Problem Set #5 Part C Official Solutions

Inna Zakharevich

October 31, 2003

(10) 1. First let k = min{f(0), f(1)}, and let x ∈ [0, 1] be such that f(x) = 0. Then by the intermediate value theorem, there exists x_{1} ∈ [0, x] and x_{2} ∈ [x, 1] such that f(x_{1}) = k/2 and f(x_{2}) = k/2. However, since f is a bijection we know that k_{1} = k_{2}. The only way that this is possible is if k = 0, so either f(0) = 0 or f(1) = 0. A similar argument with k = max{f(0), f(1)} shows that f(0) = 1 or f(1) = 1.

First assume that f(0) = 0. We will show that f is strictly increasing by contradiction. Suppose that there exist x, y ∈ [0, 1] such that x < y but f(x) ≥ f(y). Then by the intermediate value theorem there exist x_{1} ∈ [0, x] and x_{2} ∈ [x, y] such that f(x_{1}) = (f(x) + f(y))/2 and f(x_{2}) = (f(x) + f(y))/2. However, this contradicts the fact that f is a bijection. Thus such x, y do not exist, so f must be strictly increasing.

Now suppose that f(0) = 1. Let g(x) = f(1 − x). Then g is a continuous bijection with g(0) = 0, so it is strictly increasing. Thus f is strictly decreasing. So we are done.

(12) 2.

(4) (1) First we will show that f is surjective. Let x ∈ [0, 1], and let y = f^{n_{x}−1}(x). Clearly, y ∈ [0, 1]. Then, by definition f(y) = x, so f is surjective. Now we will show that f is injective. Suppose that f(x) = f(y). Then f^{n_{x}}(x) = x, so f^{n_{x}a}(x) = x for all a ∈ N. Similarly, f^{n_{y}a} = y. Thus f^{n_{x}n_{y}}(x) = x and f^{n_{x}n_{y}}(y) = y so x = y. Thus f is injective. Since f is injective and surjective it is a bijection.

(8) (2) We will show that f(x) = x. If f is continuous and f(0) < 1 we know (by the previous problem) that f(0) = 0 and f(1) = 1 and f is strictly increasing. Now suppose that there is some x ∈ [0, 1] such that f(x) < x. Consider the sequence {x, f(x), f^{2}(x), …, f^{n_{x}−1}(x), x, …}. We know that there must be some element y of this sequence such that f(y) > y (otherwise it would be a strictly decreasing, which would be a contradiction). Consider the first such element in the sequence. Thus we have three elements y_{1}, y_{2}, y_{3} such that y_{1} > y_{2} and y_{2} < y_{3} and y_{2} = f(y_{1}) and y_{3} = f(y_{2}). Then we have that y_{2} < y_{1} but f(y_{1}) < f(y_{2}).

This contradicts the fact that f is strictly increasing. Thus there must not exist such an x. Thus for all x ∈ [0, 1], f(x) ≥ x. A similar argument shows that for all x ∈ [0, 1], f(x) ≤ x. Thus f(x) = x.

(18) 3.

(4) (1) Note that if f(x + λ) = f(x) then f(x + nλ) = f(x) for all n ∈ Z. Let α = p/q. Then g(x + qα) = g(x + p) and f(x + p · 1) = f(x). Thus (f + g)(x + p) = f(x + p) + g(x + p) = f(x) + g(x) = (f + g)(x) so f + g is periodic.

(7) (2) We will show that if α ∈ R\Q then for all y ∈ R and all ϵ > 0 there exists x ∈ R such that h(x) = h(0) and |x − y| < ϵ. Since α is irrational we know that there exist p, q ∈ Z such that |qα − p| < ϵ. Let β = qα − p. Then we know that h(β + x) = h(x − β) = h(x). Then we have h(⌊y/β⌋β) = h(0) and |y − β⌊y/β⌋| < β < ϵ. Then, since h is continuous we will have that h(y) = h(0). Thus h is constant.

(7) (3) We have already show that if α ∈ Q then f + g is periodic. Thus all we have left to show is that if α ∉ Q then f +g is not periodic. Suppose that it is. Then there is some λ ∈ R such that (f + g)(λ + x) = (f + g)(x). Thus f(x + λ) − f(x) = g(x)−g(x+λ). Note that this function is periodic with two periods: 1 and α. Thus it is constant. Let a = f(x + λ) − f(x). Then we see that f(x + nλ) = f(x) + na. Let M = ∑_{x∈R} |f| (which exists because f is periodic), and let n be such that na > 2M. Then f(x + nλ) ≤ M ≤ f(x) + na = f(x + nλ), a contradiction. The only way this can be avoided is if a = 0. However, if a = 0 then f and g must both have period λ, which means that λ is both rational and irrational, another contradiction. So f + g cannot be periodic if α ∈ R\Q. So f + g is periodic if and only if α ∈ Q.