Show that $l$ is a continuouse linear functional on $C([0,1]) \ $.
$l(f)=\int_0^1f(t)h(t)dt$ where $h\in\mathscr{L^1}([0,1])$

Question

Show that $l$ is a continuouse linear functional on $C([0,1]) \ $.
 $l(f)=\int_0^1f(t)h(t)dt$  where $h\in\mathscr{L^1}([0,1])$

zzr0ck3r · Answer

I already showed that $|||l|||\le ||f||_{[0,1]}||h||_1$  where $||h||_1$is the max norm; I'm not sure if that's standard notation. I assume that this is the norm of the linear functional, but none of the methods we were playing with worked for me so far. I took it to my teacher and he said that I should not worry about it, and that he should not have assigned it. But I have played with it to long to let it go.

zzr0ck3r · Answer

@eliassaab

zzr0ck3r · Answer

if its a pain in the but don't worry about it.

anonymous · Answer

$$
|l(f)|=\left|\int_0^1f(t)h(t)dt\right|\le\int_0^1|f(t)h(t)|dt\le \sup_{0\le t \le 1 }|f(t)|\int_0^1| h(t)|dt=||f||_{C[0,1]} ||h||_1

$$
That is what you did. That is enough to show that l is contnious.
If $ f_n  \to f  $ in C[0,1] then$$
|l(f_n ) -l(f)|=|l(f_n-f)| \le ||f_n-f||_{C[o,1]} ||h||_1

$$
You are done

anonymous · Answer

A linear operator T is continuous if and only if $$
||T(x)|| \le M ||x||$$

anonymous · Answer

Where  M is positive constant

zzr0ck3r · Answer

I'm confused on how this shows what the norm is.