Is differentiation as a map discontinuous?

Solution 1:

For a counterexample, take the sequence $$\frac {\sin nx} n$$ These are all continuously differentiable, but the sequence converges to $0$ and the sequence of derivatives doesn't converge at all. The derivative of the limit is not equal to the limit of the derivatives, so it is not continuous.

Solution 2:

It depends on your understanding of $C^1([0,1])$, the space of all differentiable functions whose derivative is continuous. It is a linear subspace of $C([0,1])$, where $C([0,1])$ is is equipped with the supremum norm $\lVert f \rVert = \sup_{x \in [0,1]} \lvert f(x) \rvert$. If you give $C^1([0,1])$ the norm inherited from $C([0,1])$, i.e. the supremum norm, then $A$ is not continuous (see Matt Samuel's answer). But you can also give $C^1([0,1])$ the norm $$\lVert f \rVert^{(1)} = \lVert f \rVert + \lVert f' \rVert .$$ Then $A : (C^1([0,1]), \lVert - \rVert^{(1)}) \to (C([0,1]), \lVert - \rVert)$ is trivially continuous.

Edited:

$(C^1([0,1]), \lVert - \rVert^{(1)})$ is a Banach space. See Prove that $C^1([a,b])$ with the $C^1$- norm is a Banach Space. In contrast, $(C^1([0,1]), \lVert - \rVert)$ is not. You can generalize this to the sets $C^k([0,1])$ of $k$-times continuously differentiable functions. They are Banach spaces if equipped with $$\lVert f \rVert^{(k)} = \lVert f \rVert + \lVert f' \rVert + \ldots + \lVert f^{(k)} \rVert$$ and $$A : (C^{(k)}([0,1]), \lVert - \rVert^{(k)}) \to (C^{(k-1)}([0,1]), \lVert - \rVert^{[k-1)}), A(f) = f' ,$$ is continuous.