uniform norm: using $\sup(A+B)=\sup A+\sup B$ to proof $|||f(x)|+|g(y)|||=||f(x)||+||g(y)||$ Let $f:D\rightarrow \mathbb{R} $ and $g:D\rightarrow \mathbb{R} $ be bounded functions.
The uniform norm ist defined by $||f(x)||=\sup\{|f(x)|| x\in D\}$.
Now my question:
I know this theorem: If $A,B$ are non-empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$ then it follows that $\sup(A+B)=\sup A+\sup B$.
Define $A=\{|f(x)|| x\in D\}, B = \{|g(y)|| y\in D\}$ and $A+B=\{|f(x)|+|g(y)| | x,y\in D\}$
Can I conclude from that $|||f(x)|+|g(y)|||=\sup(A+B)=\sup A+\sup B=||f(x)||+||g(y)||$ for every $x,y\in D$?
 A: So there are a few confusions here let me try to eleborate on all of them.
The first one is that you should not be notating this by $||f(x)||$, that implies you are looking at a specific value. You want to say something like $||f||$, which implies you are looking at the whole function.
Now as you have observed, if we let $f(D)$ be the image of $f$, the norm of $f$ is the supremum of the set $f(D)$. Ditto for $g$ and the set $g(D)$. But here is the important thing:
$$(f+g)(D) \neq f(D)+g(D)$$
Let me give an example of such maps. Let $f=g: \mathbb{Z}\longrightarrow \mathbb{Z}$ be the identity maps. then $(f+g)(n) = n+n = 2n$, so the image is precisely the set of even integers $2\mathbb{Z}$. However the image of both $f$ and $g$ are the entirity of $\mathbb{Z}$ so that when you do a set sum, it also ends up being the whole thing.
In general, you can only say $$(f+g)(D) \subseteq f(D)+g(D)$$ as the former contains elements of the form $f(x)+g(x)$ for $x\in D$, while the latter contains elements like $f(x)+g(y)$ for $x,y\in D$. So taking sup you would say $$||f+g|| \leq ||f||+||g||$$ which is the triangle inequality.
A: The conclusion does not hold in this case, because
$$
\| |f(x)| + |g(y)| \| = 
\sup\{|f(x)| + |g(x)| : x\in D\}$$
Is not equivalent to $ \sup\{|f(x)| + |g(y)| : x,y\in D\} $ in general. Here's a counterexample: Let $D = [-1,1]$, and define the indicator function $\chi_E(x)$ to be $1$ for $x\in E$ and $0$ for $x\notin E$. Notice that $|\chi_E(x)| = \chi_E(x)$ for all $x\in D$. Then we have
\begin{equation}
||\chi_{[-1,0)}(x)|| = 1, \\
||\chi_{[0,1)}(x)|| = 1, \\
||\chi_{[-1,0)}(x) + \chi_{[0,1)}(x)|| = ||\chi_{[-1,1]}(x)|| = 1.
\end{equation}
So here we see that $||\chi_{[-1,0)}(x) + \chi_{[0,1)}(x)|| = 1$ but
$||\chi_{[-1,0)}(x)|| + ||\chi_{[0,1)}(x)|| = 2$.
