Let f be a bounded function on a set A, and set $S = \sup\{f(x) : x ∈ A\}, I = \inf \{f(x) : x ∈ A\},$ $S' = \sup\{|f(x)| : x ∈ A\}, I' = \inf \{|f(x)| : x ∈ A\}.$ Show that $S - I ≥ S' - I'$.
1. Intuition please? I believed because $-|r| \le r \le |r|$ for all real numbers $r$, ergo $\sup f(x) \le \sup \quad \color{seagreen}{|f(x)| \ge 0} \quad \implies S < S'$ and $\inf f(x) \ge \inf \quad \color{seagreen}{|f(x)| \ge 0} \quad \implies I > I'$. But question is opposite! What did I flub?
Let e > 0 be arbitrary and choose $x_1$ and $x_2$ so that $S' − e/2 < |f(x_1)|$ and $I' + e/2 > |f(x_2)|$.
2. Where did these two inequalities issue from? How can we presage them?
3. Modus operandi of the proof please? This feels fey and occult.
Then using Exercise 1.2.5 (Reverse Triangle Inequality),
4. I don't understand the last red inequality. How does it engender $\le S - I$?
This question is ahead of Prove that if $f$ is integrable on $[a,b]$ then so is $|f|$?.
Note that $I \le f(x) \le S$, and $-S \le -f(y) \le -I$, hence $I-S \le f(x) -f(y) \le S-I$, in particular, $|f(x)-f(y) | \le S-I$, since $S \ge I$.
The triangle equality gives $|f(x)| -|f(y)| \le |f(x)-f(y) | \le S-I$ for all $x,y \in A$.
Taking the $\sup_x$ gives $S'-|f(y)| \le S-I$ for all $y \in A$.
Then take the $\sup_y$ gives $S'+\sup_y(-|f(y)|) \le S-I$, and since $\sup (-A) = - \inf A$ we have $S'-\inf_y|f(y)| = S'-I' \le S-I$.