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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!

Alexander Shamov
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ccc
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2 Answers2

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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, $g_n(x)=f_n(x)-f_n(0)$.

Jonas Meyer
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  • What did you bring new with the example you provided? – Mhenni Benghorbal Jan 08 '13 at 05:05
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    @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. – Jonas Meyer Jan 08 '13 at 05:12
  • His example is correct and you can see why? – Mhenni Benghorbal Jan 08 '13 at 05:26
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    @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. – Jonas Meyer Jan 08 '13 at 05:32
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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.

ncmathsadist
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