# Functional Analysis Proof: complete metric space

• Apr 10th 2013, 07:53 PM
randi
Functional Analysis Proof: complete metric space
Define the space
β1([a,b])
as the space of functions
f:[a,b]R
which are everywhere differentiable and whose derivative
f
is a bounded function. One equips this space with the metric
d(f,g)=sup|f(x)g(x)|+sup|f(x)g(x)|
Prove that this turns
β1([a,b])
into a complete metric space.
Please help. I know what a complete metric space is, but I'm having trouble getting started with this proof. I would prefer a hint that would point me in the right direction as I'd like the opportunity to work through the proof myself, but I'll be thankful for any help I receive.
Thanks
• Apr 10th 2013, 07:55 PM
randi
Re: Functional Analysis Proof: complete metric space
I can't make things look pretty yet and I've spent way to long trying to put this question up :(
• Apr 10th 2013, 08:02 PM
Gusbob
Re: Functional Analysis Proof: complete metric space
To show completeness in this metric, all Cauchy sequences converge to a point in your metric space. Take an arbitrary cauchy sequence and show that this is the case.

EDIT: That is, show that the limit is a differentiable function from [a,b] to R with bounded derivative.
• Apr 11th 2013, 01:05 PM
randi
Re: Functional Analysis Proof: complete metric space
So, let {fn} be a cauchy sequence in b. To show that b is complete in [a,b] we must show that given a Cauchy sequence fn of everywhere differentiable functions there exists h contained in b such that lim fn equals h with respect to d(f,g). We must prove this limit is a differentiable function from [a,b] and has a bounded derivative.

How do I prove this limit is differentiable, except that it's contained in b, so it must be everywhere differentiable (same being true for bounded derivative)? Do I need to prove this limit is contained in b, or can I just assume that?
Thanks for the help
• Apr 12th 2013, 03:15 AM
Gusbob
Re: Functional Analysis Proof: complete metric space
This proof may be greatly simplified depending on how much analysis you know. The proof I gave is long, but should be able to be understood by most people with even a passing knowledge of analysis.

I'll change notation a bit. Let $\displaystyle C_b^1([a,b]})$ be the space differentiable functions on $\displaystyle [a,b]$ with bounded derivative. Also, denote $\displaystyle C_b([a,b])$ to be the space of continuous function on $\displaystyle [a,b]$. We make use of the fact that $\displaystyle C_b([a,b])$ is complete. I will also refer to the given norm as $\displaystyle \|\cdot\|_{C^1_b}$

The space in your question is acutally $\displaystyle C_b^1([a,b]})$ since $\displaystyle [a,b]$ is compact and every continuous function defined on a real valued function on a compact set is bounded.

Now take a Cauchy sequence $\displaystyle \{f_n\}$ in $\displaystyle C_b^1([a,b]})$. Then certainly $\displaystyle \{f_n\}$ is Cauchy in $\displaystyle C_b([a,b]})$, so there is a continuous function $\displaystyle f$ on $\displaystyle [a,b]$ such that $\displaystyle f_n \ to f$ uniformly. Now

$\displaystyle \|f'_m -f'_n\|_\infty \leq \|f_m - f_n\|_\infty + \|f'_m -f'_n\|_\infty = \|f_m -f_n\|_{C^1_b}$

which shows that $\displaystyle \{f'_n\}$ is Cauchy with respect to the uniform norm. In particular, $\displaystyle \{f'_n\}$ is Cauchy in $\displaystyle C_b([a,b])$. Using the completeness of $\displaystyle C_b([a,b])$, there is a $\displaystyle g\in C_b([a,b])$ such that $\displaystyle f'_n \to g$ uniformly. It remains to show that $\displaystyle g=f'$.

Now here comes the fun part. Fix any $\displaystyle x\in [a,b]$ and $\displaystyle \epsilon >0$. THere is some $\displaystyle N>0$ such that $\displaystyle \|f'_m-f'_n\|_\infty < \epsilon$ for $\displaystyle m,n > N$. Given any $\displaystyle y\in [a,b]$ with $\displaystyle y\not=x$ and suppose $\displaystyle m,n > N$. Applying the Mean value theorem to $\displaystyle f_m-f_n$ shows that there is some point $\displaystyle c$ between $\displaystyle x$ and $\displaystyle y$ such that

$\displaystyle (f_m-f_n)(y)-(f_m-f_n)(x)=(y-x)(f'_m-f'_n)(c)$

which is the same thing as saying

$\displaystyle \left| \frac{f_m(y)-f_m(x)}{y-x}-\frac{f_n(y)-f_n(x)}{y-x} \right|$ = $\displaystyle |f'_m(c)-f'_n(c)| \leq \|f'_m-f'_n\|_\infty < \epsilon$

Taking one of the variable to infinity, say $\displaystyle m\to \infty$, we obtain for $\displaystyle n>N$
$\displaystyle \left| \frac{f(y)-f(x)}{y-x}-\frac{f_n(y)-f_n(x)}{y-x} \right|\leq \epsilon$

Using the definition of differentiability (of $\displaystyle f_n$), there is some $\displaystyle \delta >0$ such that
$\displaystyle |x-y|<\delta \implies \left| \frac{f(y)-f(x)}{y-x}-f'_n(x)} \right|$
Furthermore, the fact that $\displaystyle f'_n \ to g$ uniformly implies that there is some $\displaystyle \tilde{N}$ such that $\displaystyle \|f'_n-g\|_\infty < \epsilon$ for $\displaystyle n>\tilde{N}$.

Here is where I ask you to do some work. If you truly understood what I just said, and the implications thereof, it should be doable. Choosing $\displaystyle n>N,\tilde{N}$ and $\displaystyle |x-y|<\delta$, show that
$\displaystyle \left| g(x)-\frac{f(y)-f(x)}{y-x} \right|\leq 3\epsilon$

This would imply that
$\displaystyle g(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=f'(x)$