Are there any properties of sup?

Question

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

Answer 1

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

$$\sup_{x \in S} |f(x) - g(x)| \leq \sup_{x \in S}|f(x) - h(x)| + \sup_{x \in S} |h(x) - g(x)|$$