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Let $A \subset \mathbb R$ and consider the space $C^1(A)$. I am asked to prove that $( C^1(A), \Vert \cdot \Vert_{C^1(A)})$ is a Banach space, where $ \Vert f(x) \Vert_{C^1(A)} = \sup_{x \in A} \vert f(x) \vert + \sup_{x \in A} \vert f'(x) \vert $

First question: $A$ should be compact (or at least closed set), shouldn't it?

Secondly, how would you prove this? I've taken a Cauchy sequence, $(f_n)_{n \in \mathbb N} \subseteq C^1(A)$: if I fix $x \in A$, then I obtain two Cauchy sequences $(f_n(x))$ and $(f'_n(x))$ in $\mathbb R$ (?) so they converge to two numbers, $f(x)$ and $f'(x)$. The function $f$ that I obtain is the pointwise limit: how can I prove that this gives me exactly the $C^1(A)$ limit?

I've still one more question: is $( C^1(A), \Vert \cdot \Vert_{\infty})$ still a Banach space? I did not manage to find a counterexample...

Thanks for your help.

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    It is true that $A$ an interval is the case to understand first. Whitney defined $C^1(A)$ for an arbitrary $A$, and proved (for his definition) that if $f$ is $C^1$ on a closed set $A \subseteq \mathbb R^n$, then it extends to a $C^1$ function on $\mathbb R^n$. But of course a $C^1$ function on an open interval may be unbounded.2012-06-09

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If you already know that $C^0$ is a Banach space, and if $f_n$ is Cauchy in $C^1$, then you already do know that $f_n$ and $D_i f_n$ converge uniformly to continuous functions $f, g_i, 1\le i\le n$. You then only have to show that in this case $f$ is differentiable and $D_i f = g_i$. Which theorems do you know about sequences of differentiable functions and the differentiability of the limit?

$(C^1, ||.||_\infty)$ is not a Banach space, consider a sequence of smooth functions converging to $x\mapsto |x|$ on $[-1,1]$ (e.g. polynomials, which are known to be dense in $C^0$).

Since this is homework I deliberately ignore your question about compactness of $A$ and suggest you check corresponing statements for $C^0$ and see whether they carry over. (Note, though, that you usually define differentiability on open sets).

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    It is certainly possible but rather tedious to explicitly write down a sequence which provides a counterexample. For this reason I suggested to use a more abstract result which simply guarantees the existence of such sequences (i.e. the fact that the polynomials are a dense set in $C^0$). I have to admit I did not check you example, but you, of course, have to only check the values of $f_n$ and $f_n' $ to see whether they are $C^1$. The sequence cannot be Cauchy in $C^1$ if $C^1$ is a Banach space, otherwise it would follow that $x\mapsto |x|$ is $C^1$ as well, and that is certainly not true.2012-06-09
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I think I have found a counterexample which shows that $C^1([-a;a], \mathbb R)$ ($a \gt 0$) with the sup-norm is not a Banach space (at the endpoints I mean the one-sided derivatives).

Let's define $f_n \colon [-a,a] \to \mathbb R$ by $ f_n(x)=\vert x \vert^{1+\frac{1}{n}} $

By direct calculation, we can see that $f'_n(0)=0$ for every $n \in \mathbb N$. Therefore, we can say that $(f_n)_{n \in \mathbb N}$ is a sequence in $C^1([-a,a])$. The pointwise limit is clearly the function $f \colon x \mapsto \vert x \vert$.

Indeed, the limit is also uniform: to prove this, we observe that $(f_n)_{n \in \mathbb N}$ is increasing. Since the limit function is continuous and the space $[-a,a]$ is compact then, by Dini's theorem, we can conclude that the convergence is uniform, i.e. sup-norm convergence. To conclude, just note that the limit function $f \notin C^1([-a,a])$.