Chapter 3: Problem 1
Suppose \(f:[0,1] \rightarrow \mathbb{R}\) is monotone. Prove \(f\) is bounded.
Short Answer
Expert verified
Since \( f \) is monotone, it is bounded by its endpoint values.
Step by step solution
01
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.
02
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.
03
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 \( f(1) \).
04
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 \( f(0) \).
05
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 \([0, 1]\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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 \([0, 1]\).