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Can someone refer me to a verification of the completeness of $C^{n}[0,1]$ under the norm $\|f\|_{C^{n}} = \sum_{k=0}^{n}\|f^{(k)}\|_{\infty}$?

I tried to follow the same approach as the standard proof the $C[0,1]$ is complete under the supremum norm, but I run into the problem that the limits I end up with do not necessarily define a function in $C^n[0,1]$.

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Some hints: as a first step, just start by considering the case $n=1$. Then, given a Cauchy sequence $(f_n)$ in $C^1[0,1]$, what can you say about the sequence of continuous functions $(f_n')$...? – user16299 Jan 10 '12 at 6:15
Lemma: If $f_n' \to g$ uniformly, then $f_n$ converges uniformly to some function $f$, $f$ is differentiable, and $f' = g$. (Hint: $f_n(x) = f_n(0) + \int_0^x f_n'(y) dy$.) – Nate Eldredge Jan 10 '12 at 6:36
@Nate: You also need pointwise convergence at some $x_0$ ($0$ in your comment) to ensure that your hint works. – kahen Jan 10 '12 at 7:45
@kahen: Yes, you're right. Of course, in this case we already know that $f_n$ converges uniformly to some $f$. – Nate Eldredge Jan 11 '12 at 16:03
up vote 3 down vote accepted

Let me do it for $C^1([0,1])$ then you can adapt the argument for $C^n ([0,1])$.

We want to show that $C^1([0,1])$ is complete with respect to $\|f \| := \| f \|_\infty + \| f^\prime \|_\infty$. So let $f_n$ be a Cauchy sequence. Then

(i) $f_n$ converges pointwise to some $f: [0,1] \to \mathbb{R}$

For this you observe that $f_n(x_0)$ is a Cauchy sequence in $\mathbb{R}$ for all $x_0$ in $[0,1]$. $\mathbb{R}$ is complete so $f(x_0)$ is also in $\mathbb{R}$. By the same argument $f_n^\prime$ converges pointwise to some $g(x) := \lim_{n \to \infty} f_n(x)$.

(ii) Next you want to show that the pointwise limit function $f$ is in $C^1([0,1])$. Here you want to show two things: that $f$ is continuous and that it's differentiable.

Let's do continuous first: Let $\varepsilon > 0$ and $x_0 \in [0,1]$. Then $|f(x_0) - f(x)| \leq |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f(x) - f_n(x)|$

Now you can argue that each term can be made less than $\frac{\varepsilon}{3}$. For $n$ big enough you get $|f(x_0) - f_n(x_0)| < \frac{\varepsilon}{3}$ and $|f(x) - f_n(x)| < \frac{\varepsilon}{3}$ and because $f_n$ is continuous you can find a $\delta$ such that $|x-x_0| < \delta$ implies $|f_n(x_0) - f_n(x)| < \frac{\varepsilon}{3}$.

Use the same argument to show that $g = \lim_{n \to \infty} f^\prime_n$ is continuous.

Next you want to show that $f$ is differentiable. In particular you claim that $f^\prime = \lim_{n \to \infty} f^\prime_n = g$. To do this you can use the fundamental theorem of calculus to rewrite $f_n$ as $f_n(x) = f_n(0) + \int_0^x f^\prime_n (t) dt$.

Also using the fundamental theorem your claim translates into $f(x) = f(0) + \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt$. To show this you want to show $|f(x) - f(0) - \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt| < \varepsilon$ as follows:

Choose an $n$ such that $\| f_n - f\| < \varepsilon$ then you have $\| f_n - f\|_\infty < \varepsilon$ and $\| f_n^\prime - g\|_\infty < \varepsilon$. Then use the triangle inequality:

$$\begin{align} &|f(x) - f(0) - \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt| \\ &\leq |f(x) - f_n(x)| + |f(0) - f_n(0)| + |f_n(x) - f_n(0) - \int_0^x f_n^\prime(t) dt | + | \int_0^x f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t) dt| \end{align}$$

The first two terms can be made less than $\frac{\varepsilon}{4}$ for $n$ large enough, the third term is $0$ and for last term we can choose $n$ so large that $| f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t)| < \frac{\varepsilon}{4}$ and hence

$$ \left | \int_0^x f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t) dt \right | \leq \int_0^x | f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t)| dt \leq \int_0^x \frac{\varepsilon}{4} dt = x \frac{\varepsilon}{4} \leq \frac{\varepsilon}{4}$$

(iii) The last thing you need to show is that $\| f - f_n \| \xrightarrow{n \to \infty} 0$:

$$ \| f - f_n \| = \| f - f_n \|_\infty + \| f^\prime - f_n^\prime \|_\infty \leq 2 \max ( \| f - f_n \|_\infty, \| f^\prime - f_n^\prime \|_\infty )$$

So to finish the proof you want to show that $f_n$ converges in the sup norm (and by the same argument that $f^\prime_n$ does, too). Let's do it for $f_n$:

$| f(x) - f_n(x) | \leq |f(x) - f_N(x)| + |f_N(x) - f_n(x)|$

Now choose $N$ such that for $n,m\geq N$ you have $|f_N(x) - f_n(x)| < \frac{\varepsilon}{2}$ and $|f_m(x) - f_N(x)| < \frac{\varepsilon}{2}$. Then $\lim_{m \to \infty} |f_m(x) - f_N(x)| \leq \frac{\varepsilon}{2}$ for all $x$ in $[0,1]$. Hence $\| f - f_n \|_\infty \xrightarrow{n \to \infty} 0$.

You can find a proof of this for the norm $\|f\| := \max(\|f\|_\infty, \|f^\prime\|_\infty)$ on $C^1([0,1])$ and for the norm $\|f\| := \max_{\operatorname{deg}{(\partial_\alpha)} \leq k} \|\partial_\alpha f\|_\infty$ on $C^n_b(\Omega)$ where $\Omega$ is an open set in $\mathbb{R}^d$ here starting on page 35.

Hope this helps.

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Small question/nitpick: does $C^n_b(\Omega)$ mean that the $n$th derivative is *uniformly continuous*? I think you need uniform continuity to get round the fact that $\Omega$ might not be compact. – user16299 Jan 10 '12 at 18:13
@YemonChoi The $n$-th derivative is continuous and bounded. Where would you use uniform continuity in the proof? – Rudy the Reindeer Jan 10 '12 at 18:25
Also: no offence meant, since I'm relatively new here, but is the preferred practice on MSE to give complete solutions rather than hints, or attempts to direct the OP to work things out themselves? See e.g.… – user16299 Jan 10 '12 at 18:26
@YemonChoi The OP was not about $C^1([0,1])$ so this is not a "full" solution. As for general practice: I think a full solution is more helpful than a one liner pretending to be a hint. : ) – Rudy the Reindeer Jan 10 '12 at 18:36
Let me emphasize that both the hint and the full solution were very appreciated. After seeing how complex the "full solution" (at least for $n=1$) is, I'm especially grateful that it was supplied as I may have invested far too much time in this. For the record this was in fact a remark in my instructor's notes, not a homework problem. – roo Jan 11 '12 at 3:02

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