Extreme Value Theorem

From ProofWiki
Jump to navigation Jump to search


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:

$\forall z \in X: \norm {\map f x} \le \norm {\map f z} \le \norm {\map f y}$

where $\norm {\map f x}$ denotes the norm of $\map f x$.

Moreover, $\norm f$ attains its minimum and maximum.

Extreme Value Theorem for a Real Function

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


$\forall x \in \closedint a b: \exists x_n \in \closedint a b: \map f {x_n} \le \map f x$
$\forall x \in \closedint a b: \exists x_n \in \closedint a b: \map f {x_n} \ge \map f x$

Extreme Value Theorem for Normed Vector Spaces

Let $X$ be a normed vector space.

Let $K \subseteq X$ be a compact subset.

Suppose $f : X \to \R$ is a continuous mapping at each $x \in K$.


$\ds \exists c \in K : \map f c = \sup_{x \mathop \in K} \map f x = \max_{x \mathop \in K} \map f x$
$\ds \exists d \in K : \map f d = \inf_{x \mathop \in K} \map f x = \min_{x \mathop \in K} \map f x$


By Continuous Image of Compact Space is Compact, $f \sqbrk X \subseteq Y$ is compact.

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

Let $\ds A = \inf_{x \mathop \in X} \norm {\map f x}$.

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

$\ds \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$

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

So there exists a convergent subsequence $\sequence {x_n}$ of $\sequence {y_n}$.

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

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

$\ds \norm {\map f x} = \norm {\map f {\lim_{n \mathop \to \infty} x_n} } = \norm {\lim_{n \mathop \to \infty} \map f {x_n} } = \lim_{n \mathop \to \infty} \norm {\map f {x_n} } = A$

So $\norm f$ attains its minimum at $x$.

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


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.