Multi-Dimensional Extreme Value Theorem Difficulty An excerpt from lecture notes:
$$
\begin{array}{l}{\text { Proposition 4.20 Let X be a closed bounded set in } \mathbb{R}^{m}, \text { and let } f: X \rightarrow \mathbb{R}^{n}} \\ {\text { be a continuous function mapping } X \text { into } \mathbb{R}^{n} . \text { Then there exists a point }\mathbf{w}} \\ {\text { of } X \text { such that }|f(\mathbf{x})| \leq|f(\mathbf{w})| \text { for all } \mathbf{x} \in X .}\end{array}
$$
$$
\begin{array}{l}{\text { Theorem } 4.21 \text { (The Multidimensional Extreme Value Theorem) }} \\ {\text { Let } X \text { be a closed bounded set in } \mathbb{R}^{m} \text { , and let } f: X \rightarrow \mathbb{R} \text { be a continuous }} \\ {\text { real-valued function defined on } X \text { . Then there exist points } \mathbf{u} \text{ and } \mathbf{v} \text { of } X} \\ {\text { such that } f(\mathbf{u}) \leq f(\mathbf{x}) \leq f(\mathbf{v}) \text { for all } \mathbf{x} \in X .}\end{array}
$$
Now begins the proof:
$$
\begin{array}{l}{\text { Proof It follows from Proposition } 4.20 \text { that the function } f \text { is bounded on }} \\ {X . \text { It follows that there exists a real number } C \text { large enough to ensure that }} \\ {f(\mathbf{x})+C>0 \text { for all } \mathrm{x} \in X \text { . It then follows from Proposition } 4.20 \text { that there }} \\ {\text { exists some point } \mathbf{v} \text { of } X \text { such that }} \\ {\qquad f(\mathbf{x})+C \leq f(\mathbf{v})+C}\end{array}
$$
Shouldn't there be absolute value bars on $f(\bf{v})$ since we only know that $f(\bf{x}) \le$ $|$$f(\bf{v})$$|$ for some $\bf{v}$?
Having said that, here's the rest of the proof, which I don't really understand either, since there is another point $\bf{u}$ (possibly distinct from $\bf{v}$), that is there are two different points coming from a proposition that only provides one point '$\bf{w}$'.
$$
\begin{array}{l}{\text { for all } \mathbf{x} \in X . \text { But then } f(\mathbf{x}) \leq f(\mathbf{v}) \text { for all } \mathbf{x} \in X . \text { Applying this result }} \\ {\text { with } f \text { replaced by }-f, \text { we deduce that there exists some } \mathbf{u} \in X \text { such that }} \\ {-f(\mathbf{x}) \leq-f(\mathbf{u}) \text { for all } \mathbf{x} \in X . \text { The result follows. }}\end{array}
$$
 A: The first part of the proof applies Proposition 4.20 to a function that is constructed to take only positive values on $X$ (it would work the same if $C$ were chosen to make $f(x) + C \geqslant 0$ for all $x$ instead of requiring $f(x) + C > 0$) because that allows one to drop the absolute value.
The underlying property of the order on $\mathbb{R}$ that is used in the construction is the equivalence
$$ r \leqslant s \iff (r + t) \leqslant (s + t)$$
that holds for all real $r,s,t$. Thus for a $v \in X$ we have $f(x) \leqslant f(v)$ if and only if $f(x) + C \leqslant f(v) + C$ first for each individual $x \in X$, and then the universally quantified equivalence
$$\bigl(\forall x \in X\bigr)\bigl(f(x) \leqslant f(v)\bigr) \iff \bigl(\forall x \in X\bigr)\bigl(f(x) + C \leqslant f(v) + C\bigr)\,. \tag{$\ast$}$$
This is independent of the chosen $C$.
A first application of Proposition 4.20 yields the existence of a $C \in \mathbb{R}$ such that $f(x) + C > 0$ for all $x \in X$. We can take $C = \lvert f(w)\rvert + 1$, where $w$ is the element whose existence Proposition 4.20 asserts.
Then Proposition 4.20 is applied a second time, not to $f$, but to the function $g$ defined by $g(x) = f(x) + C$ for $x \in X$. Since $g$ attains only positive values, we have $\lvert g(x)\rvert = g(x)$ for all $x$, hence we may write or omit the absolute value as we desire. Omitting it is useful, since then we can invoke $(\ast)$. So by Proposition 4.20 there is a $v \in X$ with $\lvert g(x)\rvert \leqslant \lvert g(v)\rvert$ for all $x \in X$. Dropping the absolute value and expanding the definition of $g$ we obtain the existence of $v \in X$ such that
$$f(x) + C \leqslant f(v) + C$$
holds for all $x \in X$. Now invoking $(\ast)$ yields $f(x) \leqslant f(v)$ for all $x \in X$, in other words
$$f(v) = \max \: \{ f(x) : x \in X\}\,.$$
At the end, it is shown that $f$ also attains its minimum on $X$ by applying the first part to the function $h \colon x \mapsto -f(x)$. And we have
$$h(x) \leqslant h(u) \iff -f(x) \leqslant -f(u) \iff f(u) \leqslant f(x),$$
thus $f$ attains a minimum at $u$ if and only if $h$ attains a maximum at $u$.
