# Extreme Value Theorem

## Theorem

Let $X$ be a compact metric space and $Y$ a normed vector space.

Let $f: X \to Y$ be a continuous mapping.

Then $f$ is bounded, and there exist $x, y \in X$ such that:

- $\forall z \in X: \left\Vert{f \left({x}\right)}\right\Vert \le \left\Vert{f \left({z}\right)}\right\Vert \le \left\Vert{f \left({y}\right)}\right\Vert$

where $\left\Vert{f \left({x}\right)}\right\Vert$ denotes the norm of $f \left({x}\right)$.

Moreover, $\left\Vert{f}\right\Vert$ attains its minimum and maximum.

### Extreme Value Theorem for a Real Function

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

Then:

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

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

## Proof

By Continuous Image of Compact Space is Compact, $f \left({X}\right) \subseteq Y$ is compact.

Therefore, by Compact Subspace of Metric Space is Bounded, $f$ is bounded.

Let $\displaystyle A = \inf_{x \mathop \in X} \left\Vert{f \left({x}\right)}\right\Vert$.

It follows from the definition of infimum that there exists a sequence $\left\langle{y_n}\right\rangle$ in $X$ such that:

- $\displaystyle \lim_{n \mathop \to \infty} \left\Vert{f \left({y_n}\right)}\right\Vert = A$

By Sequence of Implications of Metric Space Compactness Properties, $X$ is sequentially compact.

So there exists a convergent subsequence $\left\langle{x_n}\right\rangle$ of $\left\langle{y_n}\right\rangle$.

Let $\displaystyle x = \lim_{n \mathop \to \infty} x_n$.

Since $f$ is continuous and a norm is continuous, it follows by Composite of Continuous Mappings is Continuous that:

- $\displaystyle \left\Vert{f \left({x}\right)}\right\Vert = \left\Vert{f \left({\lim_{n \mathop \to \infty} x_n}\right)}\right\Vert = \left\Vert{\lim_{n \mathop \to \infty} f \left({x_n}\right)}\right\Vert = \lim_{n \mathop \to \infty} \left\Vert{f \left({x_n}\right)}\right\Vert = A$

So $\left\Vert{f}\right\Vert$ attains its minimum at $x$.

By replacing the infimum with the supremum in the definition of $A$, we also see that $\left\Vert{f}\right\Vert$ attains its maximum by the same reasoning.

$\blacksquare$

## 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.