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)
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.
