Functional Analysis Proof: complete metric space

Define the space *β*1([*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 *β*1([*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

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 :(

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.

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

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 be the space differentiable functions on with bounded derivative. Also, denote to be the space of continuous function on . We make use of the fact that is complete. I will also refer to the given norm as

The space in your question is acutally since is compact and every continuous function defined on a real valued function on a compact set is bounded.

Now take a Cauchy sequence in . Then certainly is Cauchy in , so there is a continuous function on such that uniformly. Now

which shows that is Cauchy with respect to the uniform norm. In particular, is Cauchy in . Using the completeness of , there is a such that uniformly. It remains to show that .

Now here comes the fun part. Fix any and . THere is some such that for . Given any with and suppose . Applying the Mean value theorem to shows that there is some point between and such that

which is the same thing as saying

=

Taking one of the variable to infinity, say , we obtain for

Using the definition of differentiability (of ), there is some such that

Furthermore, the fact that uniformly implies that there is some such that for .

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 and , show that

This would imply that