Solution
= Solution
Take $f_n(t)=\sin(2\pi nt)$. Its domain <norm> is $1$, whereas its image <norm> is $1+2\pi n$. Hence \b[$T$ is unbounded], so it has no finite <operator norm> and is not a <compact operator>.
= Solution
Take $f_n(t)=\sin(2\pi nt)$. Its domain <norm> is $1$, whereas its image <norm> is $1+2\pi n$. Hence \b[$T$ is unbounded], so it has no finite <operator norm> and is not a <compact operator>.