Given the following operator in $L_2[0,1]$ $$(Tf)(x):=sin(x)\cdot f(x)$$
Prove or Disprove that the opertor is Compact.
I thought it is compact and used arzelà–Ascoli theorem, but apparently I am wrong. I can't really find my mistake so if you can please help me it would be very much appreciated :)
My attemp: To prove $T$ is compact it is enough to prove that it's image on the unit ball is relatively compact (remark from lecture). I will denote $B$ to be the unit ball, so I must show that $TB$ is relatively compact.
Using arzelà–Ascoli theorem it is enough to prove that $TB$ is uniformly bounded and equicontinous.
Note: arzelà–Ascoli theorem is for $C[0,1]$ but because the continouse function are dense in $L_2[0,1]$ I thought my use of the theorem is correct from this post: https://math.stackexchange.com/questions/1126888/arzela-ascoli-theorem-in-lp0-1#:~:text=Let%20S%20be%20a%20subset,the%20set%20S%20is%20compact.
Uniformly Bounded: $$g\in TB \Rightarrow ||g||=||sin(x)f(x)||\leq ||f||\leq 1$$
Equicontinouse: $$||g(x_1)-g(x_2)||=||Tf(x_1)-Tf(x_2)||\leq||T||f||x_1-x_2||\leq||x_1-x_2||\leq \epsilon$$
I showed in a different part that $||T||=1$ (and this is correct).
Thank you for the naswers :)