Why do the factors 1/|z-a| and |z-a| cancel in the displayed estimate?

Conditions

• The integration path is the straight line segment from a to z.
• The difference quotient has a denominator of $z-a$, whose absolute value is $|z-a|$.
• The ML inequality is applied to the integral over this segment.

Reasoning, step by step

1. Write down the difference quotient error term: $\frac{1}{z-a} \int_{[a,z]} f(w) dw - f(a)$.
2. Rewrite the expression to combine the terms under a single integral: $\frac{1}{z-a} \int_{[a,z]} (f(w) - f(a)) dw$.
3. Take the absolute value of the expression: $\left| \frac{1}{z-a} \int_{[a,z]} (f(w) - f(a)) dw \right|$.
4. Apply the ML inequality to the integral part: $\left| \int_{[a,z]} (f(w) - f(a)) dw \right| \le \sup_{w \in [a,z]} |f(w) - f(a)| \cdot |z-a|$.
5. Substitute this bound back into the absolute value expression: $\frac{1}{|z-a|} \cdot \sup_{w \in [a,z]} |f(w) - f(a)| \cdot |z-a|$.
6. Observe that the $|z-a|$ in the numerator and the $|z-a|$ in the denominator cancel out, leaving $\sup_{w \in [a,z]} |f(w) - f(a)|$.

Example

The lecturer points to the factor $\frac{1}{|z-a|}$ and then to the factor $|z-a|$, stating that they cancel. The board shows $\frac{1}{|z-a|} \sup_{w \in [a,z]} |f(w)-f(a)| \cdot |z-a| \to 0$.

Common misconceptions

• Assuming the cancellation is a mathematical error or a typo.
• Forgetting that the length of the path $[a,z]$ is exactly $|z-a|$.

Watch the explanation

Explore next

• What is the ML inequality as used in this complex-analysis lecture?
• How is the estimation lemma applied to bound the integral of $f(w) - f(a)$?