Real Analysis, Math 821.

Instructor: Dmitry Ryabogin

Assignment IX.

1. Problem 1.

a) Let f(x) be a monotonic function satisfying

∀x, y ∈ R,  f(x) + f(y) = f(x + y),  f(1) = 1.

Prove that f(x) = x. Hint: Prove that f(p/q) = p/q, then consider two monotonic sequences (r_{k})_{k=1}^{∞}, (m_{k})_{k=1}^{∞}, r_{k}, m_{k} ∈ Q, converging to x ∈ R, x > 0. For x < 0, observe that f(x) = −f(−x).

b) Change the word monotonic by measurable in a), and answer the same question. Hint: Observe that

∫_{0}^{1} f(x + y)dy = ∫_{0}^{1}(f(x) + f(y))dy = f(x) + ∫_{0}^{1} f(y)dy,

and conclude that f is continuous. c)* Is it possible to drop the measurability assumption in b)?

2. Problem 2. a) Let f(x), g(x) be increasing functions on the real line. Does it follow that f(x)g(x) is increasing?

b) Construct the monotonic function on the real line which is not continuous only in rational points. Hint: Consider

f(x) := ∑_{k∈N: r_{k}<x} \frac{1}{2^{k}} , r_{k} ∈ Q.

To show that f(x) is not continuous in rational points observe that ∀x > r ∈ Q we have

f(x) = ∑_{r_{k}<x} \frac{1}{2^{k}} = ∑_{r_{k}<r} \frac{1}{2^{k}} + ∑_{r≤r_{k}<x} \frac{1}{2^{k}} ,  ∑_{r≤r_{k}<x} \frac{1}{2^{k}} > \frac{1}{2^{n}} ,

provided r has n-th place in enumeration of rationals. To show that f is continuous in R \ Q, observe that

sup_{R} f(x) = 1,  inf_{R} f(x) = 0,

hence the sum of jumps may not be bigger than 1.

3. Problem 3. a) Let E ⊂ [0, 1], and let f(x) be a bounded function on E, satisfying f(x_{1}) ≤ f(x_{2}) ∀x_{1}, x_{2} ∈ E, x_{1} < x_{2}. Is it possible to extend this function in such a way that the extension ϕ(x) would be nondecreasing on the whole segment [0, 1]?

Hint: Let x_{0} = inf E. Put ϕ(x) := sup_{y<x} f(y) for x ∈ [0, 1] \ E, x > x_{0}, and ϕ(x) :=

inf_{E} f(y) for x ∈ [0, 1] \ E, x ≤ x_{0}.

b) Change the word bounded by unbounded in a), and answer the same question.

Hint: Look at “ends”.

4. Problem 4. a) Let E ⊂ [0, 1] be nowhere dense closed set of a positive measure. Construct an increasing continuously differentiable f(x) on [0, 1] such that f′(x) = 0 ∀x ∈ E.

Hint: Take

f(x) := ∫_{0}^{x} ϕ(t)dt, ϕ(t) := inf_{y∈E} |y − t|.

Then observe that ϕ is continuous, and f′(x) = 0 ∀x ∈ E. To show that f is increasing take 0 ≤ x_{1} < x_{2} ≤ 1, and observe that there exists an interval (α, β) ⊂ (x_{1}, x_{2}), such that (α, β) ∩ E = ∅.

b) Is it possible to drop the assumption nowhere dense in a)?

Hint: There exists (α, β) ⊂ E such that f′(x) = 0 ∀x ∈ (α, β).

5. Problem 5. Prove that ∀E ⊂ [0, 1], |E| = 0, there exists a continuous increasing function f(x) on [0, 1] such that ∀x ∈ E we have f′(x) = +∞.

Hint:

a) ∀n ∈ N there exists an open bounded set G_{n} such that E ⊂ G_{n}, and m(G_{n}) < 2^{−n}.

b) Put f_{n}(x) := m(G_{n} ∩ [0, x]). Then f_{n} is increasing, continuous, nonnegative, and satisfying f_{n}(x) < 2^{−n}.

c) A function f(x) := ∑_{n=1}^{∞} f_{n}(x) is also increasing, continuous, nonnegative.

d) Let x_{0} ∈ E. Then [x_{0}, x_{0} + h] ⊂ G_{n} for some n fixed, provided |h| is small. Prove that f_{n}(x_{0} + h) = f_{n}(x_{0}) + h.

e) Use d) to conclude that ∀k ∈ N and |h| small we have

\frac{f(x_{0} + h) − f(x_{0})}{h} ≥ ∑_{n=1}^{k} \frac{f_{n}(x_{0} + h) − f_{n}(x_{0})}{h} = k.