Triangle Inequality for Integrals

From ProofWiki
Jump to navigation Jump to search





Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f: X \to \overline \R$ be a $\mu$-integrable function.


Then:

$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$


Corollary

Let $f: X \to \overline \R$ be a $\mu$-integrable function be such that:

$\ds \int \size f \rd \mu = 0$


Then:

$\ds \int f \rd \mu = 0$


Real Number Line

On the real number line, the Triangle Inequality for Integrals takes the following form:

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.


Then:

$\ds \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$


Complex Plane

In the complex plane, the Triangle Inequality for Integrals takes the following form:

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 1

Let $\ds z = \int_X f \rd \mu \in \C$.

By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\cmod \alpha = 1$ such that:

$\alpha z = \cmod z \in \R$

Let $u = \map \Re {\alpha f}$, where $\Re$ denotes the real part of a complex number.

By Modulus Larger than Real Part, we have that:

$u \le \cmod {\alpha f} = \cmod f$


Thus we get the inequality:

\(\ds \cmod {\int_X f \rd \mu}\) \(=\) \(\ds \alpha \int_X f \rd \mu\)
\(\ds \) \(=\) \(\ds \int_X \alpha f \rd \mu\) Integral of Integrable Function is Homogeneous
\(\ds \) \(=\) \(\ds \int_X u \rd \mu\)
\(\ds \) \(\le\) \(\ds \int_X \cmod f \rd \mu\) Integral of Integrable Function is Monotone

$\blacksquare$


Proof 2

We have:

\(\ds \size {\int f \rd \mu}\) \(=\) \(\ds \size {\int f^+ \rd \mu - \int f^- \rd \mu}\) Definition of Integral of Integrable Function
\(\ds \) \(\le\) \(\ds \size {\int f^+ \rd \mu} + \size {-\int f^- \rd \mu}\) Triangle Inequality for Real Numbers, since $f$ is $\mu$-integrable both integrals are certainly real
\(\ds \) \(=\) \(\ds \int f^+ \rd \mu + \int f^- \rd \mu\)
\(\ds \) \(=\) \(\ds \int \paren {f^+ + f^-} \rd \mu\) Integral of Positive Measurable Function is Additive
\(\ds \) \(=\) \(\ds \int \size f \rd \mu\) Sum of Positive and Negative Parts

$\blacksquare$


Also see


Sources