# Thread: showing a functional is a norm

1. ## showing a functional is a norm

having trouble showing whether
J(y)=suplyl is a norm on the space B[0,infinity) where B is the space of bounded functions on interval [0,infinity)

I have no trouble showing llfll=0 iff f=0 but the other three conditions have me stuck any help would be really appreciated

2. Hi,

let f and g be 2 functions in B[0,infinity].
Then, $\forall x \in [0,\infty [, |f(x)+g(x)| \leq |f(x)|+|g(x)| \leq J(f)+J(g)$
Finally, $J(f+g) \leq J(f)+J(g)$

moreover,

let f be a function in B, let k be a real number.
Then $\forall x \in [0,\infty [, |kf(x)| \leq J(k.f)$ and $\forall x \in [0,\infty [, |kf(x)| \leq |k|.J(f)$, so $J(k.f)=|k|.J(f)$