Question

(a) Use the Mean Value theorem to prove |sin x-sin y| ≤ |x-y| for all x, y in ℝ; see the proof of Theorem 19.6.

(b) Show sin x is uniformly continuous on ℝ.

Answer 1

To prove |sin x-sin y| ≤ |x-y| for all x, y in ℝ using the Mean Value theorem, we first define a function f(x) = sin(x) and apply the Mean Value theorem to f(x) on the interval [x,y]. Then, we substitute the result into the inequality and conclude |sin(x) - sin(y)| ≤ |x - y| for all x, y in ℝ. Additionally, we show that sin(x) is uniformly continuous on ℝ by proving that for any ε > 0, there exists a δ > 0 such that |x - y| < δ implies |sin(x) - sin(y)| < ε for all x, y in ℝ.

Step-by-step explanation:

To prove |sin(x) - sin(y)| ≤ |x - y| for all x, y in ℝ using the Mean Value theorem, we will first define a function f(x) = sin(x). We will then apply the Mean Value theorem to the function f(x) on the interval [x,y].

1. Apply the Mean Value theorem to f(x) on the interval [x,y]. This will give us a value c between x and y where f'(c) = (f(x) - f(y))/(x - y).
2. Since f'(c) = cos(c) and cos(c) is between -1 and 1 for all c in ℝ, we can conclude that |f'(c)| ≤ 1.
3. Substituting f'(c) = (f(x) - f(y))/(x - y) into the inequality from step 2, we have |sin(x) - sin(y)|/|x - y| ≤ 1.
4. Thus, we can conclude that |sin(x) - sin(y)| ≤ |x - y| for all x, y in ℝ.

To show that sin(x) is uniformly continuous on ℝ, we need to prove that for any ε > 0, there exists a δ > 0 such that |x - y| < δ implies |sin(x) - sin(y)| < ε for all x, y in ℝ.

1. Let ε > 0 be given.
2. Choose δ = ε.
3. Then, for any x, y in ℝ such that |x - y| < δ, we have |sin(x) - sin(y)| ≤ |x - y| < ε.

Therefore, sin(x) is uniformly continuous on ℝ.