# Thread: Riemann integration and Lipschitz

1. ## Riemann integration and Lipschitz

I need to show that, given f:[a,b]-->R where f is Riemann integrable, and F:[a,b]-->R defined by F(x)=$\displaystyle \int^x_a f(t)dt$ prove that F is Lipschitz... kind of stuck on how to approach this.
Any help would be much appreciated, thanks

2. Hello.

Recall that $\displaystyle F$ is Lipschitz continuous if there exists a fixed $\displaystyle K>0$ such that for every $\displaystyle x,y\in \mathbb{R}$, $\displaystyle |F(x)-F(y)|\leq K |x-y|$.

Can you think of a way to find such a $\displaystyle K$?

3. well, I know that for some c in [a,b], F'(c)=f(c), not sure where this gets me though...

4. $\displaystyle F$ is differentiable as long as $\displaystyle f$ were given as continuous, but $\displaystyle f$ was given as an integrable function, so it's bounded.

5. I recommend using the Mean Value Theorem.

For every $\displaystyle x,y\in [a,b]$ there exists $\displaystyle c\in (a,b)$ such that $\displaystyle \frac{F(x)-F(y)}{x-y}=F'(c)$. What happens when you maximize $\displaystyle F'$?

6. not sure what you mean by maximizing F'. Do you mean set F'(c)=0?

7. Originally Posted by roninpro
I recommend using the Mean Value Theorem.

For every $\displaystyle x,y\in [a,b]$ there exists $\displaystyle c\in (a,b)$ such that $\displaystyle \frac{F(x)-F(y)}{x-y}=F'(c)$. What happens when you maximize $\displaystyle F'$?
As Krizalid said, this works if $\displaystyle f$ is continous, but it's only assumed integrable. Use the fact that it's bounded and that $\displaystyle \vert \int_{a}^{b} f(t)dt \vert \leq \int_{a}^{b} \vert f(t) \vert dt$

8. I still dont see it,
so from your point, I gathered that |F(x)|<= integral(f(t)) but I dont see how this will lead me to the lipschitz condition

9. $\displaystyle \vert F(x)-F(y)\vert =\vert \int_{x}^{y} f(t)dt \vert \leq M\vert x-y \vert where \vert f(t)\vert \leq M$.

10. Thank you Jose, sorry to make you spell it all out like that, I guess I was ignoring the boundedness.