Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Solutions to Exercises 4.5.7, 5.2.1, 5.2.2, 5.3.7, and 7.3.5 in Analysis, Assignments of Business and Labour Law

The solutions to various exercises in analysis, including proofs for the existence of a fixed point for a continuous function, the derivative of a specific function, the contractivity of a differentiable function, and the uniqueness of a fixed point for a differentiable function. It also covers the relationship between an increasing function and its derivative.

Typology: Assignments

Pre 2010

Uploaded on 08/31/2009

koofers-user-lig-1
koofers-user-lig-1 🇺🇸

5

(1)

10 documents

1 / 6

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
MT320 Homework 6
Solutions
17 April, 2009
Exercise 4.5.7. Let fbe a continuous function on the closed interval [0,1] with range also
contained in [0,1]. Prove that fmust have a fixed point; that is, show f(x) = xfor at least
one value of x[0,1].
Proof: Define g: [0,1] Rby g(x) = f(x)x. Then gis continuous, and one has
g(0) = f(0) 00 and g(1) = f(1) 111 = 0. By the Intermediate Value Theorem,
there exists c[0,1] such that 0 = g(c) = f(c)c, and hence f(c) = c.
Exercise 5.2.2. (a) Use Definition 5.2.1 to produce the proper formula for the derivative of
f(x) = 1/x.
Proof: We simply compute:
f0(c) = lim
xc
1
x1
c
xc= lim
xc
cx
xc
xc= lim
xc
cx
xc
1
xc
= lim
xc1
xc =1/c2.
Exercise 5.3.2. Recall from Exercise 4.3.9 that a function fis contractive on a set Aif
there exists a constant 0 <s<1 such that |f(x)f(y)| s|xy|for all x, y A. Show
that if fis differentiable and f0is continuous and satisfies |f0(x)|<1 on a closed interval,
then fis contractive on this set.
Proof: Since we have assumed that f0is continuous and Ais a closed interval, the Extreme
Value Theorem provides us with cAsuch that |f0(x)| |f0(c)|for all xA. We show
that fis contractive on Afor the constant s=|f0(c)|. Given any x, y Awith (without
loss of generality) x < y, the Mean Value Theorem guarantees the existence of d[x, y]
with f(x)f(y) = f0(d)(xy). We then compute that
|f(x)f(y)|=|f0(d)| |xy| |f0(c)| |xy|=s|xy|.
We complete the proof by noting that, by hypothesis, s=|f0(c)|<1.
Exercise 5.3.5. Afixed point of a function fis a value xwhere f(x) = x. Show that if f
is differentiable on an interval with f0(x)6= 1, then fcan have at most one fixed point.
Proof: Suppose x, y were distinct points of the domain with f(x) = xand f(y) = y. Without
loss of generality, assume that x < y. By the Mean Value Theorem, there exists c(x, y)
1
pf3
pf4
pf5

Partial preview of the text

Download Solutions to Exercises 4.5.7, 5.2.1, 5.2.2, 5.3.7, and 7.3.5 in Analysis and more Assignments Business and Labour Law in PDF only on Docsity!

MT320 Homework 6

Solutions

17 April, 2009

Exercise 4.5.7. Let f be a continuous function on the closed interval [0, 1] with range also contained in [0, 1]. Prove that f must have a fixed point; that is, show f (x) = x for at least one value of x ∈ [0, 1]. Proof: Define g : [0, 1] → R by g(x) = f (x) − x. Then g is continuous, and one has g(0) = f (0) − 0 ≥ 0 and g(1) = f (1) − 1 ≤ 1 − 1 = 0. By the Intermediate Value Theorem, there exists c ∈ [0, 1] such that 0 = g(c) = f (c) − c, and hence f (c) = c. 

Exercise 5.2.2. (a) Use Definition 5.2.1 to produce the proper formula for the derivative of f (x) = 1/x. Proof: We simply compute:

f ′(c) = lim x→c

1 x −^

1 c x − c

= lim x→c

c−x xc x − c

= lim x→c

c − x x − c

xc = lim x→c

xc

= − 1 /c^2. 

Exercise 5.3.2. Recall from Exercise 4.3.9 that a function f is contractive on a set A if there exists a constant 0 < s < 1 such that |f (x) − f (y)| ≤ s|x − y| for all x, y ∈ A. Show that if f is differentiable and f ′^ is continuous and satisfies |f ′(x)| < 1 on a closed interval, then f is contractive on this set. Proof: Since we have assumed that f ′^ is continuous and A is a closed interval, the Extreme Value Theorem provides us with c ∈ A such that |f ′(x)| ≤ |f ′(c)| for all x ∈ A. We show that f is contractive on A for the constant s = |f ′(c)|. Given any x, y ∈ A with (without loss of generality) x < y, the Mean Value Theorem guarantees the existence of d ∈ [x, y] with f (x) − f (y) = f ′(d)(x − y). We then compute that

|f (x) − f (y)| = |f ′(d)| |x − y| ≤ |f ′(c)| |x − y| = s|x − y|.

We complete the proof by noting that, by hypothesis, s = |f ′(c)| < 1. 

Exercise 5.3.5. A fixed point of a function f is a value x where f (x) = x. Show that if f is differentiable on an interval with f ′(x) 6 = 1, then f can have at most one fixed point. Proof: Suppose x, y were distinct points of the domain with f (x) = x and f (y) = y. Without loss of generality, assume that x < y. By the Mean Value Theorem, there exists c ∈ (x, y)

such that f (x) − f (y) = f ′(c)(x − y). But then

f ′(c) =

f (x) − f (y) x − y

x − y x − y

contradicting our assumption that f ′^6 = 1. 

Exercise 5.3.7. (a) Recall that a function f : (a, b) → R is increasing on (a, b) if f (x) ≤ f (y) whenever x < y in (a, b). Assume f is differentiable on (a, b). Show that f is increasing on (a, b) if and only if f ′(x) ≥ 0 for all x ∈ (a, b). Proof: Assume first that f is increasing, and let x, y ∈ (a, b) be distinct. If x < y, then

f (x) ≤ f (y) and 0 ≤ f (y) − f (x), so 0 ≤ f^ (y y)−−fx^ ( x)= f^ (x x)−−fy^ (y). On the other hand, if x > y,

then f (x) ≥ f (y) and f (x) − f (y) ≥ 0, so f^ (x x)−−fy^ (y) ≥ 0. We see that, in either case, the

difference quotient is always nonnegative. For any c ∈ (x, y), the Order Limit Theorem then gives

f ′(c) = lim x→c

f (x) − f (c) x − c

≥ lim x→c

as was desired. Assume, conversely, that f ′^ ≥ 0. Given x, y ∈ (a, b) with x < y, the Mean Value Theorem guarantees the existence of c ∈ (x, y) with f (y) − f (x) = f ′(c)(y − x). Since f ′(c) ≥ 0 and y > x, we deduce that f (y) − f (x) ≥ 0, or f (x) ≤ f (y). 

Exercise 5.3.8. Assume g : (a, b) → R is differentiable at some point c ∈ (a, b). If g′(c) 6 = 0, show that there exists a δ-neighborhood Vδ(c) ⊆ (a, b) for which g(x) 6 = g(c) for all x ∈ Vδ(c). Proof: (Note an error in the problem statement: one wants “... for all x ∈ V (^) δ^0 (c).”)

Applying the definition of the limit limx→c g(x x)−−gc (c)= g′(c) to the choice of  = |g′(c)|, we find that there exists δ 1 > 0 such that x ∈ V (^) δ^01 (c) ∩ (a, b) implies

∣ ∣ ∣ ∣

g(x) − g(c) x − c

− g′(c)

∣ < ^ =^ |^0 −^ g

′(c)|.

In particular, for such x one has g(x x)−−gc (c) 6 = 0, and hence g(x) − g(c) 6 = 0, so that g(x) 6 = g(c). Since (a, b) is open, we may choose δ ≤ δ 1 such that Vδ(c) ⊆ (a, b). Thus for any x ∈ V (^) δ^0 (c) we have x ∈ V (^) δ^01 (c) ∩ (a, b) and hence g(x) 6 = g(c), as was desired. 

Exercise 7.3.5. Let

f (x) =

1 if x = 1/n for some n ∈ N 0 otherwise.

Show that f integrable on [0, 1] and compute

0 f^. Proof: Let  > 0 be given. We will produce a partition P satisfying U (f, P ) − L(f, P ) < . First, let N = [[2/]] + 1 ∈ N, and notice for later use that 1/N < /2. Next, let δ = min( (^) 4(N −1) , (^3) N (N^1 −1) ). The significance of 1/ 3 N (N − 1) is that the shortest distance between

As noted earlier, 1/N < /2. Also, by definition, δ ≤ /4(N − 1). Thus, we have

U (f, P ) =

N

  • 2(N − 1)δ <

+ 2(N − 1)

4(N − 1)

as was desired. Finally, we noted above that L(f, P ) = 0 for every P. This shows that

0 f^ = 0.^ 

Exercise 7.4.1. (a) Let f be a bounded function on a set A, and set

M = sup{f (x) : x ∈ A}, m = inf{f (x) : x ∈ A}, M ′^ = sup{|f (x)| : x ∈ A}, and m′^ = inf{|f (x)| : x ∈ A}.

Show that M − m ≥ M ′^ − m′. Proof: Most students assumed, intuitively, and correctly, that M ′^ = max(|M |, |m|). Assum- ing this for the moment, we solve the problem. Suppose first that M ′^ = |M |. Then M ≥ 0, hence M ′^ = M. On other other hand, either one has m ≥ 0, in which case |f | = f and m′^ = m, or one has m < 0, whence m′^ ≥ 0 > m. In both cases, one has m′^ ≥ m. Together with M ′^ = M , this implies M − m ≥ M ′^ − m′, as was desired. Now suppose that M ′^ = |m|. Let g be the bounded function g = −f on A, so that g satisfies the same hypotheses as does f. However, one has (in the obvious notation) M (g) = −m(f ), m(g) = −M (f ), M ′(g) = M ′(f ), and m′(g) = m′(f ). In particular, M ′(g) = |M (g)|, so we have already treated g in the first case of our proof above, and therefore

M (f ) − m(f ) = (−m(g)) − (M (g)) = M (g) − m(g) ≥ M ′(g) − m′(g) = M (f ) − m′(f ),

as was desired. It remains to prove that M ′^ = max(|M |, |m|). Let x ∈ A. If f (x) ≥ 0, then since f (x) ≤ M we have |f (x)| ≤ |M |. Otherwise, if f (x) ≤ 0, then since f (x) ≥ m we have |f (x)| ≤ |m|. Thus, for any x ∈ A, at least one of |f (x)| ≤ |M | or |f (x)| ≤ |m| holds, so that |f (x)| ≤ max(|M |, |m|). This last inequality, together with the definition of the least upper bound, shows that M ′^ ≤ max(|M |, |m|). On the other hand, since |f (x)| ≤ M ′^ we have −M ′^ ≤ f (x) ≤ M ′. This last inequality, together with the definition of inf and sup, implies that −M ′^ ≤ m and M ≤ M ′. But m ≤ M , so we have −M ′^ ≤ m ≤ M ≤ M ′. This shows that |M |, |m| ≤ M ′. Or, in other words, max(|M |, |m|) ≤ M ′. Since both M ′^ ≤ max(|M |, |m|) and max(|M |, |m|) ≤ M ′, we get the desired claim. 

(b) Show that if f is integrable on the interval [a, b], then |f | is also integrable on this interval. Proof: Let P = {a = x 0 < · · · < xN = b} be a partition of [a, b]. We observe that

U (|f |, P ) − L(|f |, P ) =

∑^ N

i=

M (^) i′ (xi − xi− 1 ) −

∑^ N

i=

m′ i(xi − xi− 1 ) ≤

∑^ N

i=

(M (^) i′ − m′ i)(xi − xi− 1 )

∑^ N

i=

(Mi − mi)(xi − xi− 1 ) =

∑^ N

i=

Mi(xi − xi− 1 ) −

∑^ N

i=

mi(xi − xi− 1 )

= U (f, P ) − L(f, P ),

where the inequality is by applying part (a) to each interval [xi− 1 , xi]. Let  > 0 be given. Since f is integrable on [a, b] we may choose a partition P such that U (f, P ) − L(f, P ) < . By our above observation, this partition P also satisfies

U (|f |, P ) − L(|f |, P ) ≤ U (f, P ) − L(f, P ) < ,

as was desired. 

(c) Provide the details for the argument that in this case we have |

∫ (^) b a f^ | ≤^

∫ (^) b a |f^ |. Proof: One has −|f | ≤ f ≤ |f |, and so

|f | =

−|f | ≤

f ≤

|f |.

This implies that |

f | ≤

|f |. 

Exercise 7.4.4. Decide which of the following conjectures is true and supply a short proof. For those that are not true, give a counterexample.

(a) If |f | is integrable on [a, b] then f is also integrable on this set. Answer: False. Let f : [0, 1] → R be defined by

f (x) =

1 if x is rational, − 1 if x is irrational

Then |f | is constant with value 1, hence intgrable, but f is not integrable (by a modification of the proof that Dirichlet’s function is not integrable).

(b) Assume g is integrable and g ≥ 0 on [a, b]. If g(x) > 0 for an infinite number of points x ∈ [a, b], then

g > 0. Answer: False. Exercise 7.3.5, above, provides a counterexample.

(c) If g is continuous on [a, b] and g ≥ 0 with g(x 0 ) > 0 for at least one point x 0 ∈ [a, b],

then

∫ (^) b a g >^ 0. Answer: True. Proof: By work done in class, for fixed g, the function L(g, P ) of partitions is an increasing function. Therefore, it suffices to provide a single partition P with L(g, P ) > 0. Since L(g, P ) =

∑N

i=1 mi(xi^ −^ xi−^1 ), and each of terms of this sum are nonnegative, in order for^ P to satisfy L(g, P ) > 0 it suffices that a single mi be positive. Suppose we can find a subinterval [c, d] ⊂ [a, b] and a constant  > 0 with g(x) ≥  for all x ∈ [c, d]. Then the partition P = {a, c, d, b} satisfies mi ≥  > 0, where i is the index for which [xi− 1 , xi] = [c, d], so we will be done. We will find [c, d] with  = g(x 0 )/2. By the definition of continuity, there exists δ > 0 such that x ∈ Vδ(x 0 ) ∩ [a, b] implies |g(x) − g(x 0 )| < g(x 0 )/2. But the latter inequality implies that g(x) > g(x 0 )/2. Thus, any closed interval [c, d] (with c < d) contained in Vδ(x 0 ) ∩ [a, b] works. 

(d) If

∫ (^) b a f >^ 0, there is an interval [c, d]^ ⊆^ [a, b] and a^ δ >^ 0 such that^ f^ (x)^ ≥^ δ^ for all x ∈ [c, d]. Answer: True.