PROBLEM SET 5 (DUE IN LECTURE ON OCT 16 (FRIDAY))
(All Theorem and Exercise numbers are references to the textbook by Apostol; for instance
“Exercise 1.15-3” means Exercise 3 in section 1.15.)
Problem 1. Do Exercise 3.11-1.
Problem 2. Do Exercise 3.11-5.
Problem 3. Do Exercise 3.20-7.
Problem 4. Do Exercise 3.20-8. (Hint: Try to apply the previous problem.)
Problem 5. Assume that f : [0, 1] → R is a continuous function satisfying f (0) = f (1).
(a) Prove that there is a unique continuous function F : R → R that extends
F (i.e. F (x) = f (x) for x ∈ [0, 1]) and satisfies
F (x + 1) = F (x)
for all x ∈ R.
(b) Let F be the function constructed in part (a). Prove that for any c ∈ R,
there exists p ∈ R such that
F (p + c) = F (p).
(You might find Theorem 3.15 helpful for this.)
Problem 6. (a) Let f, g : [a, b] → R be continuous functions satisfying
f (x) > g(x) for all x ∈ [a, b].
Prove that there exists a positive real number C > 0 such that the
inequality can be strengthened to
f (x) > g(x) + C for all x ∈ [a, b].
(b) Give a counterexample to part (a) if f and g are not required to be
continuous.
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