What is ML inequality property in complex integral which says $|\int_{c}f(z)dz| \leq ML$. I can't understand a thing from this expression. I want to understand it conceptually(if that helps).
How can we find the upper bound of a complex integral

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    $\begingroup$ The area of a fence is no larger than (highest point)(length of fence). $\endgroup$ – user61527 Feb 6 '14 at 16:13
  • $\begingroup$ highest point? is it f(Z), is it making a perpendicular point on argand plane, forming a 3d structure $\endgroup$ – mahes Feb 6 '14 at 16:22
  • $\begingroup$ By "highest point," I mean the maximum value in $|f(z)|$. $\endgroup$ – user61527 Feb 6 '14 at 16:23
  • $\begingroup$ so area is $|\int f(Z)|$ and highest point is $\int |f(Z)|$, what can be upper bound $\endgroup$ – mahes Feb 6 '14 at 16:28

$$\left|\int_c f(z) \, dz \right| \leq \int_c |f(z)| \cdot |dz|$$

Now assume that $|f(z)| \leq M$ that means the function is bounded on the curve

$$\int_c |f(z)| \cdot |dz| \leq M \int_c |dz|$$

Now assume that the $c=\gamma(t)$ is a parametrization of the curve then

$$\int_c |dz| = \int^b_a |\gamma'(t)| \, dt$$

Now by the definition of the length $L$ of a curve we have

$$\int^b_a |\gamma'(t)| \, dt=L$$


$L$ is the arc length of $c$, $M$ is an upper bound for the absolute value of $f$ on $c$.

Let's compare the result to real integrals: Let $f$ be defined on $[a,b]$ and $|f|$ bounded by $M$. Then:

$\left| \int_a^b f(x) dx \right| \leq \int_a^b |f(x)| dx \leq \int_a^b M dx = M \int_a^b dx = M(b-a)$.

Here, $b-a$ is the arc length of $[a,b]$ and $M$ is an upper bound for $|f|$. The ML inequality is just a generalization of this to complex integrals.

  • $\begingroup$ if a and b are complex what would be the arc length $\endgroup$ – mahes Feb 6 '14 at 16:40
  • $\begingroup$ $a$ and $b$ were assumed to be real in what I wrote, but if you integrate along the straight line segment from a complex number $a$ to a complex number $b$, then the length of the segment would be $|b-a|$. $\endgroup$ – Ulrik Feb 6 '14 at 16:46
  • $\begingroup$ can a and b be a straight line since it cannot bound to form a closed area. I am asking this from reading the first comment by T.Bongers $\endgroup$ – mahes Feb 6 '14 at 16:53

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