Let $U$ be the set of al bounded functions $f: \Bbb R \to \Bbb R$.
Define $|| \cdot || : U \to \Bbb R$ as $ f \mapsto ||f||:=\sup\{|f(x)| : x \in \Bbb R\}$
$U$ is a vector space with origin $0_U \equiv 0$
How can I check if $||\cdot||$ is a norm on $U$? It's not the same as just checking that
right?
On
Since for any $x\in \mathbb R$, we always have $$|f(x)+g(x)| \le |f(x)|+|g(x)|$$
which leads that $$|f(x)+g(x)| \le |f(x)|+|g(x)| \le \sup\{|f(x)| x \in \Bbb R\}+ \sup\{|g(x)| x \in \Bbb R\}=||f||+||g||.$$
Note that the above inequality holds for any $x\in \mathbb R$. Therefore,
$$\sup\{|f(x)+g(x)| x \in \Bbb R\} \le ||f||+||g||,$$
and hence $$||f+g||\le ||f||+||g||$$.
It is exactly the same as checking the veracity of the axioms of a norm.
You do have $\vert \vert f\vert\vert \geqslant 0$ because of your definition with the $\sup$ of an absolute value.
It is homogenous with the properties of the supremum, and same goes for the triangle inequality.