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

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:

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

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

How is this looking?

• 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). – mathematician Sep 5 '14 at 4:29
• 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). – David Peterson Sep 5 '14 at 4:33
• 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 $g = f'$. – copper.hat Sep 5 '14 at 4:38
• I have updated the original post with my new attempt at the proof... how is it looking? – user3784030 Sep 5 '14 at 5:28
• Anyone able to take a look at the update, and let me know how it is looking? – user3784030 Sep 5 '14 at 6:47

## 2 Answers

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 $f'=g$.

• Wouldn't you need to show that $f$ is differentiable first, so that $Df$ is well-defined? – Vincent Boelens Sep 5 '14 at 11:13
• @VincentBoelens: Since $D$ is continuous and $f_n \to f$ (in $\|\cdot\|$), we have $Df_n \to Df$. – copper.hat Sep 5 '14 at 13:33
• @VincentBoelens: You are right, I think my argument is circular. – copper.hat Sep 5 '14 at 14:47
• @VincentBoelens: I fixed my rather egregious mistake. Thanks for catching that. – copper.hat Sep 5 '14 at 18:24
• @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 :-). – copper.hat Sep 5 '14 at 19:10

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.