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?
Best Answer
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.