# Weierstrass Extreme Value Theorem

## Theorem

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

Then:

- $\exists x_M: \forall x \in \closedint a b: \map f {x_M} \ge \map f x$

- $\exists x_m: \forall x \in \closedint a b: \map f {x_m} \le \map f x$

## Proof

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We will prove the case for $x_M$, the maximum.

The case for $x_m$, the minimum, is similar.

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

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

Then:

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

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 $\closedint a b$

Hence by Continuity of Mapping between Metric Spaces by Convergent Sequence:

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

But by the Axiom of Archimedes, there exists $N_d \in \N$ such that:

- $N_d > \map f d$

By assumption, for every $N \ge N_d$:

- $x_N \ge N \ge N_d$

Therefore, there are at most $N_d$ terms of $x_N$ satisfying:

- $\size {x_N - \map f d} < N_d - \map f d$

But as $g_n$ is a subsequence of $x_N$, that contradicts the fact that:

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

$\Box$

We have shown that:

- $\set {\map f x : x \in \closedint a b}$

is bounded above.

Therefore, by Least Upper Bound Property:

- $M = \sup_{x \in \closedint a b} \map f x$

It remains to be proved that:

- $\exists x_M \in \closedint a b : \map f {x_M} = M$

We will construct $\sequence {x_n}$ such that:

- $\forall n \in \N_{> 0}: M - \dfrac 1 n < \map f {x_n} \le M$

as follows:

Let $n \in \N_{> 0}$ be arbitrary.

Then, $\dfrac 1 n > 0$.

Therefore:

- $M - \dfrac 1 n < M$

Thus, by Characterizing Property of Supremum of Subset of Real Numbers:

- $\exists y_n \in f \closedint a b: M - \dfrac 1 n < y_n \le M$

That is:

- $\exists x_n \in \closedint a b: M - \dfrac 1 n < \map f {x_n} \le M$

by definition of image.

Therefore, $\sequence {x_n}_{n \in \N_{> 0}}$ satisfies:

- $\forall n \in \N_{> 0}: M - \dfrac 1 n < \map f {x_n} \le M$

$\Box$

For every $n \in \N_{> 0}$:

- $\size {M - x_n} < \dfrac 1 n$

By Sequence of Reciprocals is Null Sequence, it follows that:

- $\ds \lim_{n \mathop \to \infty} x_n = M$

Consider $\sequence {x_n}$.

Because it is bounded as before, 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:

- $\ds \lim_{n \mathop \to \infty} \map f {s_n} = \map f l$

But $\sequence {\map f {s_n}}$ is a subsequence of $\sequence {\map f {x_n}}$.

So, by Limit of Subsequence equals Limit of Real Sequence:

- $\ds \lim_{n \mathop \to \infty} \map f {s_n} = M$

Therefore:

- $\map f l = M$

$\blacksquare$

## Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass.

## 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 ($\text {1815}$ – $\text {1897}$) - 2021: Richard Earl and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(6th ed.) ... (previous) ... (next):**Weierstrass' theorem**