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The classical Jensen integral inequality says:

Let $\mu$ be a probabilistic measure defined on some $\sigma$-algebra subsets of $\Omega$. If $f\colon \Omega \rightarrow \mathbb R$ be an integrable function with $f(x)\in (a,b)$ and $\phi\colon (a,b) \rightarrow \mathbb R$ a convex function then

$$ \phi\left(\int_{\Omega}f \mathrm d \mu\right) \leq \int_{\Omega} \phi \circ f\mathrm d \mu. $$

Are there generalizations of this result in the case when $f$ has values in a convex subset $D$ of $\mathbb R^n$ and $\phi$ is a a convex function on $D$ ?

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Yes, for example Theorem 0.7 here. Note that we can push forward $\mu$ by $f$, thus reducing the inequality $$\phi\left(\int_{\Omega}f \,\mathrm d \mu\right) \leq \int_{\Omega} \phi \circ f \,\mathrm d \mu \tag1$$ to $$\phi\left(\int_{D}x \, \mathrm d \nu (x)\right) \leq \int_{D} \phi(x)\, \mathrm d \nu(x)\tag2$$ where $\nu=f_\#\mu$. That is, the value of $\phi$ at the baricenter of $(D,\nu)$ is at most the mean of $\phi$.

Anyway, the proof of (1) is similar to the standard Jensen's inequality. Let $\ell$ be a supporting function for $\phi$ at the baricenter $x_0=\int_{\Omega}f \,\mathrm d \mu$. Then $$\ell\left(\int_{\Omega}f \,\mathrm d \mu\right) =\int_{\Omega} \ell \circ f \,\mathrm d \mu \tag3$$ If we replace $\ell$ with $\phi$ in (3), the left side stays the same, while the other side can only go up. (Recall that $\ell(x_0)=\phi(x_0)$ and $\ell\le\phi$ everywhere). The existence of $\ell$ is a somewhat technical issue if $D$ is ugly and $\phi$ isn't continuous: see the notes linked above.

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  • $\begingroup$ Many thanks for answer and reference. $\endgroup$ – Alex Jun 24 '13 at 18:12

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