Question

Let \(I=[0,1]\) be the closed unit interval. Suppose \(f\) is a continuous mapping of \(I\) into \(I\). Prove that \(f(x)=x\) for at least one \(x \in I\).

Short answer

Question: Show that for a continuous function f mapping the closed unit interval I into itself, there exists at least one x in I such that f(x) = x. Answer: By applying the Intermediate Value Theorem to a new function g(x) = f(x) - x, we showed that there must exist a point x in I such that g(x) = 0, which implies that f(x) = x for at least one x in I.

Step-by-step solution

Define a new function g

Define a new function \(g: I \rightarrow \mathbb{R}\) by setting \(g(x) = f(x) - x\) for each \(x \in I\).

Determine the endpoints values of g

Let us evaluate \(g\) at endpoints of the interval \(I\): - For \(x = 0\), we have \(g(0) = f(0) - 0 = f(0)\). Since \(f\) maps \(I\) into itself, \(f(0) \in I\). Therefore, \(g(0) \geq 0\). - For \(x = 1\), we have \(g(1) = f(1) - 1\). Again, since \(f\) maps \(I\) into itself, \(f(1) \in I\). Therefore, \(g(1) \leq 0\).

Show that g is continuous

Since \(f\) is continuous and \(x\) is continuous on \(I\), their difference, \(g(x) = f(x) - x\), is also continuous on \(I\).

Apply the Intermediate Value Theorem to g

Now, we know that \(g(0) \geq 0\) and \(g(1) \leq 0\), so the function \(g\) achieves values greater than or equal to 0 and less than or equal to 0 in the interval \(I\). Since \(g\) is continuous on \(I\), by the Intermediate Value Theorem, there must exist a point \(x \in I\) such that \(g(x) = 0\).

Conclude that f(x) = x for at least one x in I

Finally, if there exists an \(x \in I\) such that \(g(x) = 0\), then \(f(x) - x = 0 \Rightarrow f(x) = x\) for at least one \(x \in I\). This proves the claim.

Key concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental principle in calculus and real analysis. It states that for any continuous function \( f \) that is defined on a closed interval \([a, b]\), if \( f(a) \) and \( f(b) \) have different signs, then there is at least one point \( c \) in the interval \( (a, b) \) where \( f(c) = 0 \). This means the function must cross the x-axis at least once.

When you apply the IVT to our exercise, we look at a new function \( g(x) = f(x) - x \). It's continuous and evaluates to different signs at the endpoints \( x = 0 \) and \( x = 1 \). Therefore, by the IVT, \( g \) must have a root in the interval \([0, 1]\). This is what tells us there exists an \( x \) such that \( f(x) = x \). The IVT is powerful because it guarantees the existence of a solution without actually finding the solution itself.

Continuous Functions

Continuous functions are those that have no breaks, jumps, or holes in their graphs. Mathematically, a function \( f \) is continuous at a point \( a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \). In simpler terms, you can draw the curve of a continuous function without lifting your pen from the paper.

For our exercise, the function \( f \) is continuous over the closed unit interval \([0, 1]\). This is crucial because the Intermediate Value Theorem, which we use to prove the existence of a fixed point, only applies to continuous functions. If \( f \) had discontinuities, we could not apply the IVT, and the problem would become much more complex, possibly without such a simple solution.

Closed Unit Interval

The closed unit interval \([0, 1]\) is a fundamental concept in real analysis. It consists of all real numbers \( x \) such that \( 0 \leq x \leq 1 \). It includes both endpoints and forms a compact set, which means it is both closed and bounded. These properties are essential in the context of functions, particularly when applying the Intermediate Value Theorem.

Using \([0, 1]\), we know exactly where our function \( f \) is evaluated, including the critical endpoints. The bounded nature of the interval ensures that \( f \) is mapped within this interval, as stipulated by the problem. The fact that \([0, 1]\) is closed helps us apply certain theorems and guarantees, such as achieving minimum and maximum values within the interval, which are key in demonstrating the function's behavior as described in our proof.