Extreme Value Theorem Proof

To prove the extreme value theorem, we use the contradiction and boundedness theorem. We will prove the function f has its maximum value in the interval [a, b]. Similarly, the function f has its minimum value in the interval [a, b] can be proved.

f is continuous on [a, b] means f is bounded on [a, b] such that there exists m and M i.e., m ≤ f(x) ≤ M by the Boundedness theorem.

Let M be the least upper bound of f(x) and at point, c lies in [a, b] i.e., f(c) = M, f(x) attains its maximum value on the interval [a, b]. Hence proved.

Let there is no such value c in [a, b] then, f(x) < M ∀x in [a, b].

A function g(x) is defined as g(x) = 1 / [M – f(x)] on [a, b]. We know that g(x) > 0 as f(x) < M ∀x in [a, b] and g is also continuous on [a, b].

By the boundedness theorem g(x) is also bounded on [a, b] which implies there exists k > 0 such that g(x) ≤ k, ∀x in [a, b].

⇒ 1 / [M – f(x)] ≤ k

⇒ M – f(x) ≥ 1 / k

Adding f(x) – (1 / k) on both sides of the inequality

⇒ M – (1/k) ≥ f(x)

⇒ f(x) ≤ M – (1/k)

The above expression contradicts that M is the least upper bound of f(x). Therefore, the assumption that there exists no such c in [a, b] so that f(c) = M is incorrect. Hence, f has its maximum on [a, b].

Extreme value theorem proves the existence of the maximum and minimum value of the function if the function is continuous in the closed interval [a, b]. In this article, we will discuss the extreme value theorem in depth along with the extreme value theorem statement, extreme value theorem proof and how to use the extreme value theorem. We will also solve some examples related to the extreme value theorem.

Let’s start our learning on the topic “Extreme Value Theorem.”