# Complex variables: ML inequality

1. Oct 13, 2007

### futurebird

I'm studying the proof of the $$ML$$ inequlity from complex analysis. I don't know what they did in one step of the proof and I was wondering if anyone can explain the step to me.

First of all the theorem says:

Let f(z) be continuous on a contour C. Then

$$|\int_{c}f(z)dz| \leq ML$$

Where $$L$$ is the length of C and $$M$$ is an upper bound for |f| on C.

PROOF:
After a few steps we have:

$$|\int_{c}f(z)dz| \leq \int^{b}_{a} |f(z(t))||z'(t)|dt$$

And the text says " |f| is bounded on C, $$|f(z)| \leq M$$ on C, where $$M$$ is constant, then:

$$|\int_{c}f(z)dz| \leq M\int^{b}_{a} |z'(t)|dt$$

How did they get the $$M$$ out of the integral?

Last edited: Oct 13, 2007
2. Oct 13, 2007

### morphism

$$\int^{b}_{a} |f(z(t))||z'(t)|dt \leq \int_a^b M|z'(t)|dt = M \int_a^b |z'(t)|dt$$

3. Oct 14, 2007

### futurebird

So it's okay that M is an upper bound for |f(z)|, that's the same thing as an upper bound for |f(z(t))|? Okay thanks, I think I get it now.