The ML Inequality
Cauchy's
theorem tells us when a contour integral is exactly zero. Often we need something
softer: not the precise value, but a guarantee that an integral is small — small enough
to ignore, or small enough to vanish in a limit. The tool for that is the estimation
lemma, universally known as the ML inequality:
\left| \int_\gamma f(z)\,dz \right| \le M\,L,
where M = \max_{z \in \gamma} |f(z)| is the largest the integrand gets
on the contour, and L is the contour's length. A modest bound, but the
workhorse behind nearly every limit argument in the subject.
Deriving the bound
It follows from the triangle inequality for integrals — the integral of a sum is no bigger in
size than the sum of sizes — applied to the parametrised form.
Step 1 — write the integral over the parameter. With
z(t),\ t \in [a, b]:
\int_\gamma f(z)\,dz = \int_a^b f\bigl(z(t)\bigr)\,z'(t)\,dt.
Step 2 — pull the modulus inside. The size of an integral is at most the
integral of the size:
\left| \int_a^b f\bigl(z(t)\bigr)\,z'(t)\,dt \right| \le \int_a^b \bigl| f(z(t)) \bigr|\,\bigl| z'(t) \bigr|\,dt.
Step 3 — bound the integrand by M. Since
|f(z(t))| \le M everywhere on the contour, replace it by its
maximum:
\int_a^b \bigl| f(z(t)) \bigr|\,\bigl| z'(t) \bigr|\,dt \le M \int_a^b \bigl| z'(t) \bigr|\,dt.
Step 4 — recognise the length. The remaining integral
\int_a^b |z'(t)|\,dt is exactly the arc length of
\gamma, i.e. L:
\left| \int_\gamma f(z)\,dz \right| \le M\,L.
For a contour \gamma of length L:
-
\displaystyle\left| \int_\gamma f(z)\,dz \right| \le M\,L, where
M is any upper bound for |f| on
\gamma;
-
it never gives the integral's value — only a ceiling on its size;
-
its power is in limits: if M\,L \to 0 along a family of contours,
the integral is forced to 0 as well.
A worked bound
Estimate \displaystyle\oint_{|z| = 2} \frac{dz}{z^2 + 1} over the
circle |z| = 2.
Step 1 — find L. The contour is a circle of radius
2, so its length is
L = 2\pi (2) = 4\pi.
Step 2 — bound |f| to get M.
We need |z^2 + 1| to be at least as small as possible to make
1/|z^2+1| as large as possible. By the reverse triangle inequality,
on |z| = 2:
|z^2 + 1| \ge |z|^2 - 1 = 4 - 1 = 3, \qquad\text{so}\qquad \frac{1}{|z^2 + 1|} \le \frac{1}{3} = M.
Step 3 — multiply. Combining the two:
\left| \oint_{|z| = 2} \frac{dz}{z^2 + 1} \right| \le M\,L = \frac{1}{3} \cdot 4\pi = \frac{4\pi}{3}.
We never computed the integral — yet we know its magnitude cannot exceed
4\pi/3. That is exactly the kind of leverage the ML inequality gives.
The real reason to learn this lemma: it is how we kill off unwanted pieces of a contour. To
evaluate a real integral by complex methods, one often closes a path with a big semicircular
arc of radius R and then sends R \to \infty.
On such an arc L = \pi R grows, but if |f|
shrinks faster — say M \sim 1/R^2 — then
M\,L \sim \pi/R \to 0, and the arc contributes nothing in the limit.
The ML inequality is what licenses that "the arc vanishes" step, and it is also the workhorse
in the proof of Cauchy's integral formula coming next.
See the bound tighten
For f(z) = 1/(z^2 + 1) on the circle |z| = R,
slide the radius R. The readout tracks
L = 2\pi R, the bound M = 1/(R^2 - 1), and
their product M\,L — the ceiling on the integral's size. Watch how a
growing contour and a shrinking integrand trade off.