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I start with the following lemma

Lemma 1.(Fundamental inclusion for convex sets) Let $C$ be a closed subset of $\mathbb{R}^n$. Then, $C$ is convex if and only if \begin{equation*} \int_\Omega f\;d\mu \in C \end{equation*} for all measures spaces $(\Omega,\mu)$ with $\mu(\Omega) = 1$ and all $\mathbb{R}^n$-valued integrable functions $f$ on $\Omega$ satisfying $f(x)\in C$ for $\mu$-a.e. $x\in\Omega$.

A proof can be given based on Hahn-Banach theorem (strictly separable):

Proof: Assume the latter condition holds, we show that $C$ is convex. Let $x,y\in C$ and $\lambda\in (0,1)$, we consider $\Omega = \{0,1\}$ with the discrete $\sigma$-algebra, $\mu$ be the discrete measure $\mu(\{0\}) = \lambda$ and $\mu(\{1\}) = 1-\lambda$, and $f:\Omega\longrightarrow C$ defined by $f(0) = x$ and $f(1) = y$, then clearly $\int_\Omega f\;d\mu = \lambda x+ (1-\lambda)y \in C$ and therfore $C$ is convex. Conversely, if $C$ is convex, we prove by contradiction. Assume that there exists $(\Omega,\mu)$ and $f$ as above but \begin{equation*} x_0 = \int_\Omega f\;d\mu \notin C. \end{equation*} Using Hahn-Banach theorem\footnote{In $\mathbb{R}^n$ then every two nonempty disjoint convex sets are separable by a closed hyperplane.}, there exists $\xi\in \mathbb{R}^n$ and $\alpha\in \mathbb{R}$ such that \begin{equation*} \xi\cdot x < \alpha < \xi\cdot x_0 \qquad\text{for all}\qquad x\in C. \end{equation*} But since $f(z)\in C$ for $\mu$-a.e. $z\in \Omega$, we also have \begin{equation*} \alpha<\xi\cdot x_0 = \xi\cdot\left(\int_\Omega f\;d\mu\right) = \int_\Omega \xi\cdot f(z)\;d\mu(z) \leq \alpha \end{equation*} which is a contradiction.

Using that lemma we can prove the Jensen inequality with the assumption $\Phi$ is lower semicontinuous as following (we need $\mathrm{epi}(\Phi)$ to be closed so that we can use strictly separation of Hahn Banach theorem)

Theorem [Jensen inequality for convex l.s.c. functions] Let $(\Omega,\mathcal{A},\mu)$ be a probability space, i.e., $\mu(\Omega) = 1$. If $f$ is a function $\Omega\longrightarrow\mathbb{R}^n$ such that it is $\mu$-integrable and if $\Phi$ is a l.s.c. convex function $\mathbb{R}^n\longrightarrow\mathbb{R}$ then \begin{equation*} \Phi\left(\int_\Omega f\;d\mu\right) \leq \int_\Omega\Phi\circ f\;d\mu. \end{equation*}

Proof. Let us define $C = \mathrm{epi}\;\Phi = \{(x,\lambda)\in \mathbb{R}^n\times\mathbb{R} :\Phi(x)\leq \lambda\}$. Since $\Phi$ is convex l.s.c, $C$ is a closed convex subset of $\mathbb{R}^{n+1}$. Let us define \begin{equation*} \mathcal{F}:\Omega\longrightarrow\mathbb{R}^n\times\mathbb{R} \qquad\text{by}\qquad \mathcal{F}(z) = \big(f(z),\Phi(f(z))\big). \end{equation*} It is clear that $\mathcal{F}(z)\in \mathrm{epi}\;\Phi = C$ for $\mu$-a.e. $z\in \Omega$, hence by Lemma 1 we deduce that \begin{equation*} \int_\Omega \mathcal{F}(z)\;dz = \left(\int_\Omega f\;d\mu, \int_\Omega \Phi\circ f\;\mu\right) \in \mathrm{epi}\;\Phi \end{equation*} which gives us our desired claim.

My question is,

We have a version of Hahn Banach theorem saying that in $\mathbb{R}^n$, any two nonempty disjoint convex sets can be separated by a closed hyperplane (not strictly seperated). Can we use that to relax the condition $\Phi$ is lower semicontinuous in Lemma 1 so that we can generalize Jensen inquality without l.s.c. assumption?

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If $\Phi\colon \mathbb{R}^n\to\mathbb{R}$ is a convex function, then it is continuous.

You can have problems if $\Phi$ is extended-valued.

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  • $\begingroup$ It is not correct, it is only correct for $\mathbb{R}^1$. $\endgroup$ – Sean Sep 8 '18 at 16:47
  • $\begingroup$ $\Phi$ is locally Lipschitz continuous in $\mathbb{R}^n$. $\endgroup$ – Rigel Sep 8 '18 at 16:50
  • $\begingroup$ Can you give a reference? I don't think it is true in dimension bigger than 1. $\endgroup$ – Sean Sep 8 '18 at 16:52
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    $\begingroup$ You can check on a book of convex analysis (e.g. Rockafellar) or here: math.stackexchange.com/questions/216778/… $\endgroup$ – Rigel Sep 8 '18 at 16:54

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