Weierstrass Extreme Value Theorem

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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

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