X Sin 1/x Is Absolute Continuous

Proving the Absolute Continuity of x sin(1/x)

This article will delve into the proof demonstrating that the function f(x) = x sin(1/x) for x ≠ 0 and f(0) = 0 is absolutely continuous on any closed interval [a, b]. Understanding absolute continuity is crucial in various areas of analysis, including the study of differentiation and integration. This particular function provides a compelling example of a function that is continuous but not continuously differentiable, highlighting the nuances of absolute continuity.

What is Absolute Continuity?

Before jumping into the proof, let's define absolute continuity. A function f(x) is absolutely continuous on an interval [a, b] if for every ε > 0, there exists a δ > 0 such that for any finite disjoint collection of intervals (xᵢ, yᵢ) in [a, b] with Σᵢ (yᵢ - xᵢ) < δ, we have Σᵢ |f(yᵢ) - f(xᵢ)| < ε. Essentially, this means that the total variation of the function over a set of small intervals can be made arbitrarily small by choosing sufficiently small intervals. This is a stronger condition than simply being continuous.

Understanding the Function f(x) = x sin(1/x)

The function f(x) = x sin(1/x) presents an interesting case. At x = 0, the function is defined as f(0) = 0, making it continuous at x = 0. However, the derivative of f(x) for x ≠ 0 is given by:

f'(x) = sin(1/x) - cos(1/x)/x

Notice that as x approaches 0, the derivative oscillates wildly and does not have a limit. This means f(x) is not continuously differentiable at x=0.

The Proof of Absolute Continuity

To prove the absolute continuity of f(x) on [a, b], we will leverage the fact that f(x) is continuous on [a, b] and its derivative is bounded almost everywhere on this interval.

1. Continuity: We already established that f(x) is continuous on [a, b], including at x = 0. This is a necessary but not sufficient condition for absolute continuity.

2. Bounded Derivative Almost Everywhere: While f'(x) is unbounded near x = 0, it's bounded almost everywhere on [a,b] excluding a single point. The oscillations of the derivative do not impact the absolute continuity. We can use the fact that the function is continuously differentiable on (a, b) excluding the point x=0. On any subinterval that does not include 0, the function is Lipschitz continuous, a stronger condition implying absolute continuity on that subinterval.

3. Applying the Fundamental Theorem of Calculus: Because f(x) is differentiable and the derivative is integrable (although unbounded near 0), we can use the Fundamental Theorem of Calculus to relate the function to its integral:

   ∫ₐᵇ f'(x) dx = f(b) - f(a)

   This relationship, combined with the boundedness of f'(x) almost everywhere, allows us to control the total variation of f(x) over small intervals. This control, as outlined in the definition of absolute continuity, ultimately proves the absolute continuity of f(x) on [a, b].

Conclusion

Therefore, despite the unbounded derivative near x = 0, the function f(x) = x sin(1/x) is absolutely continuous on any closed interval [a, b]. This example demonstrates that absolute continuity is a more relaxed condition than continuous differentiability, allowing for functions with wilder behavior in their derivatives to still possess this important property. This property has significant consequences in the context of integration theory and the study of function spaces.