MATH 172 Homework 2 Solutions 1. Prove that the following series ...

MATH 172 Homework 2 Solutions
1. Calculate the following limit (prove your work) :
Solution : For n ≥ 1, put
∫ 3
nx 2 + 3
lim
n→+∞
1 x 3 + nx dx
f n : [1, 3] → R : x → nx2 + 3
x 3 + nx
Put f(x) = x for all x ∈ [1, 3]. We prove that {f n } n≥1
converges uniformly to f on [0, 1]. So we have to show :
∀ɛ > 0 : ∃N ∈ N : ∀n ≥ N, ∀x ∈ [1, 3] : |f n (x) − f(x)| < ɛ
Let ɛ > 0. Pick N ∈ N with N > 84 ɛ
. Let n ≥ N and x ∈ [1, 3]. Note that
|3 − x 4 | ≤ |3| + |x 4 | ≤ 3 + 3 4 = 84 and |x 3 + nx| = x(x 2 + n) ≥ 1 · n = n
Hence ∣ ∣∣∣ nx 2 + 3
x 3 + nx − x ∣ ∣∣∣
=
3 − x 4
∣x 3 + nx∣ ≤ 84
n ≤ 84
N < ɛ
u
Since f n is Riemann integrable over [1, 3] for all n ≥ 1 (because f n is continuous over [1, 3]) and {f n } n≥1
−→ f
on [1, 3], we have by Theorem 1.4 that
Hence
∫ 3
lim
n→+∞
1
f n (x) dx =
∫ 3
1
lim f n(x) dx =
n→+∞
∫ 3
1
f(x) dx
∫ 3
nx 2 ∫
+ 3
3
lim
n→+∞
1 x 3 + nx dx =
1
[ ] 3 1
x dx =
2 x2 = 4 ✷
1
2. Let f, g : [a, b] → R be bounded on the closed interval [a, b] and let P be a partition of [a, b]. Prove that
U(f + g, P) ≤ U(f, P) + U(g, P)
Proof :
First, we prove a general statement:
Let f, g : D → R be bounded on D. Then sup
x∈D
(f + g)(x) ≤ sup
x∈D
f(x) + sup g(x).
x∈D
Indeed, put α = sup f(x) and β = sup g(x). Note that α, β ∈ R since f and g are bounded on D. Then for
x∈D
x∈D
x ∈ D, we have that
(f + g)(x) = f(x) + g(x) ≤ α + β
So α + β is an upper bound for {(f + g)(x) : x ∈ D}. Thus sup(f + g)(x) ≤ α + β, which proves the claim.
x∈D
Next, we prove the exercise. Let P be given by a = x 0 < x 1 < · · · < x n−1 < x n = b. Put D i = [x i−1 , x i ] for
1 ≤ i ≤ n. Using the claim with ‘D = D i ’, we get that
1

U(f + g, P) =
≤
=
n∑
(x i − x i−1 ) sup (f + g)(x)
i=1
x∈D i
n∑
(
)
(x i − x i−1 ) sup f(x) + sup g(x)
i=1
x∈D i x∈D i
n∑
n∑
(x i − x i−1 ) sup f(x) + (x i − x i−1 ) sup g(x)
x∈D i x∈D i
i=1
i=1
= U(f, P) + U(g, P) ✷
3. Let I be an open interval, x 0 ∈ I, and {f n } n≥1 a sequence of functions such that f n is differentiable on I for all
n ≥ 1 and the sequence {f n (x 0 )} n≥1 converges and the sequence {f ′ n} n≥1 converges uniformly on I. Prove that
the sequence {f n } n≥1 converges on I to some f such that f is differentiable on I and f ′ = lim
n→∞ f ′ n on I.
Proof : Fix x 1 ∈ I. Let J be a bounded open interval such that x 0 , x 1 ∈ J ⊆ I. Note that f n is differentiable
on J for all n ≥ 1 and that the sequence {f n} ′ n≥1 converges uniformly on J since J ⊆ I. It follows from Theorem
1.5 that the sequence {f n } n≥1 converges (uniformly) to some function g J (this function seemingly depends on
J) such that g J is differentiable on J and g J ′ = lim f n ′ on J.
n→∞
Since this is true for all x 1 ∈ I, it follows that the sequence {f n (x 1 )} n≥1 converges for every x 1 ∈ I. Hence the
sequence {f n } n≥1 converges pointwise to some function f on I.
Again fix x 1 ∈ I. Then f = g J on J (from above). So f is differentiable at x 1 and f ′ (x 1 ) = lim f n(x ′ 1 ). Since
n→∞
this is true for all x 1 ∈ I, we get that f is differentiable on I and f ′ = lim f n ′ on I.
✷
n→∞
4. Let D ⊆ R and {f n } n≥1 a sequence of functions such that f n is uniformly continuous on D for all n ≥ 1 and the
sequence {f n } n≥1 converges uniformly on D (to some function f). Prove that f is uniformly continuous on D.
Proof : Let ɛ > 0. Since {f n } n≥1 converges uniformly to f on D, we have :
∃N ∈ N : ∀n ≥ N, ∀t ∈ D : |f n (t) − f(t)| < ɛ 3
(1)
Since f N is uniformly continuous on D, we have
∃δ > 0 : ∀x, y ∈ D : |x − y| < δ ⇒ |f N (x) − f N (y)| < ɛ 3
(2)
Pick x, y ∈ D with |x − y| < δ. From (1) and (2), it follows that
|f(x) − f(y)| ≤ |f(x) − f N (x)| + |f N (x) − f N (y)| + |f N (y) − f(y)| < ɛ 3 + ɛ 3 + ɛ 3 = ɛ
Hence f is uniformly continuous.
✷
5. Let b n ∈ [0, 1] and f n : [0, 1] → R be integrable over [0, 1] for all n ≥ 1. Suppose that the sequence {f n } n≥1
converges
{
uniformly to some function f on [0, 1] and that the sequence {b n } n≥1 converges to 1. Prove that the
∫ }
bn
sequence f n (x) dx converges to
∫ 1
0
0
n≥1
f(x) dx.
Proof :
Note that f is integrable over [0, 1] and that
∫ 1
0
f(x) dx = lim
n→∞
∫ 1
0
f n (x) dx by Theorem 1.4.
2

Since 0 ≤ b n ≤ 1, we have that
∫ bn
∫ 1
∫ 1
f n (x) dx = f n (x) dx − f n (x) dx for all n ≥ 1
0
0
b n
{∫ 1
}
Hence it suffices to prove that the sequence f n (x) dx converges to 0.
b n
Since f n is bounded on [0, 1] for all n ≥ 1 (f n is integrable) and the sequence {f n } n≥1 converges uniformly on
[0, 1], it follows from HW 1 #3 that the sequence {f n } n≥1 is uniformly bounded on [0, 1]:
there exists some constant M > 0 such that |f n (x)| ≤ M for all n ≥ 1 and all x ∈ [0, 1]
n≥1
Let ɛ > 0. Since the sequence {b n } n≥1 converges to 1, we have:
∃N ∈ N : ∀n ≥ N : |b n − 1| < ɛ M
Let n ≥ N. Since |b n − 1| = 1 − b n , we get
∫ 1
∫ 1
∣ f n (x) dx − 0
∣ ≤ |f n (x)| dx ≤
b n b n
∫ 1
b n
M dx = (1 − b n )M < ɛ M M = ɛ
✷
3