I can't make things look pretty yet and I've spent way to long trying to put this question up
Define the spaceβ1([a,b])as the space of functionsf:[a,b]↦Rwhich are everywhere differentiable and whose derivativef′is a bounded function. One equips this space with the metricProve that this turnsd(f,g)=sup|f(x)−g(x)|+sup|f′(x)−g′(x)|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
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.
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
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