges uniformly on [a, ∞).
We also know that sn ⇉ f by part (i). By Theorem 7.17, f is differentiable and s′n converges
uniformly to f ′ (x) over [a, ∞). Since for any x > 0 we can find an a > 0 such that x > a, it follows
that f is differentiable over the entire interval (0, ∞).
(iv) We’ll prove f is infinitely many times differentiable by induction. Fix a > 0. Precisely, we show
(`)
that f is `-times differentiable and that the derivatives of the partial sums sn converge to f (`)
uniformly over [a, ∞) for all `.
We showed in part (iii) that f can be differentiated once over [a, ∞), which is the inductive base
case.
(`)
We now carry out the inductive step. Suppose that f is `-times differentiable, and that sn
converges to f (`) uniformly over [a, ∞). We look at the ` + 1-derivatives of the partial sums sn .
The absolute value of the (` + 1)st derivative of 1+kk 4 x is
∣
(` + 1)! (` + 1)!
(−1)`+1 (` + 1)!k 4`+5
∣ ≤ 3 `+2 ≤ 3 `+2 .
4
`+2
(1 + k x)
k x
k a
Since ∑k≥1 k3 a`+2 = a`+2 ∑k≥1 k13 converges, the Weierstrass M-test implies that the derivatives
(`+1)
(`)
sn
converge uniformly. By the induction hypothesis, we already know that sn converges to
(`)
(`)
f . By Theorem 7.17, it follows that f is differentiable, so f is (` + 1) times differentiable, and
(`+1)
that sn
converges (uniformly, as we showed) to f (`+1) . This completes the induction step.
(`+1)!
(`+1)!
Therefore, f is infinitely differentiable on [a, ∞), hence also on the union of these intervals which
is (0, ∞).
Problem 5
See online solutions.
Problem 6
Fix ε > 0 and let δ = ε/M . Then for any f ∈ F and any x, y ∈ R such that ∣x − y∣ < δ there exists
c ∈ (x, y) such that
ε
∣f (x) − f (y)∣ = ∣f ′ (c)∣ ⋅ ∣x − y∣ < M
= ε.
M
Therefore F is equicontinuous. For the particular sequence of functions given,
∣fn′ (x)∣ = ∣ cos(nx) − sin(x + n)∣ ≤ 2
for any choice of n and x. Since the derivatives are uniformly bounded, {fn } is an equicontinuous
family.
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140B HOMEWORK 5 SOLUTIONS
Problem 7
Note that the function f is bounded by 1. By the fundamental theorem of calculus, we compute
1
∣Fn′ (x)∣ = ∣ f (an x)an ∣ = ∣f (an x)∣ ≤ 1.
an
Thus, the family {Fn } has uniformly bounded derivatives. By Problem 6, {Fn } is an equicontinuous
family.