Extreme Value Theorem/Normed Vector Space

Theorem
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 $x \in K$.

Then:


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


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

Proof
Let $K$ be compact.

By Continuous Mappings preserve Compact Subsets, $\map f K$ is compact.

By Compact Subset of Normed Vector Space is Closed and Bounded, $K$ is bounded.

Hence, $\map f K$ is bounded.

$K$ is nonempty, so $\map f K$ is non-empty.

By Characterizing Property of Supremum of Subset of Real Numbers, non-empty bounded subset of $\R$ has least upper bound.

Let $\ds M := \sup_{x \mathop \in K} \map f x$.

Then $M \in \R$.

Let $n \in \N$.

We have that:

$\ds M - \frac 1 n < M$

However, $M$ is the least upper bound.

Hence, $\ds M - \frac 1 n$ cannot be an upper bound for $\map f K$.

Therefore:


 * $\ds \exists x_n \in K : \map f {x_n} > M - \frac 1 n$

Thus we have a sequence $\sequence {x_n}_{n \mathop \in \N}$ contained in $K$.

$K$ is compact.

By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ with a limit $c \in K$.

$f$ is continuous.

By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N}$ converges with a limit $\map f c$.

But from $\ds \map f {x_n} > M - \frac 1 n$ it follows that $\map f c \ge M$.

On the other hand, by definition of $M$, $\map f c \le M$.

Therefore, $\map f c = M$.

Finally, since $c \in K$:


 * $\ds \map f c \in \set {\map f x : x \in K}$.

Hence, $M$ is a maximum.

For the second part, let $g : X \to \R$ be such that:


 * $\forall x \in X : \map g x := - \map f x$

Then above arguments apply, if infima, lower bounds and minimum are replaced with suprema, upper bounds and maximum.