Problem with showing a normed vector space is complete

**Kazimierz Twardowski** · November 25th 2009, 09:26 AM

In my homework I am asked to show that:

1) $\| - \|_{C^1}$ is a norm on $C^{1}$ [a,b] where $\| - \|_{C^1}$ is defined by $\| f \|_{C^1}$ = $\| f \|_{\infty} + \| f' \|_{\infty}$ (the second is meant to be 'f dash' i.e. the derivative of f )

2) $C^{1}$ [a,b] is complete in this norm.

My attempt so far:

1) Fairly easy, just show that the norm satisfies the three conditions in the definition of a norm.

2) This is what I'm stuck with.

$\left( C^1[a,b] , \| - \|_{C^1} \right)$ is complete if :

Given a Cauchy sequence $(f_{n})$ in $\left( C^1[a,b] , \| - \|_{C^1} \right) \exists f$
such that the limit of the sequence $(f_{n})$ is $f.$

Choose a sequence of functions $(f_{n})$ in
$\left( C^1[a,b] , \| - \|_{C^1} \right)$ converging uniformly to $f:[a,b] \rightarrow \mathbb{R}$
such that $f \in C^1[a,b]$ . Then by definition $\left( C^1[a,b] , \| - \|_{C^1} \right)$ is complete.

PROBLEM: Is there any reason that there is such a sequence? Obviously I would have to show somehow why such a sequence exists ... and unfortunately I don't know how to do this because in my notes there isn't any examples to show why, for example, any finite-dimensional normed vector space is complete over $\mathbb{R}$ .

N.B. Sorry about the format, it's the first time I've used latex!

**Jose27** · November 25th 2009, 10:48 AM

Notice that by your definition of norm, any Cauchy sequence $(f_n)\in C^1[a,b]$ will induce two Cauchy sequences in $(C^0[a,b], \Vert \cdot \Vert _{\infty } )$ namely $(f_n)$ and $(f'_n)$ . Since $(C^0[a,b], \Vert \cdot \Vert _{\infty } )$ is complete this implies that there exists $f$ , $g$ such that $f_n \rightarrow f$ and $f'_n \rightarrow g$ . Now it is a standard result that if $(h_n)$ is a sequence of $C^1$ functions such that the seguence of derivatives converges uniformly and $(h_n)$ converges in one point (you have uniform convergence for this!) then $h=\lim_{n\rightarrow \infty } h_n$ is $C^1$ and $h'=\lim_{n\rightarrow \infty } h'_n$ . It follows that your space is complete.

To show that a finite dim. normed space is comlete use the fact that all norms are equivalent.

**Kazimierz Twardowski** · November 25th 2009, 11:56 AM

Thanks this has helped a lot. I didn't think about breaking a Cauchy sequence in $C^1[a,b]$ into two Cauchy sequences based on the fact that the norm is defined by two norms.

Whilst this certainly helps with my homework, I, myself don't know how to prove that $\left( C^0[a,b] , \| - \|_{\infty} \right)$ is complete. In fact I don't know, assuming there is one, a general technique to show that a certain normed vector space is complete.

I can show that any finite dimensional normed space is complete in $\mathbb {R}^n$ due to the theorem that any two norms on $\mathbb {R}^n$ are equivalent. Obviously this theorem breaks down in infinite dimensions because $inf \left\{ \| - \| : x \in \mathbb {R}^ n , \| x \|_{\infty} = 1 \right\}$ can equal 0 which becomes a problem when you start dividing!

Also I think my problem is I don't understand why there has to be a Cauchy sequence in a vector space. What result guarantees this?

Any more help would be greatly appreciated, I was impressed with the speed of a reply!

**Jose27** · November 25th 2009, 01:36 PM

To prove that $(C^0[a,b], \Vert \cdot \Vert _{\infty } )$ is complete just remember that $f_n\rightarrow f$ in this norm iff $f_n\rightarrow f$ uniformly. Now pick a Cauchy sequence $f_n$ then since the sequence of real numbers $f_n(x)$ is Cauchy for all $x\in [a,b]$ we have a pointwise limit $f(x):= \lim_{n\rightarrow \infty } f_n(x)$ and notice that $\vert f_n(x)-f_m(x) \vert < \epsilon$ if $n,m>N$ and letting $n\rightarrow \infty$ we have for all $x\in [a,b]$ that $\vert f(x)-f_m(x) \vert < \epsilon$ if $n>N$ so $f_n$ converges uniformly to $f$ .

Usually when trying to prove a space is complete you "break up" the Cauchy sequence into other such sequences in sapces you already know are complete. As for the other question, a Cauchy sequence always exists, take the constant for example.

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