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So I am a bit stuck on where to begin with this one...

Show that $C^1[0,1]$ with the norm defined as $\|f\|=\|f\|_\infty + \|f^\prime\|_\infty$ is a Banach space.

I started with an arbitrary Cauchy sequence, $f_n \to f$, so we know that for every $\epsilon > 0, \exists M \in \mathbb{N}$ such that $\|f_n - f\| \leq \epsilon , \forall n \geq M.$

Where I am getting stuck, is how do we then show that $f \in C^1[0,1]$, and that therefore $C^1[0,1]$ with the norm defined, is complete (and therefore a Banach space).


So here is my attempted solution, based on the comments below:

Assume $f_n$ is a Cauchy sequence in $C^1[0,1]$, then for every $\epsilon > 0, \exists M \in \mathbb{N}$ such that

$$\|f_n - f_m\|=\|f_n - f_m\|_\infty + \|f_n^\prime - f_m^\prime \|_\infty \leq \epsilon, \forall m,n\geq M.$$

So for any fixed $x \in C^1[0,1], \|f_n - f_m\|_\infty \leq \epsilon$ and $\|f_n^\prime - f_m^\prime \|_\infty \leq \epsilon$ and thus for this fixed $x$ $f_n$ and $f_n^\prime$ are Cauchy sequences, and therefore converge: $f_n \to f$ and $f_n^\prime \to f^\prime.$

So let $m \to \infty$ then $\|f_n - f\|\leq \epsilon, \forall n\geq M, \forall x\in [0,1]$. Then:

\begin{align}\|f_n - f\|&= \|f_n - f\|_\infty + \|f_n^\prime - f^\prime \|_\infty\\ &=\sup_{x\in [0,1]}|f_n(x)-f(x)| + \sup_{x\in [0,1]}|f_n^\prime (x)-f^\prime (x)|\\&\leq \epsilon\quad \forall n\geq M. \end{align}

Therefore $\lim_{n\to \infty} \|f_n - f\| = 0 \implies$ $C^1[0,1]$ is complete.

How is this looking?

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  • $\begingroup$ Where'd you get $f$ from? You can use the fact that $C[0,1]$ is complete to show $f\in C^1[0,1]$ (and if you haven't proven that $C[0,1]$ is a Banach space already you should start there). $\endgroup$ Sep 5, 2014 at 4:29
  • $\begingroup$ You want to assume that you have a sequence of functions $f_1, f_2,..$. Such that as $n,m$ get large $$||f_n-f_m|| = ||f_n-f_m||_{\infty}+||f'_n-f'_m||_{\infty} \to 0$$ So the point here is that individual norms go to zero, too (and uniformly so). $\endgroup$
    – David P
    Sep 5, 2014 at 4:33
  • $\begingroup$ Well, you start with a Cauchy $f_n$ and the standard route gives $f_n \to f$ and $f'_n \to g$ for some $f,g$. You need to show that $g = f'$. $\endgroup$
    – copper.hat
    Sep 5, 2014 at 4:38
  • $\begingroup$ I have updated the original post with my new attempt at the proof... how is it looking? $\endgroup$ Sep 5, 2014 at 5:28
  • $\begingroup$ Anyone able to take a look at the update, and let me know how it is looking? $\endgroup$ Sep 5, 2014 at 6:47

2 Answers 2

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Define $S:C[0,1] \times C[0,1] \to C[0,1]$ by $S(f,g)(t) = f(0)+\int_0^t g(\tau) d \tau$. It is straightforward to verify that $S$ is continuous in the product norm on $C[0,1] \times C[0,1]$.

If $f_n$ is $ \|\cdot \|$ Cauchy, then $f_n $ and $f'_n$ are $\|\cdot\|_\infty$ Cauchy and so we have $f_n \to f$ and $f'_n \to g$ (in the $\|\cdot\|_\infty$ norm).

We have $S(f_n,f'_n) = f_n$ for all $n$. Since $S$ is continuous, we have $S(f,g) = f$, from which it follows that $f$ is differentiable and $f'=g$.

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  • $\begingroup$ Wouldn't you need to show that $f$ is differentiable first, so that $Df$ is well-defined? $\endgroup$ Sep 5, 2014 at 11:13
  • $\begingroup$ @VincentBoelens: Since $D$ is continuous and $f_n \to f$ (in $\|\cdot\|$), we have $Df_n \to Df$. $\endgroup$
    – copper.hat
    Sep 5, 2014 at 13:33
  • $\begingroup$ @VincentBoelens: You are right, I think my argument is circular. $\endgroup$
    – copper.hat
    Sep 5, 2014 at 14:47
  • $\begingroup$ @VincentBoelens: I fixed my rather egregious mistake. Thanks for catching that. $\endgroup$
    – copper.hat
    Sep 5, 2014 at 18:24
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    $\begingroup$ @VincentBoelens: Sad part is that I have answered the same question somewhere else in the side after correcting someone for making the error I just made :-). $\endgroup$
    – copper.hat
    Sep 5, 2014 at 19:10
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Since $f_n,f'_n\in C([0,1])$ and both sequences converge in the sup-norm, there exist $f,g\in C([0,1])$ such that $$\|f_n-f\|_\infty \rightarrow 0 \text{ and } \|f'_n-g\|\rightarrow0.$$ It needs to be shown that $f$ is differentiable in $f'=g$. This means showing that $$\lim_{h\to 0}\frac{f(x+h)-f(x)-g(x)h}{h}=0 \ \forall x\in [0,1].$$ Fix $x_0\in [0,1]$. Let $\epsilon>0$ be given. Since $g$ is continuous, there exists $\delta>0$ such that $|x_0-x|<\delta$ implies $|g(x_0)-g(x)|<\frac\epsilon4$. Let $h$ be arbitrary but fixed, such that $|h|<\delta$. Choose $n$ such that $\|f_n-f\|_\infty<\frac{|h|\epsilon}{4}$ and $\|f'_n-g\|<\frac\epsilon4$. By the mean value theorem, there exists $\xi$ such that $|x_0-\xi|<|h|$ and $\frac{f_n(x_0+h)-f_n(x_0)}{h}=f'_n(\xi).$ It follows that \begin{align} \left|\frac{f(x_0+h)-f(x_0)-g(x_0)h}{h}\right|&\le \left| \frac{f(x_0+h)-f_n(x_0+h)}{h}\right|+\\ &+\left|\frac{f_n(x_0+h)-f_n(x_0)}{h}-g(x_0)\right|+\left|\frac{f_n(x_0)-f(x_0)}{h}\right|\le\\ &\le\frac{\epsilon}{2}+\left|\frac{f_n(x_0+h)- f_n(x_0)}{h}-g(x_0)\right|\le\\ &\le\frac\epsilon2+\left|\frac{f_n(x_0+h)-f_n(x_0)}{h}-g(\xi)\right|+\left|g(\xi)-g(x_0)\right|=\\ &=\frac\epsilon2+|f'_n(\xi)-g(\xi)|+|g(\xi)-g(x_0)|<\epsilon. \end{align} Since $h,x_0$ and $\epsilon$ were arbitrary, the conclusion follows.

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