Well my argument was was that the i've got is a continuous linear functional, and is in the kernel for all , except when . However hence is not in the kernel. The kernel is closed by continuity, hence any sequence in it, in particular, a sequence of the sin functions, cannot converge to an element outside the closed set. The only problematic term is which is not in the kernel, but that sequence is just which converges to .
I was just wondering where I have gone wrong here? Did you mean I should use Arzelą to find a fallacy in the above argument?