# 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)= $\int^x_a f(t)dt$ prove that F is Lipschitz... kind of stuck on how to approach this.
Any help would be much appreciated, thanks

2. Hello.

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

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

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

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

5. I recommend using the Mean Value Theorem.

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

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

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

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