Problem 65 - Thomas Calculus

Question

A fixed point theorem Suppose that a function \(f\) is continuous on the closed interval \([0,1]\) and that \(0 \leq f(x) \leq 1\) for every \(x\) in \([0,1] .\) Show that there must exist a number \(c\) in \([0,1]\) such that \(f(c)=c(c\) is called a fixed point of \(f)\)

Step-by-Step Solution

Answer

A continuous function on \([0,1]\) has a fixed point \(c\) where \(f(c) = c\).

Step 1: Understanding the Problem

We need to show that a continuous function \( f \) that maps the interval \([0,1]\) into itself has a fixed point \( c \). A fixed point \( c \) is where \( f(c) = c \). This is known as a fixed point theorem.

Step 2: Define a New Function

Let's define a new function \( g(x) = f(x) - x \). We aim to show that \( g(x) = 0 \) for some \( x \) in \([0,1]\), equivalent to \( f(x) = x \).

Step 3: Evaluate Endpoints of the Interval

Calculate \( g(0) = f(0) - 0 \) and \( g(1) = f(1) - 1 \). Since \( 0 \leq f(x) \leq 1 \) for all \( x \), we know that \( g(0) = f(0) \geq 0 \) and \( g(1) = f(1) - 1 \leq 0 \).

Step 4: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if \( g(x) \) is continuous on \([0,1]\) and \( g(0) > 0 \) while \( g(1) < 0 \), or vice versa, there must be some point \( c \) in \([0,1]\) where \( g(c) = 0 \). Therefore, since \( g(x) = f(x) - x \) is continuous, there is some \( c \) such that \( f(c) = c \).

Step 5: Conclusion

Thus, there exists a point \( c \in [0,1] \) where \( f(c) = c \), proving that \( f \) has a fixed point in the interval.

Key Concepts

Continuous Function

In mathematics, a continuous function is one that smoothly evolves without any sudden jumps or breaks. Think of it as a gentle curve drawn on a piece of paper without lifting your pencil.
A function is continuous on an interval if its graph doesn't have holes or vertical asymptotes within that interval.
In the case of the fixed point theorem, the function \( f \) is continuous on the closed interval \([0,1]\).
Here are some essential characteristics of continuous functions:

• Every point within the interval has a well-defined value.
• The approach to a point from either direction (left or right) yields the same function value.
• They can be drawn without lifting the pencil (metaphorically speaking).

In our particular exercise, the continuity of the function \( f \) on \([0,1]\) is crucial because it allows us to use powerful tools like the Intermediate Value Theorem.

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a pivotal concept in calculus and analysis. It states that for any continuous function, if it takes on two different values at points \( a \) and \( b \) within an interval, then it must also take on any value between those two values at some point within that interval.
The theorem is essential because it gives us insight into the behavior of continuous functions and their ranges.For our fixed point problem, suppose we define a new function \( g(x) = f(x) - x \).
By analyzing \( g(x) \), we use IVT as follows:

• Evaluate at the endpoints: \( g(0) = f(0) \) and \( g(1) = f(1) - 1 \).
• If \( g(0) \) is positive and \( g(1) \) is negative (or vice versa), then there exists a \( c\) in \([0,1]\) such that \( g(c) = 0 \).

This conclusion, \( f(c) = c \), is precisely the fixed point we are trying to find through the exercise.

Closed Interval

A closed interval, denoted as \([a, b]\), includes all the numbers between \(a\) and \(b\), including the endpoints \(a\) and \(b\) themselves.
This concept is fundamental in mathematical analysis because closed intervals have properties that open intervals don't.Why closed intervals matter in the fixed point theorem:

• The inclusion of endpoints ensures the continuity argument applies throughout the domain, including the end points \(0\) and \(1\).
• Closed intervals help maintain boundary conditions, which are critical when applying the Intermediate Value Theorem.
• If the interval were open, the continuity test at the endpoints would not hold, prohibiting the application of IVT in such an intuitive manner.

Overall, in problems involving continuous functions and the Intermediate Value Theorem, closed intervals provide a robust framework for analysis.