Are there any properties of sup? Can I say: 
$$\sup |f(x)-h(x)+g(x)-g(x)| = \sup|f(x)-g(x)|+ \sup|g(x)-h(x)|$$
I can't seem to find any properties of $\sup$ , so i am wondering if there is any properties that we can generalize to sup of some functions restricted on domain $x \in [0,1]$ for example
 A: As said inequality holds . simple case when equality doesn't hold 
$f(x) = x^2, g(x) = x^2+1 , h(x) = x^2+2$ 
$\sup_{[0,1]} |f(x) - g(x)| = 1 $ ,
$\sup_{x \in [0,1]} |f(x) - h(x)| = 2$
$\sup_{x \in [0,1]} |h(x) - g(x)| = 1$ 

let $f:S \subset \mathbb{R} \rightarrow \mathbb{R}$ and $g:S \subset \mathbb{R} \rightarrow \mathbb{R}$ , $h:S \subset \mathbb{R} \rightarrow \mathbb{R}$ are bounded functions on $S$ which is common domain then

$$\sup_{x \in S} |f(x)  - g(x)| \leq \sup_{x \in S}|f(x) - h(x)| + \sup_{x \in S} |h(x) - g(x)|$$
proof is again based on traingle inequality:
$$\begin{align*} |f(x) - g(x)| &=  |f(x) - h(x)+ h(x) -g(x)| \\ &\leq |f(x) - h(x)| + |h(x)-g(x)|\\ &\leq \sup_{x \in S}|f(x) - h(x)| + \sup_{x \in S} |h(x) -g (x)| \\ \end{align*}  $$
so this holds for arbitrary $x$ that is 
$$ \forall x , |f(x) - g(x)| \leq \sup_{x \in S}|f(x) - h(x)| + \sup_{x \in S} |h(x) -g (x)|$$
hence we have that $ \sup_{x \in S}|f(x) - g(x)| + \sup_{x \in S} |h(x) -g (x)|$ is upperbound of $|f(x) - g(x)|$ and since supremum is the least upperbound
$$\sup_{x \in S} |f(x)  - g(x)| \leq \sup_{x \in S}|f(x) - h(x)| + \sup_{x \in S} |h(x) - g(x)|$$ 
