Jensen's Inequality (Measure Theory)

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This proof is about Jensen's Inequality in the context of Measure Theory. For other uses, see Jensen's Inequality.

Theorem

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

Let $f: X \to \R$ be a $\mu$-integrable function such that $f \ge 0$ pointwise.


Convex Functions

Let $V: \hointr 0 \infty \to \hointr 0 \infty$ be a convex function.


Then for all positive measurable functions $g: X \to \R$, $g \in \map {\mathcal M^+} \Sigma$:

$\map V {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} } \le \dfrac {\int \paren {V \circ g} \cdot f \rd \mu} {\int f \rd \mu}$

where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.


Concave Functions

Let $\Lambda: \hointr 0 \infty \to \hointr 0 \infty$ be a concave function.


Then for all positive measurable functions $g: X \to \R$, $g \in \map {\MM^+} \Sigma$:

$\dfrac {\int \paren {\Lambda \circ g} \cdot f \rd \mu} {\int f \rd \mu} \le \map \Lambda {\dfrac {\int g \cdot f \rd \mu} {\int f \rd \mu} }$

where $\circ$ denotes composition, and $\cdot$ denotes pointwise multiplication.


Also see


Source of Name

This entry was named for Johan Ludwig William Valdemar Jensen.


Sources