Consider the derivation

$ D: \mathcal{C}^1([0,1]) \to \mathcal{C}([0,1]), \ \ f \mapsto f'$

Both spaces are given the norm $\| \ .\|_\mathcal{C}$. Is $D$ continues?

I think that $D$ is not continuous, but I have issues to find a good counter-example. Would be happy about some help.


I believe you work with the supremum norm. If that's the case then the operator is indeed not continuous. For each $n\in\mathbb{N}$ let $f_n(x)=\sin(nx)$. Its norm is clearly bounded by $1$. However $f_n'(x)=n\cos(nx)$ has norm $n$. So it follows that the operator $D$ is not bounded.