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.

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    $\begingroup$ What it means is that $A$ is an unbounded linear operator from $C^1([a,b])$ to $C([a,b])$ (both spaces having the supremum norm). $\endgroup$ Commented Sep 21, 2019 at 15:03

2 Answers 2


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.


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.


$(C^1([0,1]), \lVert - \rVert^{(1)})$ 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.

  • $\begingroup$ I got it, thank you. $\endgroup$ Commented Sep 22, 2019 at 5:15
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    $\begingroup$ This (along with @MattSamuel's answer) is such a great pair -- it addresses the question, but also shows that there's something more to look at even though the basic answer is relatively straightforward. Best thing I've read so far today (although it is only 9 AM). $\endgroup$ Commented Sep 22, 2019 at 13:02
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    $\begingroup$ It might be worth mentioning the name Sobolev space/Sobolev norm for readers interested in the general setting this is a special case of. $\endgroup$ Commented Sep 23, 2019 at 14:04

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