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.
The Maximum-Minimum Theorem
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 $s_* = f(x_*) ≤ f(x)$. $\blacksquare$