Extreme Value Theorem
If \(I\) is a closed bounded interval and \(f:I \rightarrow \mathbb{R}\) is continuous on \(I\), then \(f\) is bounded on \(I\). Moreover, if $$ \begin{align*} M = \sup_{x \in I} f(x) \quad \text{ and } \quad m = \inf_{x \in I} f(x), \end{align*} $$ then there exists points \(x_m, x_m \in I\) such that $$ \begin{align*} f(x_M) = M \quad \text{ and } \quad f(x_m) = m. \end{align*} $$

Proof

Suppose for the sake of contradiction that \(f\) is unbounded on \(I\). Then by definition of unbounded, for every \(M > 0 \in \mathbb{R}\), we can find some \(x \in I\) such that \(|f(x)| > M\). In particular (if we discretize this), then for every natural number \(n \in \mathbb{N}\), we can find a point \(x_n \in I\) such that

$$ \begin{align*} |f(x_n)| > n \quad \quad \quad (1) \end{align*} $$

Thus, we obtain a sequence \(\{x_n\}\) such that \(|f(x_n)| \to \infty\) as \(n \to \infty\). But recall that \(\{x_n\} \in I\) and we’re given that \(I\) is bounded. Hence, by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence \(x_{n_k}\) such that

$$ \begin{align*} x_{n_k} \to x_0 \quad \text{ as } \quad k \to \infty \end{align*} $$

But note that \(I\) is a closed interval so \(a \leq x_{n_k} \leq b\) for all \(x_{n_k}\). Hence by the Comparison Theorem, we must have that

$$ \begin{align*} a \leq x_0 \leq b \end{align*} $$

But we are also given that \(f\) is continuous on \(I\) so \(f\) is continuous at \(x_0\). Thus, since \(x_{n_k} \to x_0\) and \(x_{n_k} \in I\), then by the Sequential Characterization of Continuity Theorem

$$ \begin{align*} f(x_{n_k}) \to f(x_0) \quad \text{ as } \quad k \to \infty \end{align*} $$

which also implies that

$$ \begin{align*} |f(x_{n_k})| \to |f(x_0)| \quad \text{ as } \quad k \to \infty \end{align*} $$

But this means by definition that when taking \(\epsilon = 1\), that there exists a \(K_0 \in \mathbb{N}\) such that for all \(k \geq K_0\)

$$ \begin{align} |f(x_{n_k})| - |f(x_0)| &\leq |f(x_{n_k}) - f(x_0)| < 1 \end{align} $$

This implies

$$ \begin{align} |f(x_{n_k})| \leq |f(x_0)| + 1 \quad \text{ for all } k \geq K_0 \quad \quad \quad (2) \end{align} $$

So for all sufficiently large \(k\), \(|f(x_{n_k})|\) is bounded above by \(|f(x_0)| + 1\). However, by (1), we have

$$ \begin{align*} |f(x_{n_k})| > n_k \quad \text{ for all } k \end{align*} $$

This means that as \(n_k \to \infty\), then

$$ \begin{align*} |f(x_{n_k})| \to \infty \end{align*} $$

This is a contradiction since by (2), we know that \(|f(x_{n_k})|\) is bounded above for all \(k > K_0\). Thus, \(f\) must be bounded on \(I\).

Now since \(f\) is bounded on \(I\), then by the completeness axiom, \(\sup_{x \in I} f(x)\) exists. Let \(M = \sup_{x \in I} f(x)\). By definition of the supremum, this means that for all \(n \in \mathbb{N}\), there exists an \(x_n \in I\) such that

$$ \begin{align*} M - \frac{1}{n} < f(x_n) \leq M \quad \quad \quad (3) \end{align*} $$

Since \(I\) is closed and bounded, the sequence \(\{x_n\} \subset I\) is also bounded. Hence, by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence \(x_{n_k} \in I\) and a point \(x_M \in I\) such that

$$ \begin{align*} x_{n_k} \to x_{M} \quad \text{ as } \quad k \to \infty \end{align*} $$

But \(f\) is continuous on \(I\). Then by the Sequential Characterization of Continuity Theorem

$$ \begin{align*} f(x_{n_k}) \to f(x_M) \end{align*} $$

If we take the limit of (3), then note that as \(k \to \infty\), \(n_k \to \infty\) and thus \(\frac{1}{n_k} \to 0\). Then the left hand side approaches \(M\). The middle term approaches \(f(x_M)\) and the right hand side approaches \(M\) since it’s a constant. Hence

$$ \begin{align*} M \leq f(x_M) \leq M \end{align*} $$

Thus, by the Squeeze Theorem,

$$ \begin{align*} f(x_M) = M = \sup_{x \in I} f(x). \ \blacksquare \end{align*} $$

References