The Maximum Minimum Theorem

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

We have just looked at the definition of the Absolute Maximum and Absolute Minimum of a function $f : A \to \mathbb{R}$. We have seen that $f$ may contain both an absolute maximum and an absolute minimum, one or the other, or neither. We will now look at an important theorem known as The Maximum-Minimum Theorem and sometimes as The Extreme Value Theorem which says that if we have a continuous function $f$ from a closed and bounded interval $I$ into $\mathbb{R}$, then we are guaranteed to have both an absolute maximum and absolute minimum value.
 Theorem 1 (Maximum-Minimum): If $f : I \to \mathbb{R}$ is continuous function, and $I = [a, b]$ is a closed and bounded interval, then $f$ contains both an absolute maximum and an absolute minimum on $I$. • Proof: Let $I = [a, b]$ be a closed and bounded interval and $f : I \to \mathbb{R}$ be a continuous function. Then by the Boundedness Theorem, $f$ is bounded on $I$. Therefore the set $f(I) = \{ f(x) : x \in I \}$ is bounded. Let $s^* = \sup f(I)$ and $s_* = \inf f(I)$. We will first show that $\exists x^* \in I$ producing an absolute maximum $f(x^*)$ where $s^* = f(x^*)$.
• Now since $s^* = \sup f(I)$ then for $n \in \mathbb{N}$ we have that $s^* - \frac{1}{n}$ is not an upper bound of the set $f(I)$. Therefore $\forall n \in \mathbb{N}$, $\exists x_n \in I$ such that $s^* - \frac{1}{n} < f(x_n) ≤ s^*$. Now since $I = [a, b]$ is a bounded set and $x_n \in I$ for all $n \in N$, we have that the sequence $(x_n)$ is also bounded. By The Bolzano-Weierstrass Theorem, since $(x_n)$ is a bounded sequence, then there exists a convergent subsequence $(x_{n_k})$ that converges to some $x^*$, that is $\lim_{k \to \infty} x_{n_k} = x^*$. Also, since $x_{n_k} \in I$ for all $n_k \in \mathbb{N}$ we have that $x^* \in I$. Since $x^* \in I$ and $f : I \to \mathbb{R}$ is a continuous function, then $f$ is continuous at $x^*$. Therefore by the Sequential Criterion for the Continuity of a Function, since $\lim_{k \to \infty} (x_{n_k}) = x^*$ then we have $\lim_{k \to \infty} f(x_{n_k}) = f(x^*)$.
• Now since $s^* - \frac{1}{n} < f(x_n) ≤ s^*$ for all $n \in \mathbb{N}$, then furthermore $s^* - \frac{1}{n_k} < f(x_{n_k}) ≤ s^*$ for all $k \in \mathbb{N}$. Since $\lim_{k \to \infty} s^* - \frac{1}{n_k} = s^*$ and $\lim_{k \to \infty} s^* = s^*$, then by the Squeeze Theorem we have that $\lim_{k \to \infty} f(x_{n_k}) = s^*$. Therefore $x^* \in I$ produces the absolute maximum $f(x^*) = s^*$ since for all $x \in I$ we have that $s^* = f(x^*) ≥ f(x)$.
• We will now show that $\exists x_* \in I$ producing an absolute minimum $f(x_*)$ where $s_* = f(x_*)$.
• Since $s_* = \inf f(I)$ then for $n \in \mathbb{N}$ we have that $s_* + \frac{1}{n}$ is not a lower bound to the set $f(I)$. Therefore $\forall n \in \mathbb{N}$, $\exists x_n \in I$ such that $s_* ≤ f(x_n) < s_* + \frac{1}{n}$. Now since $I = [a, b]$ is a bounded set and $x_n \in I$ for all $n \in \mathbb{N}$, we have that the sequence $(x_n)$ is also bounded. Once again by The Bolzano Weierstrass Theorem, since $(x_n)$ is a bounded sequence, then there exists a convergent subsequence $(x_{n_k})$ that converges to some $x_*$, that is $\lim_{k \to \infty} x_{n_k} = x_*$. Also, since $x_{n_k} \in I$ for all $n_k \in \mathbb{N}$ we have that $x_* \in I$. Since $x_* \in I$ and $f : I \to \mathbb{R}$ is a continuous function, then $f$ is continuous at $x_*$. Once again by the Sequential Criterion for Continuity of a Function, since $\lim_{k \to \infty} (x_{n_k}) = x_*$ then we have $\lim_{k \to \infty} f(x_{n_k}) = f(x_*)$.
• Now since $s_* ≤ f(x_n) < s_* + \frac{1}{n}$ for all $n \in \mathbb{N}$, then furthermore $s_* ≤ f(x_{n_k}) < s_* + \frac{1}{n_k}$ for all $k \in \mathbb{N}$. Since $\lim_{k \to \infty} s_* + \frac{1}{n_k} = s_*$ and $\lim_{k \to \infty} s_* = s_*$, then by the Squeeze Theorem we have that $\lim_{k \to \infty} f(x_{n_k}) = s_*$. Therefore $x_* \in I$ produces the absolute minimum $f(x_*) = s_*$ since for all $x \in I$ we have that $s_* = f(x_*) ≤ f(x)$. $\blacksquare$