Proving $\sup(|f|) - \inf(|f|) \leq \sup(f) - \inf(f)$ I appreciate if you could give me some hints on how to prove that :
$$\sup(|f|) - \inf(|f|) \le \sup(f) - \inf(f)$$
 A: Here is a strategy. Since you asked for hints, I let you check the details.
This is not really a fact about functions. It is rather a property of subsets $S$ of $\mathbb{R}$. So take such a set $S$ and denote $|S|:=\{|s|\;;\;s\in S\}$. We want
$$
\sup |S|-\inf|S|\leq \sup S-\inf S.
$$
We have an alternative.
Case 1: $\sup |S|=\sup S$. Clearly $\inf S\leq \inf |S|$. So the inequality holds.
Case 2: $\sup|S|=-\inf S$. Check that $-\inf|S|\leq \sup S$. The inequality follows.
A: Let $f:E\rightarrow \mathbb{R}$ be bounded on $E$.
We want to show that
\begin{equation}
M^*-m^* \leq M-m. \tag{1}
\end{equation}
Define
\begin{equation}
m:=\inf_{x\in E}f,~~~~M:=\sup_{x\in E}f,\\
m^*:=\inf_{x\in E}|f|,~~~~M^*:=\sup_{x\in E}|f|.
\end{equation}
Note that $m^*$ and $M^*$ exist, since $|f|$ is also bounded on $E$.
Since $m\leq f(x)\leq M$ for all $x\in E$, we have $|f(x)-f(y)| \leq M-m $ for all $x,y \in E$ and then the reverse triangle inequality gives
\begin{equation}
|f(x)|-|f(y)| \leq |f(x)-f(y)| \leq M-m.
\end{equation}
So, $|f(x)| \leq M-m+|f(y)|$ for all $x\in E$ and this gives
\begin{equation}
M^* \leq M-m+|f(y)|,
\end{equation}
which when rearranged gives
\begin{equation}
|f(y)|\geq M^*- M+m,~~~\text{for all}~y\in E. \tag{2}
\end{equation}
Therefore, eq. $(2)$ gives 
\begin{equation}
m^* \geq M^*-M+m
\end{equation}
from whence $(1)$ follows. This completes the proof. $\blacksquare$
Reference
Howie, J.M. Real Analysis, Springer Undergraduate Mathematics Series, Springer-Verlag London 2001.
A: Let $c = \pm 1$. Then
$$\sup(f) - \inf(f) \leq \left| \sup(f)-\inf(f) \right| \leq c\cdot[\sup(f)-\inf(f)] = c\cdot\sup(f)-c\cdot\inf(f) = \sup(c\cdot f)-\inf(c\cdot f) \leq  \sup(\left| f \right|) - \inf(\left| f \right|).$$
