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:


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

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

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

Proof
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 $\displaystyle A = \inf_{x \mathop \in X} \norm {\map f x}$.

It follows from the definition of infimum that there exists a sequence $\sequence {y_n}$ in $X$ such that:
 * $\displaystyle \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$

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

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


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

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.