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:


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

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

Proof
By Continuous Image of a 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 $\displaystyle A = \inf_{x \in X} \left\Vert{f \left({x}\right)}\right\Vert$.

It follows from the definition of infimum that there exists a sequence $\left\langle{y_n}\right\rangle$ in $X$ such that $\displaystyle \lim_{n \to \infty} \left\Vert{f \left({y_n}\right)}\right\Vert = A$.

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

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


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

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.