Extreme Value Theorem/Real Function

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Let $f$ be continuous in a closed real interval $\left[{a \,.\,.\, b}\right]$.


$\forall x \in \left[{a \,.\,.\, b}\right]: \exists x_s \in \left[{a \,.\,.\, b}\right]: f \left({x_s}\right) \le f \left({x}\right)$
$\forall x \in \left[{a \,.\,.\, b}\right]: \exists x_n \in \left[{a \,.\,.\, b}\right]: f \left({x_n}\right) \ge f \left({x}\right)$


First it is shown that $f \left({x}\right)$ is bounded in the closed real interval $\left[{a \,.\,.\, b}\right]$.

Aiming for a contradiction, suppose $f \left({x}\right)$ has no upper bound.

Then for all $x \in \left[{a \,.\,.\, b}\right]$:

$\forall N \in \R: \exists x_n \in \left[{a \,.\,.\, b}\right]: f \left({\left \langle {x_n} \right \rangle}\right) > N$

We have that:

$\N \subset \R$

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

Consider the sequence:

$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N}$

and the sequence:

$\left \langle {N} \right \rangle_{N \mathop \in \N}$


$f \left({\left \langle {x_n} \right \rangle}\right) > N$


$\left \langle {N} \right \rangle_{N \mathop \in \N} \to \infty$

it follows that:

$\left\langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \mathop \in \N} > \infty \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N} \to \infty$

Consider the sequence:

$\left \langle {x_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$

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

So by the Bolzano-Weierstrass Theorem there exists a subsequence of $\left \langle {x_n} \right \rangle $:

$\left \langle {g_n} \right \rangle \to d$

Since $\left[{a \,.\,.\, b}\right]$ is closed:

$d \in \left[{a \,.\,.\, b}\right]$

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

$f \left({x}\right)$ is continuous in $d$


$f \left({\left \langle {g_n} \right \rangle}\right) = f \left({d}\right)$

But our first conlusion indicates that:

$\left \langle {f \left({x_n}\right)} \right \rangle_{n \in \N} \to \infty$

which contradicts:

$\left \langle {f \left({g_n}\right)} \right \rangle_{n \in \N} \to d$

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


It remains to be proved that:

$\exists d \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) \le f \left({d}\right)$

where $f \left({d}\right) = N$ (the maximum).

It will be shown that:

$\forall n \in \R_{\ne 0}: N - 1/n < f \left({\left \langle {x_n} \right \rangle }\right) \le N$

as follows;

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

Let $I$ denote the codomain of $f \left[{a \,.\,.\, b}\right]$.

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

$\forall n \in \N: \exists y_n \in \left[{a \,.\,.\, b}\right]: N - 1/n < y_n < N$

But $y_n \in I$, so:

$\forall n \in \N: \exists x_n \in \left[{a \,.\,.\, b}\right]$

That means:

$f \left({\left \langle {x_n} \right \rangle}\right) = y_n \implies N - 1/n < f \left({x_n}\right) \le \ N$


It follows that:

$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to N$


$\left \langle {N - 1/n} \right \rangle \to \ N$ as $n \to \infty$


$\forall n \in \N: \left \langle {N} \right \rangle \to N$


$\left \langle {N - 1/n} \right \rangle < \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \le N \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to \left \langle {N} \right \rangle$

Consider $\left \langle {x_n} \right \rangle$.

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

$\left \langle {s_n} \right \rangle$

that converges.

Let it converge to $l$.

Because $\left \langle {s_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$ it follows that:

$l \in \left[{a \,.\,.\, b}\right]$.

Finally, $f \left({x}\right) $ is continuous in $\left[{a \,.\,.\, b}\right]$.

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

$\left \langle {s_n} \right \rangle \to d \implies f \left({\left \langle {s_n} \right \rangle}\right) \to f \left({d}\right)$


$\left \langle {f \left({x_n}\right)} \right \rangle \to N \implies \left \langle {f \left({s_n}\right)} \right \rangle_{n \in \N} \to N \iff f \left({d}\right) = N$


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.