Question

Let \(\left\\{r_{n}\right\\}\) be a sequence of all rational numbers in \((0,1)\). Define a function \(f\) on \((0,1)\) by \(f(x)=\sum_{n=1}^{\infty} 2^{-n}\left|x-r_{n}\right| .\) Show that \(f\) is a convex continuous function on \((0,1)\) that is differentiable exactly at irrational points of \((0,1)\) Hint: Use Lemma \(8.3\) to see that \(f\) is not differentiable at rational points. For irrational points, the fraction can be made arbitrarily small, first working with the tail and then with the remaining terms.

Short answer

The function \( f \) is convex and continuous on \((0,1)\) and is differentiable exactly at irrational points in \((0,1)\).

Step-by-step solution

Understanding the Function

The function is defined as follows: \( f(x) = \sum_{n=1}^{\infty} 2^{-n} \left| x - r_{n} \right| \), where \( r_n \) is a sequence of all rational numbers in \((0,1)\). This is a series where each term is weighted by \( 2^{-n} \) and it measures the distance between \( x \) and each rational number \( r_n \) in the sequence.

Prove Continuity

To show that \( f \) is continuous, observe that each term \( 2^{-n} \left| x - r_{n} \right| \) is continuous in \( x \). Since the series \( \sum 2^{-n} \) converges, and it is the product of a convergent geometric series \( 2^{-n} \) and continuous functions \( \left| x - r_n \right| \), \( f(x) \) is the sum of continuous functions, which makes \( f(x) \) continuous.

Prove Convexity

A function \( f \) is convex if for any two points \( x_1 \) and \( x_2 \) and any \( \lambda \in (0,1) \), \( f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda) f(x_2) \). Given that \( f \left( x \right) \) is constructed as a weighted sum of the convex functions \( |x - r_n| \) (absolute value function is convex), \( f \) is a convex function because a non-negative linear combination of convex functions is convex.

Differentiability at Irrational Points

For irrational points, consider the fact: \( f(x) = \sum_{n=1}^{\infty} 2^{-n} \left| x - r_n \right| \). Since for irrationals, the rational sequence \( \left| x - r_n \right| \) does not repeat, the terms in the sequence \( \left| x - r_n \right| \) can be made arbitrarily small by considering the tail terms. Hence, for irrational \( x \in (0,1) \), \( f \) is differentiable.

Non-differentiability at Rational Points

Using Lemma 8.3, we know that \( f(x) \) is not differentiable at rational points. This is because at rational points \( x = r_k \), the series \( \sum 2^{-n}\left| x - r_{n}\right| \) does not behave smoothly due to the distance term \( \left| x - r_k \right| = 0 \). Hence, \( f \) is not differentiable at any rational point.

Key concepts

Convex Function

Convex functions are fundamental in optimization and various fields of mathematics. A function is called convex if, for any two points within its domain, the line segment connecting these points remains above the graph of the function. Mathematically, this is stated as:
For any two points \( x_1 \) and \( x_2 \) and any \( \lambda \in (0,1) \), it must hold that \( f(\lambda x_1 + (1-\lambda)x_2) \leq \lambda f(x_1) + (1-\lambda) f(x_2) \).
The function \( f(x) = \sum_{n=1}^{\infty} 2^{-n} | x - r_n | \) is convex because it is a weighted sum of absolute value functions \( |x - r_n| \), which are all convex. The property of convexity is preserved under non-negative linear combinations.

Differentiability

Differentiability refers to the existence of a derivative for a function at a given point. A function is differentiable at a point if it has both a defined derivative and is smooth (without any sharp corners) at that point. In this exercise, the function \( f(x) \) is defined as \( f(x) = \sum_{n=1}^{\infty} 2^{-n} | x - r_n | \). It has a unique behavior based on whether \( x \) is a rational or an irrational number.

• For irrational points \( x \), the terms \( |x - r_n| \) in the series do not repeat, allowing for them to become arbitrarily small. This results in the function being differentiable at irrational points.
• For rational points \( x = r_k \), the series does not behave smoothly because the distance term \( | x - r_k | \) equals zero for the rational sequence. This lack of smoothness indicates non-differentiability at rational points.

Continuity

A function \( f(x) \) is continuous if small changes in \( x \) result in small changes in \( f(x) \). In other words, there are no sudden jumps or breaks in the function. For the function \( f(x) = \sum_{n=1}^{\infty} 2^{-n} | x - r_n | \), continuity is a key property. Each term \( 2^{-n} | x - r_n | \) is continuous in \( x \), and since the series \( \sum 2^{-n} \) converges, the overall function \( f(x) \) is a sum of continuous functions. Consequently, \( f(x) \) is continuous over the interval \( (0,1) \).

Rational and Irrational Numbers

Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is non-zero. Irrational numbers, on the other hand, cannot be expressed as a simple fraction; they have non-repeating, non-terminating decimal expansions. In this exercise, the function \( f(x) \) differentiates between rational and irrational points in its domain \( (0,1) \).

• At rational points \( x = r_k \), the defined sequence \( \{ r_n \} \) includes these points explicitly, causing the function to behave irregularly, leading to non-differentiability.
• At irrational points, the function \( f(x) \) is smooth as the terms in the sequence \( | x - r_n | \) diminish, making \( f(x) \) differentiable at these points.