stackexchange people.
I'd like to check whether my basic inequality is vaild or not.
Which is below:
If $\sup |{A}|, \sup |{B}|$ are different,
then $| \sup |{A}|- \sup |{B}|| =\sup |{A}|- \sup |{B}|$ (we can assume this because of the assumption)
$\le \sup (|{A}|-|{B}|)$
(Since $ \sup(A+B) \le \sup {A}+ \sup {B}$)
$\le \sup |A-B|$. (Because the right inside term is bigger or equal to the left term.)
Is this inequality right?
If $f$ and $g$ are real functions defined on a set $E$, let us write $\sup_E f = \sup \{f(x)\ |\ x\in E\}$. Then for any $x\in E$ one has $$f(x) + g(x)\le \sup_E f + \sup_E g$$ hence $$\sup_E(f+g)\le \sup_E f + \sup_E g$$
If $A$ and $B$ are bounded functions defined on $E$, one has $$\sup_E |A| \le \sup_E (|A| - |B |+ |B|)\le \sup_E(|A|-|B|) + \sup_E |B|$$ hence $$\sup_E |A| - \sup_E |B|\le \sup_E (|A|-|B|)$$ One also has $$\forall x\in E, |A(x)|-|B(x)|\le |A(x)-B(x)|\Longrightarrow \sup_E(|A|-|B|)\le \sup_E |A-B|$$