Problem 1 - Basic Analysis: Introduction to Real Analysis

Question

Suppose \(f:[0,1] \rightarrow \mathbb{R}\) is monotone. Prove \(f\) is bounded.

Step-by-Step Solution

Answer

Since \( f \) is monotone, it is bounded by its endpoint values.

Step 1: Understanding the Problem

The problem states that a function \( f \) is defined from the interval \([0, 1]\) to the real numbers \( \mathbb{R} \) and it is monotone. We need to prove that \( f \) is bounded on this interval.

Step 2: Define Monotonic Functions

A monotone function is either non-decreasing or non-increasing. A non-decreasing function satisfies \( f(x_1) \leq f(x_2) \) for all \( x_1 \leq x_2 \), and a non-increasing function satisfies \( f(x_1) \geq f(x_2) \) for all \( x_1 \leq x_2 \). We will prove the boundedness for both cases.

Step 3: Consider Non-decreasing Function

If \( f \) is non-decreasing, then \( f(0) \leq f(x) \leq f(1) \) for any \( x \in [0, 1] \). Thus, \( f \) is bounded below by \( f(0) \) and bounded above by \( f(1) \).

Step 4: Consider Non-increasing Function

If \( f \) is non-increasing, then \( f(0) \geq f(x) \geq f(1) \) for any \( x \in [0, 1] \). Here, \( f \) is bounded below by \( f(1) \) and bounded above by \( f(0) \).

Step 5: Conclusion

Since \( f \) is monotone, in either the non-decreasing or non-increasing scenario, it is bounded by its values at the endpoints \( f(0) \) and \( f(1) \). Therefore, \( f \) is bounded on \([0, 1]\).

Key Concepts

Boundedness

To understand boundedness, imagine that a function's values are like the swings of a pendulum within fixed walls. The idea of boundedness ensures these swings never move beyond certain limits. Mathematically speaking, a function \( f \) is bounded if there are real numbers \( M \) and \( N \) such that for all \( x \) in the domain, \( N \leq f(x) \leq M \).
A bounded function doesn't go off to infinity or drop to negative infinity.

• The highest point \( M \) is the upper bound.
• The lowest point \( N \) is the lower bound.

In the specific example where a monotone function is defined from \([0, 1]\) to \( \mathbb{R} \), proving boundedness involves observing its values at the endpoints \( f(0) \) and \( f(1) \). Because monotone functions, whether increasing or decreasing, don't reverse their direction, this ensures \( f \) remains confined between \( f(0) \) and \( f(1) \).

Non-decreasing Function

A non-decreasing function is one that never goes down as it progresses. In simpler terms, it either stays flat or goes up. Mathematically, if for all \( x_1 \leq x_2 \), we have \( f(x_1) \leq f(x_2) \), then the function is non-decreasing. This is like walking up a hill or moving across flat ground. You never lose altitude, but you can gain.

• At \( x = 0 \), you start with a particular value \( f(0) \).
• As you move to \( x = 1 \), your values never decrease below \( f(0) \).
• \( f(0) \leq f(x) \leq f(1) \) across the entire interval.

This inherent nature ensures that on a closed interval like [0, 1], the function is bounded between its values at \( f(0) \) and \( f(1) \). Thus, the highest possible value at the endpoint serves as a natural cap, while the lowest acts as the floor.

Non-increasing Function

Non-increasing functions operate oppositely to non-decreasing functions. They either maintain their level or decrease, but never rise. In this case, for any two points such that \( x_1 \leq x_2 \), \( f(x_1) \geq f(x_2) \) holds true. It's akin to sliding down a gentle slope or traveling over level ground without ever climbing.

• At \( x = 0 \), the function may start at \( f(0) \).
• Moving towards \( x = 1 \), values are constrained by \( f(1) \) as the lower potential limit.
• The condition \( f(1) \leq f(x) \leq f(0) \) defines its bounds on \([0, 1]\).

This natural tendency keeps the function from exceeding its initial value at \( x = 0 \) and assures it doesn't dip lower than \( f(1) \), ensuring boundedness by its endpoint values throughout the interval.