MATH 3150 — HOMEWORK 7
Problem 1 (p. 236, #43). Let f : R −→ R be continuous and set F (x) =
F 0 (x) = 2xf (x2 ). State a more general theorem.
R x2
0
f (y) dy. Prove that
Solution. F (x) = (G ◦ h)(x), where
Z
z
f (y) dy,
G(z) =
h(x) = x2 .
0
h is clearly differentiable, and G is differentiable by the fundamental theorem of calculus since f is
continuous. Then
F 0 (x) = G0 h(x) h0 (x) = f (x2 ) 2x
using the chain rule and the fundamental theorem of calculus.
A general theorem of this form states that if g is differentiable and f continuous, then
!
Z g(x)
d
f (t) dt = f g(x) g 0 (x).
dx
a
Problem 2 (p. 316, #2). Determine which of the following sequences converge (pointwise or
uniformly) as k → ∞. Check the continuity of the limit in each case.
(a)
(b)
(c)
(d)
(e)
(sin x)/k on R
1/(kx + 1) on (0, 1)
x/(kx + 1) on (0, 1)
x/(1 + kx2 ) on R
(1, (cos x)/k 2 ), as functions from R to R2
Solution.
(a) Converges uniformly to 0:
sin x
1
k − 0 ≤ k,
which goes to 0 independently of x. The limit is continuous.
(b) Converges pointwise to 0, but not uniformly. Indeed, given any 0 < ε < 1 and any k ∈ N, we
can find a sufficiently small x ∈ (0, 1) such that
kx ≤
(1 − ε)
1
=⇒
≥ ε.
ε
(kx + 1)
The limit is continuous even though the convergence is not uniform. (Note that on the closed
interval [0, 1] the sequence converges pointwise to a discontinuous function.)
(c) Converges uniformly to 0. Indeed, since x < 1,
x
1
=
kx + 1
k+
1
x
≤
1
k+1
which tends to 0 independently of x. The limit is continuous.
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(d) Converges uniformly to 0. Indeed, at x = 0 all the terms in the sequence evalutate to 0, and if
x 6= 0 then
1
x
= 1
1 + kx2
x + kx
For fixed x, this tends to 0 as k −→ ∞, thus the sequence converges pointwise to 0.
To see that the convergence is uniform on R, we consider seperately the intervals (−∞, −1),
[−1, 1] and (1, ∞). For x ∈ [−1, 1], we have the estimate
x 1
1 + kx2 ≤ 1 + k
which is independent of x. For x > 1, we have
x 1 1 1
1 + kx2 = 1 + kx ≤ kx ≤ k ,
x
and for x < −1 we have
x −x
=
1 + kx2 1 + k(−x)2 so the same estimate applies.
1
< ε. Then for k ≥ N the above estimates taken
Thus given ε > 0, choose N such that 1+N
together prove that
x
1 + kx2 − 0 < ε.
x
(e) As in part (a), the component sequence cos
converges to 0 uniformly, and of course the constant
k2
sequence 1 converges uniformly to 1. We deduce that (1, (cos x)/k 2 ) converges uniformly to the
continuous (constant) function (1, 0). Indeed
p
(1, (cos x)/k 2 ) − (1, 0) = (1 − 1)2 + ((cos x)/k 2 − 0)2 = cos2x ≤ 12
k
k
which tends to 0 independent of x ∈ R.
Problem 3 (p. 318, #18). Give an example of a sequence of discontinuous functions fk converging
uniformly to a limit function f that is continuous.
Solution. For instance, let fn : R −→ R be given by
(
1,
x<0
fn (x) =
1
1 + n , x ≥ 0.
The fn are all discontinuous, but fn −→ f uniformly, where f (x) = 1 for all x.
Problem 4 (p. 272, #4). Let
fn (x) =
1 nx
,
n 1 + nx
0 ≤ x ≤ 1.
Show that fn −→ 0 in C([0, 1]; R).
Solution. Convergence in C([0, 1]; R) is the same as uniform convergence. Thus we want to show
that fn −→ 0 uniformly. Cancelling the ns, this is the same sequence as in part (b) of Problem 2,
though the domain is now the closed interval. In any case, the same estimate applies; since x ≤ 1,
1
nx
x
1
−
0
=
n (1 + nx)
(1 + nx) ≤ 1 + n
which tends to 0 independent of x.
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Problem 5 (p. 272, #5). Let fk be a convergent sequence in Cb (A; Rm ). Prove that {fk : k = 1, 2, . . .}
is bounded in Cb (A; Rm ). Is it closed? [Hint: Don’t overthink this!]
Solution. Cb (A; Rm ) is a metric space, so any convergent sequence is bounded. The limit function
f (such that fk → f ) is an accumulation point of the set {fk }, so the latter is closed if and only if
f ∈ {fk }, i.e., f = fk for some k.
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