Let $A_n: L^2([0,1]) \to L^2([0,1])$ be
$(A_nf)(x)= \int_0^1 sin(n\pi(x-y))f(y)dy$.
Are they compact operators? Is there any kind of convergence?
They are continous since $\|A_nf\| \leq \|f\| $. They are compact since the kernel $sin(n(x-y)) \in L^2([0,1]\times[0,1])$. Probably there is weak convergence of $A_n \rightharpoonup 0$ because $\sin(n(x-y)) \rightharpoonup 0$ (If I don't go wrong). I don't know how to explain that there is no strong convergence to a compact operator.
Thanks!