The space $ C^1[0,1] $- the space of all continuously differentiable functions on $ [0,1]$ is not a Banach space with respect to the sup norm,$ \|.\|_{\infty} $ since the uniform limit of a continuously differentiable function need not be differentiable.

How can I illustate this statement using a counter example? Can I use $ f_{n}=\frac1 n \sin nx $ as a counter example?

Also, is $ C^1[0,1] $ is same as the space given by $ X=\{f\in C^1[0,1]:f(0)=0 \}$. Can I use the same example to show that this is not a Banach space?

More help is appreciated! Thanks!

  • $\begingroup$ I edited your title to remove "Hilbert space" (and thanks to Alexander for fixing my confusing "fix" and fixing the tags). In the future, please check that your title and question don't conflict much. Also, because you already include the general subject in the tags, there is no need to add it to the title. $\endgroup$ Jan 8, 2013 at 3:16
  • $\begingroup$ You have got the right counter example, since $f_n(x)=\frac{\sin(nx)}{n}$ converges to $0$ uniformly while the derivatives $ f'_n(x)=\cos(nx) $ do not approach $0$. See here. $\endgroup$ Jan 8, 2013 at 6:27
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    $\begingroup$ Mhenni's comment is incorrect. Lack of convergence of $f_n'$ is not directly relevant, and it has been pointed out why the example in the question doesn't work. (This was also pointed out in comments on Mhenni's deleted answer.) $\endgroup$ Jan 8, 2013 at 6:35
  • $\begingroup$ @JonasMeyer: So it is relevant! Thanks. $\endgroup$ Jan 8, 2013 at 6:39
  • $\begingroup$ See here. $\endgroup$ Jan 8, 2013 at 7:04

2 Answers 2


The example you gave converges uniformly to the zero function, which is continuously differentiable.

Every continuous function on $[0,1]$ is a uniform limit of polynomial functions (by the Weierstrass approximation theorem), and polynomial functions are continuously differentiable.

For an explicit example, you could also consider the sequence $f_n(x)=\left|x-\frac12\right|^{(n+1)/n}$.

I might not fully understand the last question. Those are not the same, because elements of $C^1[0,1]$ do not generally have to vanish at $0$. But you can use similar examples. E.g., you could still think about the Weierstrass approximation theorem for dramatic counterexamples, or you could modify the example above by taking, say, $g_n(x)=f_n(x)-f_n(0)$.

  • $\begingroup$ What did you bring new with the example you provided? $\endgroup$ Jan 8, 2013 at 5:05
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    $\begingroup$ @Mhenni: The example $(f_n)$ given in this answer converges uniformly to the function $f(x)=\left|x-\frac12\right|$. While each element of the sequence is in $C^1$, the limit is not. This implies that $(f_n)$ is a Cauchy sequence in $C^1$ (with the norm given in the question) that does not converge to an element of $C^1$. What it brought new to this thread is a correct (explicit) example of what the OP requested. $\endgroup$ Jan 8, 2013 at 5:12
  • $\begingroup$ His example is correct and you can see why? $\endgroup$ Jan 8, 2013 at 5:26
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    $\begingroup$ @Mhenni: I have mentioned 4 times in this thread (once in my answer and three times in comments to your answer) why it is incorrect. If you will be more specific about what you do not understand about the reason I already gave, I may be able to explain it further. As is, I do not know precisely what you have a question about. $\endgroup$ Jan 8, 2013 at 5:32

The ${\cal C}^1[0,1]$ functions are a dense subset of $\cal{C}[0,1]$ in the sup norm. However the inclusion is proper so the $\cal{C}^1$ functions are not a complete subspace of ${\cal C}[0,1]$.

The continuous functions on $[0,1]$ that vanish at 0 are a Banach space, they are the kernel of the continuous map $f\mapsto f(0)$. They form a closed subspace. But those in ${\cal C}^1[0,1]$ are not.


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