Is differentiation as a map discontinuous?

Question

I came across the statement below:

Let $C([0,1])$ be the space of all continuous functions over the interval $[0,1]$ equipped with the Supremum norm. Assume $A$ is a map on the space of all differentiable functions whose derivative is continuous into $C([0,1])$. Also, $A$ is differentiation in the sense that it maps a functions to its derivative. The map $A$ (differentiation) is discontinuous.

It's written that the last sentence is well-known but I can't make any sense of it. How can I arrive at such a conclusion? Actually, I am looking for an explicit counterexample.

Any help would be highly appreciated.

Answer 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.

Answer 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.