Question

Which of the following real-valued functions is/are uniformly continuous on \([0, \infty)\)?

• A. \( \sin^2 x \)
• B. \( x \sin x \)
• C. \( \sin(\sin x) \)
• D. \( \sin(x \sin x) \)

Hint:

Bounded and Lipschitz continuous functions on the entire real line (or any interval) are always uniformly continuous. Unbounded oscillatory functions like \( x \sin x \) fail this property as their derivatives grow without bound.

Answer 1

The Correct Option is A, C

Step 1: Understanding uniform continuity.
A function \( f(x) \) is said to be uniformly continuous on an interval if, for every \( \epsilon>0 \), there exists a \( \delta>0 \) such that for all \( x_1, x_2 \in [0, \infty) \), if \( |x_1 - x_2|<\delta \), then \( |f(x_1) - f(x_2)|<\epsilon \). Intuitively, uniform continuity means that the rate of change of the function does not become unbounded as \( x \to \infty \).

Step 2: Analyze each function.
(A) \( f(x) = \sin^2 x \): Since \( \sin x \) is bounded and continuous for all real \( x \), and squaring it does not affect its periodic bounded nature, \( \sin^2 x \) oscillates between 0 and 1. Furthermore, \( \sin^2 x \) has bounded derivative \( f'(x) = 2 \sin x \cos x = \sin 2x \), which implies that \( f(x) \) is Lipschitz continuous. Hence, \( \sin^2 x \) is uniformly continuous on \([0, \infty)\).
(B) \( f(x) = x \sin x \): Here, the term \( x \) increases without bound as \( x \to \infty \). Although \( \sin x \) oscillates between -1 and 1, the product \( x \sin x \) has amplitude that grows without bound. The derivative \( f'(x) = \sin x + x \cos x \) is unbounded, implying that small changes in \( x \) can lead to arbitrarily large changes in \( f(x) \). Therefore, \( x \sin x \) is not uniformly continuous on \([0, \infty)\).
(C) \( f(x) = \sin(\sin x) \): Since \( \sin x \) is bounded between \([-1, 1]\), the inner \( \sin x \) ensures that the input to the outer sine function is also bounded within this range. Thus, \( f(x) = \sin(\sin x) \) remains bounded and continuous for all \( x \). As both sine functions are Lipschitz continuous with constant 1, their composition \( \sin(\sin x) \) is also Lipschitz continuous and therefore uniformly continuous on \([0, \infty)\).
(D) \( f(x) = \sin(x \sin x) \): In this case, \( x \sin x \) is unbounded as \( x \to \infty \), which means the argument of the sine function increases indefinitely. The sine function remains bounded, but due to rapid oscillations of \( x \sin x \), \( \sin(x \sin x) \) does not have a uniformly small variation over large \( x \). Hence, \( \sin(x \sin x) \) is not uniformly continuous.

Step 3: Conclusion.
From the above analysis, we conclude that functions (A) \( \sin^2 x \) and (C) \( \sin(\sin x) \) are uniformly continuous on \([0, \infty)\).