# Thread: Functional Analysis Proof: complete metric space

1. ## 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

2. ## 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

3. ## 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.

4. ## 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

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

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

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

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

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

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

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

which is the same thing as saying

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

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

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

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 $n>N,\tilde{N}$ and $|x-y|<\delta$, show that
$\left| g(x)-\frac{f(y)-f(x)}{y-x} \right|\leq 3\epsilon$

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