Let $g\in C([0,1])$. Consider the linear operator $L_g:C([0,1])\rightarrow C([0,1])$ defined by $L_g(f) = fg$. Show that $L_g$ is continuous and find $|||L_g|||$.

Question

zzr0ck3r · Answer

I did the continuous part, but now I need to find $|||L_g|||$.
@eliassaab

anonymous · Answer

ah, linear operators. i remember learning linear/abstract algebra as if it were yesterday.

kainui · Answer

Tell us more @bloopman

zzr0ck3r · Answer

it was for me

zzr0ck3r · Answer

but this is for real analysis:)

anonymous · Answer

unfortunately i'm busy with other work of my own right now, but i'll try to get to a good stopping point to try to help

kainui · Answer

I feel like this hinges on the definition of continuity.

anonymous · Answer

oh. i've never looked into R.A.; just proofs it seems. either way, this is part of abstract algebra, which is how i learned linear operators
anyway i'll get back to work

zzr0ck3r · Answer

$|||L||| = \sup\{ \ ||L(f)|| \ :  ||f||\le 1\}$

zzr0ck3r · Answer

well because its continuous we know $|||L|||<\infty$

anonymous · Answer

Notice that, let f be such $ || f|| \le 1$
$$

||| L_g|||\le || g f||= \sup_{0\le t \le 1}| g(t) f(t)| \le\sup_{0\le t \le 1}| g(t)|=||g||\
||| L_g||| \ge ||L_g(1)||=||g||\
||| L_g||| =||g||
$$

zzr0ck3r · Answer

ty

anonymous · Answer

YW