Prove $\sup(|f|) - \inf(|f|) \leq \sup(f) - \inf(f)$ for bounded function $f$ on set $S$

Question

The supremums and infimums are taken on set $S$.

I managed to show that for all $x, y \in S$, we have $|f(x)| - |f(y)| \leq \sup(f) - \inf(f)$. But I didn't see a clear way to proceed.

Answer 1

If $\inf f\ge0$ then $|f|=f$ and the claim is clear. Similarly, if $\sup f\le 0$ then $|f|=-f$ and the claim is also clear (using $\sup(-f)=-\inf f$ etc.) So assume $\inf f<0<\sup f$. Clearly $\inf(|f|)\ge 0$. If $\sup(|f|)>\sup f$, then $\sup(|f|)=-\inf f$ and we have $-\inf(|f|)\le 0< \sup f$, whence the claim. If $\sup(|f|)\le\sup f$, then $\sup(|f|)-\inf (|f|)\le \sup(|f|)\le\sup f\le \sup f-\inf f$.

Answer 2

Let's have a set $T$ and a function $g:T\to\mathbb{R}$. Suppose we can prove that $$g(x)\leq a$$ holds for all $x\in T$. Let be $M = \sup g$. Is $M\leq a $ also?

There exists a sequence $\{x_n\}_n\subset T$ with the limit $M$ and $x_n\leq M$ for all $n\in \mathbb{N}$ (by definition of supremum). If $M>a$, then $\varepsilon =M-a>0 $. By definition of limit, at least one $x_{n_0}$ falls into interval $(M-\varepsilon/2,M+\varepsilon/2)$. Since $a <M-\varepsilon/2 < x_{n_0} $, we come to a contradiction. Therefore $M\leq a$. Analogically: if $g(x)\geq a'$, then $\inf g \geq a'$.

Now, you just apply these two facts to your LHS.