 Solution

By the mean value theorem, a continuously differentiable function with zero derivative on $[-1,1]$ is constant. Thus the kernel of the differentiation operator isIt is a closed set by the finite-dimensional subspace is closed theorem, applied in the normed space $(C^1([-1,1]),\Vert \cdot {\Vert }_{\infty })$. It is proper because the function $x\mapsto x$ has nonzero derivative.

For $n\ge 1$, take $u_n(x)=\sin (nx)$. Its supremum norm is at most one, whereas $u_n^{\prime }(0)=n$ and $|u_n^{\prime }(x)|\le n$. Hence