# Extreme Value Theorem/Real Function

## Theorem

Let $f$ be a real function which is continuous in a closed real interval $\closedint a b$.

Then:

$\forall x \in \closedint a b: \exists x_n \in \closedint a b: \map f {x_n} \le \map f x$
$\forall x \in \closedint a b: \exists x_n \in \closedint a b: \map f {x_n} \ge \map f x$

## Proof

First it is shown that $\map f x$ is bounded in the closed real interval $\closedint a b$.

Aiming for a contradiction, suppose $\map f x$ has no upper bound.

Then for all $x \in \closedint a b$:

$\forall N \in \R: \exists x_n \in \closedint a b: \map f {x_n} > N$

We have that:

$\N \subset \R$

Without loss of generality, we can consider $N \in \N$.

Consider the sequence:

$\sequence {\map f {x_n} }_{n \mathop \in \N}$

which by definition satisfies:

$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \infty$

Consider the sequence:

$\sequence {x_n} \in \closedint a b$

$\sequence {x_n}$ is bounded:

bounded above by $b$
bounded below by $a$.

By the Bolzano-Weierstrass Theorem there exists a subsequence of $\sequence {x_n}$, call it $\sequence {g_n}$, which converges to a point in $\closedint a b$.

So, let $g_n \to d$ with $d \in \closedint a b$.

We have by hypothesis that $f$ is continuous on $\closedint a b$

$\ds \lim_{n \mathop \to \infty} \map f {g_n} = \map f d$

But our first conclusion indicates that:

$\ds \sequence {\map f {x_n} }_{n \mathop \in \N} \to \infty$
$\ds \lim_{n \mathop \to \infty} \sequence {\map f {g_n} } = \ds \lim_{n \mathop \to \infty} \sequence {\map f {x_n} }$

$\sequence {\map f {g_n} }_{n \mathop \in \N} \to \map f d$

which is necessarily finite.

A similar argument can be used to prove the lower bound.

$\Box$

It remains to be proved that:

$\exists d \in \closedint a b: \map f x \le \map f d$

where $\map f d = N$ (the maximum).

It will be shown that:

$\forall n \in \R_{\ne 0}: N - 1/n < \map f {\sequence {x_n} } \le N$

as follows:

Without loss of generality, consider $n \in \N$.

Let $I$ denote the codomain of $f \closedint a b$.

Because $N$ is its maximum and $N - 1/n < N$:

$\forall n \in \N: \exists y_n \in \closedint a b: N - 1/n < y_n < N$

But $y_n \in I$, so:

$\forall n \in \N: \exists x_n \in \closedint a b$

That means:

$\map f {\sequence {x_n} } = y_n \implies N - 1/n < \map f {x_n} \le \ N$

$\Box$

It follows that:

$\sequence {\map f {\sequence {x_n} } } \to N$

Considering:

$\sequence {N - 1 / n} \to \ N$ as $n \to \infty$

and:

$\forall n \in \N: \sequence N \to N$

by:

$\sequence {N - 1 / n} < \sequence {\map f {\sequence {x_n} } } \le N \implies \sequence {\map f {\sequence {x_n} } } \to \sequence N$

Consider $\sequence {x_n}$.

Because it is bounded, by Bolzano-Weierstrass Theorem there exists a subsequence:

$\sequence {s_n}$

that converges.

Let it converge to $l$.

Because $\sequence {s_n} \in \closedint a b$ it follows that:

$l \in \closedint a b$.

Finally, $\map f x$ is continuous in $\closedint a b$.

$\sequence {s_n} \to d \implies \map f {\sequence {s_n} } \to \map f d$

But:

$\sequence {\map f {x_n} } \to N \implies \sequence {\map f {s_n} }_{n \mathop \in \N} \to N \iff \map f d = N$

$\blacksquare$

## Historical Note

The Extreme Value Theorem in its application to real functions is usually attributed to Karl Weierstrass, as an example of what has been referred to as Weierstrassian rigor.

Hence this result's soubriquet the Weierstrass Extreme Value Theorem.