Thread: Proof involving Lipschitz continuity with constant L

Problem:
Prove that if the function f:[a,b]-->R is Lipschitz continuous with constant L then for every partition P of [a,b]
U(f,P)-L(f,P) <= ||P||*L*(b-a)




Okay so from Lipschitz continuity I see that dY(f(a),f(b)) <= L*dx(a,b) for all a,b in X

So dx(a,b) is the same as (b-a)
Then take a partition P with ||P|| < delta for both sides

Where do i go from here .... i nearly have the right side completed .... or so i think but what should my next move be when it comes to the left side? How does dY(f(a),f(b)) become U(f,P)-L(f,P)?

PLease help

Originally Posted by derek walcott

Problem:
Prove that if the function f:[a,b]-->R is Lipschitz continuous with constant L then for every partition P of [a,b]
U(f,P)-L(f,P) <= ||P||*L*(b-a)




Okay so from Lipschitz continuity I see that dY(f(a),f(b)) <= L*dx(a,b) for all a,b in X

So dx(a,b) is the same as (b-a)
Then take a partition P with ||P|| < delta for both sides

Where do i go from here .... i nearly have the right side completed .... or so i think but what should my next move be when it comes to the left side? How does dY(f(a),f(b)) become U(f,P)-L(f,P)?

PLease help

Why are you using the metric notation?

. Now, for some and similarly for . But, by Lipschitz . Thus, .

oh so i shouldn't start with dY(f(a), f(b)) and instead start with U(f,P)-L(f,P) and work towards L*||P||*(b-a)

what exactly does metric notation mean and how does that differ from the notation you are using?

Originally Posted by derek walcott

oh so i shouldn't start with dY(f(a), f(b)) and instead start with U(f,P)-L(f,P) and work towards L*||P||*(b-a)

what exactly does metric notation mean and how does that differ from the notation you are using?

In this context it doesn't really matter. But,the real numbers have much more than a topological structure (a metric) they have an algebraic structure (it is a field). Consequently while the true definition of a Lipschitz continuous function from a metric space to a metric space is writing it like that when you have the idea of addition and subtraction can do nothing but confuse you.