# Thread: Proof involving Lipschitz continuity with constant L

1. ## 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)?

2. 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)?

Why are you using the metric notation?

$U(f,P)-L(f,P)=\left|U(f,p)-L(f,p)\right|$ $=\left|\sum_{j=1}^{n}(M_j-m_j)\Delta x_j\right|\leqslant\sum_{j=1}^{n}|M_j-m_j|\Delta x_j$. Now, $M_j=f(x*)$ for some $x*\in[x_{j-1},x_j]$ and similarly for $m_j$. But, by Lipschitz $|M_j-m_j|=|f(x*)-f(y*)|\leqslant L|x_{j}-x_{j-1}|\leqslant L\|P\|$. Thus, $\left|U(P,f)-L(P,f)\right|\leqslant\sum_{j=1}^{n}L\|P\|\Delta x_j=L\|P\|\sum_{j=1}^{n}\Delta x_j=L\|P\|(b-a)$.

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 $X$ to a metric space $Y$ is $d_Y(f(x),f(y))\leqslant \delta d_x(x,y)$ writing it like that when you have the idea of addition and subtraction can do nothing but confuse you.