Jensen's Inequality (Measure Theory)/Convex Functions

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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.

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.


Proof

Let $\d \map \nu x := \dfrac {\map f x} {\int \map f s \rd \map \mu s} \rd \map \mu x$ be a probability measure.

Let $x_0 := \displaystyle \int \map g s \rd \map \nu s$.

Then by convexity there exists constants $a, b$ such that $\forall x \in \R_{\ge 0}$:

$\map V {x_0} = a x_0 + b$
$\map V x \ge a x + b$


In other words, there is a tangent line at $\tuple {x_0, V_0}$ that falls below the graph of $V$.

Therefore:

\(\displaystyle \map V {\map g s}\) \(\ge\) \(\displaystyle a \map g s + b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \map V {\map g s} \rd \map \nu s\) \(\ge\) \(\displaystyle a \int \map g s \rd \map \nu s + b\) Integration with respect to $\map \nu s$
\(\displaystyle \) \(=\) \(\displaystyle \map V {x_0}\)
\(\displaystyle \) \(=\) \(\displaystyle \map V {\int \map g s \rd \map \nu s}\)