# 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 $x \in \closedint a b$:

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

We have that:

- $\N \subset \R$

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

Consider the sequence:

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

and the sequence:

- $\sequence N_{N \mathop \in \N}$

Because:

- $\map f {\sequence {x_n} } > N$

and:

- $\sequence N_{N \mathop \in \N} \to \infty$

it follows that:

- $\sequence {\map f {\sequence {x_n} } }_{n \mathop \in \N} > \infty \implies \sequence {\map f {\sequence {x_n} } }_{n \mathop \in \N} \to \infty$

Consider the sequence:

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

It is bounded: bounded above by $b$ and bounded below by $a$.

So by the Bolzano-Weierstrass Theorem there exists a subsequence of $\sequence {x_n}$:

- $\sequence {g_n} \to d$

Since $\closedint a b$ is closed:

- $d \in \closedint a b$

By Continuity of Mapping between Metric Spaces by Convergent Sequence, then:

- $\map f x$ is continuous in $d$

and:

- $\map f {\sequence {g_n} } = \map f d$

But our first conlusion indicates that:

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

which contradicts:

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

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 $\closedint a b$.

So, by Continuity of Mapping between Metric Spaces by Convergent Sequence:

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

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.

## Sources

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.33$: Weierstrass ($\text {1815}$ – $\text {1897}$)