Math 414: Analysis I
Homework 11
Due: April 30, 2014
Name:
The following problems are for additional practice and are not to be turned in: (All
problems come from Basic Analysis, Lebl )
Exercises: 4.2.3, 4.2.4, 4.2.5, 4.2.6, 4.2.7, 4.2.8, 5.1.1, 5.1.2, 5.1.3, 5.1.5
Turn in the following problems.
1. Exercise 4.2.8 in Basic Analysis, Lebl
2. Exercise 5.1.2 in Basic Analysis, Lebl
3. Exercise 5.1.6 in Basic Analysis, Lebl
4. (a) Suppose that f : R → R satisfies |f (x) − f (y)| ≤ |x − y|α for some α > 1.
Prove that f is a constant function.
(b) Show by example that if α = 1, then f need not be a constant function.
5. The Mean Value Theorem is also useful for establishing inequalities. Using the Mean
Value Theorem, prove that (x − 1)/x < ln x < x − 1 for x > 1. Hint: You may use
d
(without proof ) that dx
[ln x] = x1 .
6. Let f and g be differentiable on R and suppose that f (0) = g(0) and f 0 (x) ≤ g 0 (x)
for all x ≥ 0. Prove that f (x) ≤ g(x) for all x ≥ 0.
7. Prove Darboux’s Theorem: If f is differentiable on I = [a, b] and if k is a
number between f 0 (a) and f 0 (b), then there is at least one point c in (a, b) such that
f 0 (c) = k. Remark: Note that this tells us that derivatives possess an intermediate
value theorem property, even when they are not continuous!
Suggestion: Assume without loss of generality that f 0 (a) < k < f 0 (b), and define a
new function g(x) = kx − f (x). Justify the claim that g attains a maximum at some
c ∈ (a, b). Compute g 0 (x), and prove that the maximum of g cannot be attained at
a or b. Conclude that there is a c ∈ (a, b) such that g 0 (c) = 0.
8. Suppose that f : [0, 2] → R is continuous on [0, 2], differentiable on (0, 2), and that
f (0) = 0, f (1) = 1, f (2) = 1.
(a) Show that there exists c1 ∈ (0, 1) such that f 0 (c1 ) = 1.
(b) Show that there exists c2 ∈ (1, 2) such that f 0 (c2 ) = 0.
(c) Show that there exists c ∈ (0, 2) such that f 0 (c) = 1/3. Hint: Use Darboux’s
Theorem.
9. Suppose that f : [a, b] → R is bounded and let Q and P be any partitions of [a, b].
Prove that L(P, f ) ≤ U (Q, f ). Hint: If P and Q are partitions, then P ∪ Q is a
refinement of both P and Q (Justify this.).
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