Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Yes $A$ should be compact. If $A$ is open, there is a useful topology for $C^1(A)$, but it is a Fréchet space, not a Banach space. – GEdgar Jun 9 '12 at 16:37
@GEdgar Thanks; I was almost sure that $A$ should be compact but I preferred to ask, since the text did not mention it. – Romeo Jun 9 '12 at 16:41
@GEdgar I do have to admit that I do not completely agree with this statement. It was my hope that Romeo could rely on some results from his course which help him to settle this question. If you just assume $A$ to be compact you run into all kinds of nasty questions regarding the regularity of $\partial A$. For this reason the best bet is to consider bounded open sets $\Omega$ and assume in the definition of $C^1$ that for $f\in C^1$ $f$ and $f'$ extend continuously to the closure of $\Omega$. – user20266 Jun 9 '12 at 17:38
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. – GEdgar Jun 9 '12 at 21:33
up vote 1 down vote accepted

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).

share|cite|improve this answer
Thank you for your kind reply. I've understood how to prove the main statement. Concerning the counterexample I was thinking about $f_n(x) = \vert x \vert$ if $ \vert x\vert \gt \frac{1}{n}$ and $f_n(x)=\frac{n}{2}x^2 + \frac{1}{2n}$ otherwise. The sequence is $C^1$ but its uniform limit is $\vert x \vert$. What do you think? This is Cauchy in $C^0$: is it Cauchy also in $C^1$? I think yes, but I'm not sure. – Romeo Jun 9 '12 at 16:46
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. – user20266 Jun 9 '12 at 17:02

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 $f \notin C^1([-a,a])$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.