MH9100: Advanced Investigations In Calculus I.

Solutions to Homework Assignment #2.

Problem 1:

Find the supremum and infimum (if they exist) of the following sets. Alsodecide which sets have greatest and least elements (i. e. decide when thesupremum and infimum happens to belong to the set).

(a)

{1

n: n ∈ Z and n 6= 0

}.

(b) {x : 0 6 x 6√

2 and x is rational}.

(c)

{1

n+ (−1)n : n ∈ N

}.

Answer. (a) 1 and −1 are an upper and a lower bound for the given set,respectively. Indeed, for each n ∈ Z, we have 1

n 6 1; if n < 0 this istrivial, as the left-hand side is negative. If n > 0, then n > 1, hence1n 6 1. Using a similar argument, we can prove that 1

n > −1 for alln ∈ Z. But these bounds, actually belong to the set, because they canbe written as 1

1 and 1−1 . Therefore, 1 is the supremum of the given

set (and is also a maximum), and −1 is the infimum (and is also aminimum).

(b) 0 is by definition a lower bound of the given set, and it belongs to it,because it is rational. Therefore, 0 is the infimum, and also a minimumof the given set. The supremum is

√2; first, it is an upper bound, by

definition. Second, for every ε > 0, there is a rational number q with√2−ε < q 6

√2, or equivalently, there is some q in the given set with√

2− ε < q 6√

2. This shows that√

2 is the supremum of the set; itis not a maximum, because it is irrational, and it does not belong tothe given set.

(c) For every n, we have 1/n + (−1)n > (−1)n > −1, so −1 is a lowerbound for the given set. We will show that it is the infimum; let ε > 0be arbitrary. There is an odd number n, such that 1

n < ε, therefore,

1

−1 < 1n + (−1)n = 1

n −1 < −1 + ε. This shows that −1 is the infimumof this set, but it is not a minimum because it does not belong to theset, as we have seen 1/n+ (−1)n > −1 for all n.

Next, we observe that all terms corresponding to n odd satisfy 1n +

(−1)n = 1n − 1 6 1− 1 = 0. For even n, we have

1

n+ (−1)n =

1

n+ 1 6

1

2+ 1 =

3

2,

since n > 2 for all even natural n. So, 3/2 is an upper bound; it alsobelongs to the given set, as it is equal to the term corresponding ton = 2. Therefore, it is both a supremum and a maximum of the givenset.

Problem 2:

How many continuous functions are there which satisfy (f(x))2 = x2 for allreal x?

Answer. We will show that the only continuous functions satisfying (f(x))2 =x2 for all real x are precisely four; f(x) = x, f(x) = −x, f(x) = |x|,f(x) = −|x|. It is obvious that these functions are continuous and satisfythe given functional equation.

First of all, the hypothesis yields f(0) = 0. For each x, we have f(x) =±x. What we will show here, is that this sign remains constant when youconsider only positive x, and similarly when you consider only negative x;these two signs could be different in these two cases, and they indeed arewhen, for example, f(x) = |x|. In this case, we have f(x) = x for all positivex, and f(x) = −x for all negative x.

We will prove it by contradiction; suppose that the sign is not constantfor positive x, which means that there are positive x, y such that f(x) =x > 0 and f(y) = −y < 0. Since f is continuous, by the IntermediateValue Theorem, there is some x0 between x and y, such that f(x0) = 0,or x20 = (f(x0))

2 = 0, which implies that x0 = 0. But x and y are bothpositive, hence 0 cannot be between them, a contradiction. Therefore, forall positive x, we have either f(x) = x or f(x) = −x. The same conclusioncan be reached for negative x, using the same argument. Now, we can seethat we can only obtain the four functions mentioned in the beginning; if

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the signs are the same for both positive and negative x, then we have eitherf(x) = x for all x or f(x) = −x for all x. If the signs are different, then wetake f(x) = |x| for all x or f(x) = −|x| for all x.

Problem 3:

Prove that there is some number x such that

(a) x179 +163

1 + x2 + sin2 x= 119.

(b) sinx = x− 1.

Proof. (a) Consider the function

g(x) = x179 +163

1 + x2 + sin2 x− 119.

This function is continuous everywhere, so it suffices to find two valuesof g with different signs. We try the most easily computable; we haveg(0) = 44 > 0, and

g(1) = 1 +163

2 + sin2 1− 119 < 1 +

163

2− 119 = −73

2< 0.

By the Intermediate Value Theorem, there is some x in (0, 1), suchthat g(x) = 0.

(b) Now, we consider the function g(x) = sinx − x + 1, which is alsocontinuous everywhere. We have g(0) = 1 > 0, and g(π) = sinπ−π+1 = 1 − π < 0. Therefore, by the Intermediate Value Theorem, thereis some x in (0, π), such that g(x) = 0, as desired.

Problem 4:

Suppose that f and g are continuous on [a, b] and that f(a) < g(a), butf(b) > g(b). Prove that f(x) = g(x) for some x in [a, b]. (If your proof isn’tvery short, it’s not the right one.)

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Proof. Consider the function h(x) = f(x) − g(x) on [a, b]. This function iscontinuous on [a, b], as a difference of continuous functions. Furthermore, wehave h(a) = f(a)− g(a) < 0 and h(b) = f(b)− g(b) > 0. Therefore, by theIntermediate Value Theorem, there is some x in (a, b), such that h(x) = 0,or equivalently, f(x) = g(x).

(?) Problem 5:

(a) Let f(x) = sin 1/x for x 6= 0 and let f(0) = 0. Is f continuous on[−1, 1]? Show that f satisfies the conclusion of the Intermediate ValueTheorem on [−1, 1]; in other words, if f takes on two values on [−1, 1],it also takes on every value in between.

(b) Suppose that f , defined on [a, b], satisfies the conclusion of the Inter-mediate Value Theorem on any subinterval [c, d] of [a, b], and that ftakes on each value only once. Prove that f is continuous.

(c) Generalize to the case where f takes on each value only finitely manytimes.

Proof. (a) f is continuous everywhere except for x = 0; we know that thelimit limx→0 sin 1/x does not exist. On the other hand, sin 1/x is thecompositon of two continuous functions with domain R\{0}, thereforef is continuous anywhere else but zero.

f satisfies the conclusions of the Intermediate Value Theorem, indeed.For all x in [−1, 1] we have −1 6 sin 1/x 6 1. We focus on thesubinterval

[2/π, 2/(3π)] ⊆ (0, 1] ⊆ [−1, 1].

Since 0 is not an element of this interval, f is continuous on [2/π, 2/(3π)].Furthermore,

f

(2

π

)= sin

π

2= 1

and

f

(2

3π

)= sin

3π

2= −1,

hence by the Intermediate Value Theorem applied on this interval, fassumes all values between −1 and 1. So, if we take x and y in [−1, 1],

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we have −1 6 f(x), f(y) 6 1, and thus f assumes all values betweenf(x) and f(y).

(b) Without loss of generality, we assume that f(a) < f(b). We will showthat f is increasing; consider x and y such that a 6 x 6 y 6 b. Iff(x) < f(a) < f(b), then by hypothesis applied on the interval [x, b],there is some x0 in (x, b), such that f(x0) = f(a), contradicting theassumption that f assumes each value only once. Similarly, we canprove that we cannot have f(x) > f(b), using the hypothesis on theinterval [a, x], this time. Therefore, f(a) 6 f(x) 6 f(b); restricting ourattention to [x, b] and using the above argument, we also get f(a) 6f(x) 6 f(y) 6 f(b), thus f is increasing. Now, the proof proceedsexactly as in Problem 8.

(c) Let y in [a, b] be arbitrary; we want to show that limx→∞

f(x) = f(y).

For given ε > 0, f takes the values f(y) + ε or f(y)− ε finitely manytimes, so the set

B = {x ∈ [a, b]|f(x) = f(y)− ε or f(x) = f(y) + ε}

is finite. Obviously, y does not belong to B, so the minimum possibledistance |y − x|, where x ∈ B, is positive, and is denoted by δ. Now,consider x such that 0 < |y − x| < δ; by definition of δ we get that xdoes not belong to B. Assume that |f(y)− f(x)| > ε, say for examplef(x) > f(y) + ε (the case f(x) < f(y) − ε is treated similarly). Byhypothesis applied on the interval [x, y] (or [y, x], if x > y), there mustbe some x0 in (x, y) such that f(x0) = f(y) + ε, or x0 ∈ B. But|y − x0| < |y − x| < δ, hence x0 cannot belong to B, a contradiction.

In conclusion, we have proved the following: for every ε > 0, there isδ > 0 such that if 0 < |x− y| < δ then |f(y)− f(x)| < ε. Since y waschosen arbitrarily, we deduce that f is continuous.

Remark. B could be the empty set; in this case, we could pick any positiveδ, and proceed exactly as before.

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(?) Problem 6:

If f is a continuous function on [0, 1], let ‖f‖ be the maximum value of |f |on [0, 1].

(a) Prove that for any number c we have ‖cf‖ = |c| · ‖f‖.

(b) Prove that ‖f + g‖ 6 ‖f‖ + ‖g‖. Give an example where ‖f + g‖ 6=‖f‖+ ‖g‖.

(c) Prove that ‖h− f‖ 6 ‖h− g‖+ ‖g − f‖.

Proof. First of all, we note that since f is continuous, then |f | is also con-tinuous, therefore ‖f‖ is well defined.

(a) Let y in [0, 1] be such that |f(y)| > |f(x)| for all x in [0, 1], that is,‖f‖ = |f(y)|. Then, for all x in [0, 1] we have

|cf(x)| = |c| · |f(x)| 6 |c| · |f(y)| = |c| · ‖f‖,

and as we see, equality os obtained when x = y. This shows thatthe maximum value of |cf(x)| is precisely |cf(y)|, which yields ‖cf‖ =|c| · ‖f‖.

(b) Let y, z be such that ‖f‖ = |f(y)| and ‖g‖ = |g(z)|. Then, for all x in[0, 1], we have:

|f(x) + g(x)| 6 |f(x)|+ |g(x)| 6 |f(y)|+ |g(z)| = ‖f‖+ ‖g‖,

therefore the maximum value of the right-hand side must also satisfythe same inequality, thus, ‖f + g‖ 6 ‖f‖+‖g‖. Equality is not alwaysobtained, as we can see from teh following example: define f(x) = xand g(x) = 1−x for all x in [0, 1]. Then ‖f‖ = ‖g‖ = 1, but ‖f +g‖ =1 < 2 = ‖f‖+ ‖g‖.

(c) This follows easily from part (b):

‖h− f‖ = ‖(h− g) + (g − f)‖ 6 ‖h− g‖+ ‖g − f‖.

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(?) Problem 7:

Let f be any polynomial function. Prove that there is some number y suchthat |f(y)| 6 |f(x)| for all real x.

Proof. We distinguish two cases; first, let the degree of f(x), say n, be odd(so that f(x) = anx

n + . . .+ a1x+ a0, where an 6= 0). Then, we know thatf(x) has at least one real root, say y. The conclusion easily follows, since|f(y)| = 0 6 |f(x)|, for all real x.

Next, we suppose that the degree of f(x) is even. Without loss of gen-erality, we may further assume that an > 0 (otherwise, work with −f(x)and observe that |−f(x)| = |f(x)|). This means that f(x) has an absoluteminimum, say f(y), that is, f(y) 6 f(x) for all real x. If f(y) > 0, thenf(x) > 0 for all real x, and f(x) = |f(x)|, for all x, thus |f(y)| 6 |f(x)|for all x. If f(y) < 0, then f has at least one real root. Indeed, sincelimx→∞ f(x) =∞, this shows that f(x) > 0 for all x > M , for some M > 0.Since f is continuous, by the Intermediate Value Theorem there is some x0between x and y, such that f(x0) = 0, so we have |f(x0)| = 0 6 |f(x)|, forall real x.

(?) Problem 8:

Suppose that f is a function such that f(a) 6 f(b) whenever a < b.

(a) Prove that limx→a−

f(x) and limx→a+

f(x) both exist.

(b) Prove that if f satisfies the conclusions of the Intermediate Value The-orem, then f is continuous

Proof. (a) Consider the sets A− = {f(x)|x < a} and A+ = {f(x)|x > a}.By definition, f(a) is an upper bound of A− and a lower bound ofA+. This means that k = supA− and l = inf A+ are well-defined. Wewill show that limx→a− f(x) = k and limx→a+ f(x) = l. Let ε > 0 bearbitrary. Then, there is some element of A−, say f(b), where b < a,such that k−ε < f(b) 6 k. Now, let δ = a−b > 0. For all x satisfyingb = a− δ < x < a, we have

k − ε < f(b) 6 f(x) 6 k.

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This proves that limx→a− f(x) = k. The proof that limx→a+ f(x) = lis similar and is omitted.

(b) First, we will see what happens if f is discontinuous at x = a. Sinceboth side limits exist, they must be unequal, that is k 6= l, using theabove notation. Furthermore, since we have k 6 f(a) 6 l, we shouldalso have k < l. Now, suppose that f satisfies the conclusions of theIntermediate Value Theorem, and that is discontinuous at x = a. Forany x < a we have f(x) 6 k and for any y > a we have f(y) > l,so f assumes all values between k and l. But the only value that fassumes between them is f(a), and since k < l, we have established acontradiction. Therefore, f is continuous everywhere.

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