X
n=1
1 n.
(b) If P an is a convergent series of nonnegative terms, then P a2n also con-verges. This can be shown with a similar method as the proof done in Problem No. 3.
19Note: the index on the series begins at n = 2 =⇒ log n > 1 for n ≥ 2.
Claim:
P an:= convergent and an≥ 0 ∀n ∈ N =⇒P(an)2 converges.
Proof :
If we know thatP an is a convergent series, then
n→∞lim an= 0 =⇒ ∃N ∈ N : an∈ (0, 1) ∀n ≥ N =⇒ (an)2< an ∀n ≥ N (an)2< an ∀n ≥ N =⇒X
(an)2<X
an ∀n ≥ N.
Hence,P(an)2converges by the Comparison Test, as desired. ♠ (c) Consider this series:
∞
X
n=1
(−1)n
√n .
EXERCISE 17.5
(a) We will use induction to prove f (x) = xm is continuous. This will be done by first showing that when m = 1 we are dealing with f (x) = x, which will be taken as continuous (proof is provided in the Appendix). Then by assuming that f (x) = xm is continuous for some m ∈ N, we will show our induction step, f (x) = xm+1 is continuous, by noting that f (x) = xm+1= xmx is hence continuous given Theorem 17.4 (ii), which will then imply that f (x) = xm is continuous for m ∈ N.
Claim:
If m ∈ N, then the function f (x) = xmis continuous on R.
Proof :
This proof will use mathematical induction. For m = 1, f (x) = x, which will be taken as continuous (proof provided in the Appendix). f (x) = x will then serve as our induction basis. We can now assume our induction hypothesis, f (x) = xmis continuous. But we need to show continuity holds for the induction step, m + 1. But we know f(x) = xm+1≡ xm· x, which is the product of two continuous functions, our inductions hypothesis, and our induction basis, and is hence continuous by Theorem 17.4 (ii). ♠
(b) Claim:
Every polynomial function p(x) = a0+ a1x + · · · + anxn is continuous on R.
Proof :
Given the solution from part (a), p(x) is simply the sum and product of con-tinuous functions and is hence concon-tinuous, by Theorem 17.4 (i) and Theorem 17.3, as desired. ♠
EXERCISE 17.6 Claim:
Every rational function is continuous.
Proof :
A rational function is composed of constants, f (x) = c and the continuous func-tion f (x) = x by multiplicafunc-tion, addifunc-tion and division. Since f (x) = x and f (x) = c are trivially continuous =⇒ rational functions are continuous by the continuity theorems of sums, products and quotients of continuous functions, as desired. ♠
EXERCISE 17.7 (b) Claim:
|x| is a continuous function on R.
Proof :
|x| is continuous at any x0 since it coincides with x for x > 0 and −x for x < 0.
At x = 0, the function f (x) = |x| is continuous because for any δ > 0 we have:
|x − 0| < =⇒ |f (x) − f (0)| = |x| < . ♠
EXERCISE 17.8 (a) Claim:
min(f, g) =1
2(f + g) −1 2|f − g|
Proof :
Case 1 : Let f (x) ≤ g(x). Then
min(f, g) = f (x) = 1
2(f + g) −1
2(g − f ) (32)
= 1
2(f + g) −1
2|f − g| (33)
Case 2 : Let f (x) ≥ g(x). Then
min(f, g) = g(x) = 1
2(f + g) − 1
2(f − g) (34)
= 1
2(f + g) − 1
2|f − g| . (35)
Hence,
min(f, g) = 1
2(f + g) −1
2|f − g| , as desired. ♠
(b) Claim:
min (f, g) = −max (−f, −g) Proof :
Case 1 : Let f (x) ≤ g(x) =⇒ −f (x) ≥ −g(x)
=⇒ min (f, g) = f (x) = − (−f (x)) = −max(−f, −g) Case 2 : Let f (x) ≥ g(x) =⇒ −f (x) ≤ −g(x)
=⇒ min (f, g) = g(x) = − (−g(x)) = −max(−f, −g).
Hence,
min (f, g) = −max (−f, −g) , as desired. ♠
(c) Claim:
f and g continuous at x0 ∈ R =⇒ min(f, g) is continuous at x0
Proof :
Recall Theorems 17.3 and 17.4 (i):
Theorem 17.3: Let f be a real-valued function with dom(f ) ⊆ R.
If f is continuous at x0 in dom(f ), then |f | and kf , k ∈ R, are continuous at x0.
Theorem 17.4 (i): Let f and g be real-valued functions that are continuous at x0∈ R. Then f + g is continuous at x0.
In combination with the results from part (a), i.e., min(f, g) = 1
2(f + g) −1
2|f − g| ,
we see that min(f, g) is simply the sum, difference and composition of functions which are continuous at x0, and hence is itself continuous at x0, as desired. ♠.
EXERCISE 17.9 (a) Claim:
f (x) = x2 is continuous at x0= 2 Proof :
Let > 0 be given. We want to show that |x2− 4| < provided |x − 2| is sufficiently small, i.e., less than some δ. We observe that |x2− 4| = |(x + 2)(x − 2)| ≤ |x + 2| · |x − 2|. We need to get a bound for |x − 2| that doesn’t depend on x. We notice that if |x − 2| < 1, say, then |x + 2| < 5, so it suffices to get
|x − 2| · 5 < . So by setting δ = min1,5 , we see that f (x) = x2is continuous at x0= 2, as desired. ♠
(b) Claim:
f (x) =√
x is continuous at x0= 0 Proof :
Let > 0 be given. We want to show that |√
x − 0| < provided |x − 0|
is sufficiently small, i.e., less than some δ. We observe that |√
x − 0| = √ x.
Since we want this to be less than , we set δ = 2. Then |x − 0| < δ implies
√x <√
δ = , so
|x − 0| < δ =⇒ |f (x) − f (0)| < , as desired. ♠
(c) Claim:
f (x) = xsin 1 x
for x 6= 0 and f (0) = 0 is continuous at x0= 0 Proof :
Let > 0 be given. We want to show that |xsin x1 − 0| < provided |x − 0|
is sufficiently small, i.e., less than some δ. We see that |xsin 1x − 0| ≤ x ∀x.
Since we want this to be less than , we set δ = . Then |x − 0| < δ implies x < δ = , so
|x − 0| < δ =⇒
xsin 1 x
− 0
< , as desired. ♠
(d) Claim:
g(x) = x3 is continuous at x0arbitrary Proof :
By the hint given,
x3− x30= (x − x0)(x2+ x0x + x20) = (x − x0)(x2− x0)2+ 3xx0) and we know that
|x| = |x − x0+ x0| ≤ |x − x0| + |x0|, by the triangle inequality. Hence,
|g(x) − g(x0)| = |x3− x30| ≤ |x − x0|(|x − x0|2+ 3|x − x0||x0| + 3x20).
Now let > 0 be given. We show that ∃ a δ = δ(x0, ) > 0 such that
|x − x0| < δ =⇒ |g(x) − g(x0)| < .
Let δ = min
1,3,g|x
0|+1,gx2 0+1
. Then |x − x0| < δ implies
|g(x) − g(x0)| ≤ δ(δ2+ 3δ|x0| + 3x20) ≤20δ(1 + 3|x0| + 3x20)
= δ + 3δ|x0| + 3δx20<
2 + 3|x0|
g|x0| + 1+ 3x20 g|x0|2+ 1,
≤ 3 +
3+ 3 = , as desired. ♠.
Alternatively,
Assume g(x) = x2, x0arbitrary. Observe that (x3−x30) = (x−x0)(x2+x0x+x20).
Now we will make the assumption that |x − x0| < 1 =⇒ |x| < |x0| + 1, which enables us to see
|x2+ x0x + x20| ≤ |x2| + |x0x| + |x20| < |x0|2+ 2|x0| + 1 + |x0| · ||x0| + 1| + |x20| ≡ k, where the material to the right of the last inequality is all greater than 0.
We can then set δ = min1,k . Hence,
|x−x0| < δ =⇒ |g(x)−g(x0)| = |(x−x0)(x2+x0x+x20)| < |x−x0||k| < k·k = , as desired. ♠
EXERCISE 17.12
(a) Let f be a continuous real-valued function with domain (a, b).
Claim:
If f (r) = 0 for each rational number r ∈ (a.b), then f (x) = 0 ∀x ∈ (a.b).
Proof :
We are given f (x) = 0 ∀x ∈ Q. If x ∈ R \ Q, then ∃ a sequence of rational numbers, (rn), which converges to x. Hence, by continuity, rn → x =⇒ f (rn) → f (x). But f (rn) = 0 ∀n, given the conditions in the claim, so 0 → f (x) =⇒
f (x) = 0 ∀x ∈ (a, b), as desired. ♠
(b) Let f and g be continuous real-valued functions on (a, b) : f (r) = g(r) for each rational number r ∈ (a, b).
Claim:
f (x) = g(x) ∀x ∈ (a, b).
Proof :
Using the limit concept of sequences, for any x ∈ (a, b) ∃ a sequence of rational numbers, (rn) : limfn= x =⇒ f (x) = limf (rn) = limg(rn) = g(x).
20δ ≤ 1
Hence, f (x) = g(x), as desired. ♠
EXERCISE 17.13 (b)
Let h(x) = x ∀x ∈ Q and h(x) = 0 ∀x ∈ R \ Q.
Claim:
h is continuous at x = 0 and no other point.
Proof :
For any > 0, if |x − 0| < , then |h(x) − h(0)| is either 0 (if x is ∈ R \ Q) or |x|
(if x ∈ Q, and thus in both cases < ). Thus h is continuous at x = 0. ∀ other x, consider two sequences with limit x, one (rn) ∈ Q, and another, (xn) ∈ R \ Q.
Then lim h(xn) = 0 and lim h(rn) = x 6= 0. Hence, ∃ disconituity for h at x 6= 0, as desired. ♠
EXERCISE 17.14 Claim:
f is continuous at each point of R \ Q and discontinuous at each point of Q.
Proof :
For x = pq ∈ Q, define a sequence xn∈ R\Q : lim xn= x =⇒ lim f (xn) = 0, but f (x) =1q 6= 0 =⇒ f is discontinuous at x. For an irrational x and any > 0, let δ > 0 be defined as the distance from x to the closest irreducible fraction pq with denominator q ≤ 1. Then for any x0 : |x0− x| < δ =⇒ |f (x0) − f (x)| < 0 =⇒ f continuous at x ∈ R \ Q, as desired. ♠
EXERCISE 18.2
The limit x0 (or y0) of the subsequence xnk ∈ (a, b) (or ynk ∈ (a, b)) may be an endpoint a or b of the interval and thus lie outside of the domain of the function.)
EXERCISE 18.4
Let S ⊆ R and suppose there exists a sequence (xn) in S that converges to a number x0∈ S./
Claim:
∃ an unbounded continuous finction on S.
Proof :
Let x0 ∈ S. We are given a sequence (x/ n) ∈ S which converges to x0. Then
|xn− x0| := the distance to x0 is continuous and strictly positive on S. Define f := |x 1
n−x0| =⇒ f is well-defined and continuous on S =⇒ limn→∞f = ∞, as desired. ♠
EXERCISE 18.6 Claim:
x = cos x for some x ∈ 0,π2.
Proof :
We know f (x) = x−cos x is continuous ∈0,π2, < 0 at x = 0 and > 0 at x = π2. Implement the Intermediate Value Theorem. Hence, ∃ an x ∈ 0,π2 : f (x) = 0, as desired. ♠
EXERCISE 18.8
Suppose that f is a real-valued function continuous on R and that f (a)f (b) < 0 for some a, b ∈ R.
Before commencing, we will state the properties that 0 · a = 0 ∀a ∈ Z and
a · b < 0 ⇐⇒ a < 0, b > 0 or a > 0, b < 0 ∀a, b ∈ Z Claim:
∃ an x between a and b : f (x) = 0.
Proof :
Given f (a)f (b) < 0, either
Case 1 : f (a) < 0 =⇒ f (b) > 0 =⇒ f (a) < 0 < f (b), or Case 2 : f (a) > 0 =⇒ f (b) < 0 =⇒ f (b) < 0 < f (a).
In either case, the Intermediate Value Theorem tells us that ∃ x ∈ (a, b) : f (x) = 0, as desired. ♠
EXERCISE 18.10
Suppose that f is continuous on [0, 2] and f (0) = 2.
Claim:
x, y ∈ [0, 2] : |x − y| = 1 and f (x) = f (y).
Proof :
Let g(x) = f (x + 1) − f (x) =⇒ g is continuous on [0, 1] and g(0) = f (1) − f (0) = f (1) − f (2) = −g(1). Implement the Intermediate Value Theorem.
∃x ∈ [0, 1] : g(x) = 0, as desired. ♠
EXERCISE 19.2 (a) Claim:
f (x) = 3x + 1 is uniformly continuous on R.
Proof :
Let > 0 be given. We want |f (x) − f (y)| = |(3x + 1) − (3y + 1)| < for
|x − y| < δ with x, y ∈ R. We know |(3x + 1) − (3y + 1)| = |3x − 3y| = 3|x − y|.
Take δ := 3. Then
|x−y| < δ =⇒ |x−y| <
3 =⇒ 3|x−y| < =⇒ |3x−3y| < =⇒ |3x−3y+1−1| < =⇒
|3x + 1 − 3y − 1| < =⇒ |(3x + 1) − (3y + 1)| < =⇒ |f (x) − f (y)| < , as desired. ♠
(b) Claim:
f (x) = x2is uniformly continuous on [0, 3] . Proof :
Let > 0 be given. We want |f (x) − f (y)| = |x2− y2| < for |x − y| < δ with x, y ∈ [0, 3]. We know |x2− y2| = |(x − y)(x + y)| = |x − y| · |x + y|. With x, y ∈ [0, 3], let |x + y| ≤ |3 + 3| = 6. Define δ := 6. Then for any x, y ∈ [0, 3]
with |x − y| < δ,
|f (x) − f (y)| = |x2− y2| = |(x − y)(x + y)| = |x − y| · |x + y| < |x − y| · 6 = , as desired. ♠
(c) Claim:
f (x) = 1
x is uniformly continuous on 1 2, ∞
.
Proof :
If f is uniformly continuous on a bounded set S, then f is a bounded function on S.
Proof :
By contradiction. Assume that f is uniformly continuous on a bounded set S and an unbounded function on S. Then ∃ a sequence, (xn) in the domain of f such that |f (xn)| ≥ n. By the Bolzano-Weierstrass Theorem, the sequence contains a convergent subsequence (xnk), since the domain is bounded. A con-vergent subsequence is Cauchy (by definition) and hence the sequence of values f (xnk) is Cauchy by the property of uniformly continuous functions. But the sequence |f (xnk)| ≥ nk is unbounded, contradicting our assumption in the out-set of this proof. Hence, if f is uniformly continuous on a bounded out-set S, then f is a bounded function on S, as desired. ♠
(b) Claim:
1
x2 is not uniformly continuous on (0, 1).
Proof :
We want to show that if x12 is not a bounded function on (0, 1), then x12 is not uniformly continuous on (0, 1)21. So it will suffice to show that
f (x) = 1
21This is the contrapositive of what was proved in part (a).
which shows that f (x) =x12 is not a bounded function on (0, 1), as desired. ♠
EXERCISE 19.6 (a) Claim:
f (x) =√
x is uniformly continuous on (0, 1], although f0 is unbounded.
Proof :
f0(x) = 2√1x → +∞ as x → 0 =⇒ f0(x) is unbounded. f continuous on the closed interval [0, 1] =⇒ f uniformly continuous on [0, 1] =⇒ f uniformly continuous on the subset (0, 1], as desired. ♠
Let f (x) =√
x for x ≥ 1.
(b) Claim:
f is uniformly continuous on [1, ∞).
Proof :
Thus, g0 is both bounded and defined on R =⇒ g is uniformly continuous.