Triangle Inequality for Integrals/Complex
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Theorem
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \C$ be a continuous complex function.
Then:
- $\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
where the first integral is a complex Riemann integral, and the second integral is a definite real integral.
Proof
Define:
- $z \in \C$ as the value of the complex Riemann integral:
- $z = \ds \int_a^b \map f t \rd t$
- $r \in \hointr 0 \to$ as the modulus of $z$
- $\theta \in \hointr 0 {2 \pi}$ as the argument of $z$.
From Modulus and Argument of Complex Exponential:
- $z = re^{i \theta}$
Then:
\(\ds r\) | \(=\) | \(\ds z e^{-i \theta}\) | Reciprocal of Complex Exponential | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b e^{-i \theta} \map f t \rd t\) | Linear Combination of Complex Integrals | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \map \Re {e^{-i \theta} \map f t} \rd t + i \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t\) | Definition of Complex Riemann Integral |
As $r$ is wholly real, we have:
- $\ds 0 = \map \Im r = \int_a^b \map \Im {e^{-i \theta} \map f t} \rd t$
Then:
\(\ds r\) | \(=\) | \(\ds \int_a^b \map \Re {e^{-i \theta} \map f t} \rd t\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \int_a^b \size {\map \Re {e^{-i \theta} \map f t} } \rd t\) | Absolute Value of Definite Integral | |||||||||||
\(\ds \) | \(\le\) | \(\ds \int_a^b \size {e^{-i \theta} \map f t} \rd t\) | Modulus Larger than Real Part | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \size {e^{-i \theta} } \size {\map f t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \size {\map f t} \rd t\) | Modulus of Exponential of Imaginary Number is One |
As $\ds r = \size {\int_a^b \map f t \rd t}$ by its definition, the result follows.
$\blacksquare$
Also see
- Triangle Inequality for Integrals, of which this is a special case
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S 2.1$