Introduction to Analysis
Homework 5
1. (Rudin 4.11) Suppose f is uniformly continuous mapping of a metric space X into a metric space Y and prove that {f (xn)} is a Cauchy sequence in Y for every Cauchy sequence {xn} in X.
2. (Rudin 4.13) Let E be a dense subset of a metric space X, and let f be a uniformly continuous real function defined on E. Prove that there exist continuous real functions g on X such that g(x) = f (x) for all x ∈ E. (Such functions g are called continuous extensions of f from E to X.)
3. (Rudin 4.23) A real-valued function f defined in (a, b) is said to be convex if f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
whenever a < x < b, a < y < b, 0 < λ < 1. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex.
(For example, if f is convex, so is ef.)
4. Remark: In class, we discuss the contraction mapping on a metric space X into X and prove that there exists a unique fixed point if X is complete. To prove Fixed Point Theorem, the condition 0 < c < 1 is necessary. However, if X is compact, we can remove this condition.
(a) Suppose K is a compact metric space, and f : K → K is a map with the property that for every pair of distinct points x, y ∈ K one has d(f (x), f (y)) < d(x, y). Show that there exists a unique point p ∈ K such that f (p) = p.
(b) Show by example that if the “<” is weakened to “≤” in part (a), the map need not have any fixed point.
(c) Show by examples that the result of (a) is also false if we put any of the noncompact metric spaces [0, 1), [0, +∞) or R in place of the compact space K.
(d) Deduce from (a) that for K a compact metric space, there cannot exist a continuous function f : K → K such that for all distinct points x, y ∈ K one has d(f (x), f (y)) >
d(x, y).
5. (Rudin 4.19) Suppose f is a real function with domain R which has the intermediate value property: If f (a) < c < f (b), then f (x) = c for some x between a and b.
Suppose also, for every rational r, that the set of all x with f (x) = r is closed. Prove that f is continuous.
Hint: If xn→ x0 but f (xn) > r > f (x0) for some r and all n, then f (tn) = r for some tn between x0 and xn; thus tn → x0. Find a contradiction.
