A variation of Jensens inequality . The following version of Jensens inequality is used in lecture notes, But i don't seem to get it . 
if $\phi : \mathbb R^n \to R$ is a convex function and $f_i \in L^1 (\Omega) $ for all $i$ and $\Omega$ is bounded set of $\mathbb R^n$  then 
$$\phi(\bar f_i,....\bar f_N)\le \frac {1}{L^N(\Omega)} \int_\Omega \phi(f_1,.....,f_N)dx $$
where $\bar f_i=\frac {1}{L^N(\Omega)}\int f_i(x)dx$ . 
I hope i got the question correct . , 
I tried comparing directly with the way the convex function is defined . I would like to show the inequality explicitly . 
Thank you for your help . 
 A: (Note that this is exactly Jensen's inequality, not a variation thereof.)
For convenience, define the measure $\mu A = \frac{1}{L^N(\Omega)}L^N(A)$. $\mu$ is a probability measure, so $\int_\Omega d \mu = 1$.
The key point of the proof is the existence of a subdifferential for a convex function.
Let $\overline{f} = (\overline{f_1},...,\overline{f_N})$, $f(x) = (f_1(x),...,f_N(x))$. Note that $\overline{f} = \int_\Omega f(x) d \mu(x)$
Since $\phi$ is convex, it has a subdifferential at $\overline{f}$, ie, there exists $\xi \in \mathbb{R}^N$ such that $\phi(f) \geq \phi(\overline{f})+ \langle \xi, f-\overline{f} \rangle$ for all $f \in \mathbb{R}^N$. Then you have
\begin{eqnarray}
\int_\Omega \phi(f(x)) d \mu(x) & \geq & \int_\Omega \phi(\overline{f}) d \mu(x) + \int_\Omega \langle \xi, f(x)-\overline{f} \rangle d \mu(x) \\
&=& \phi(\overline{f}) \int_\Omega  d \mu(x) + \langle \xi, \int_\Omega f(x) d \mu(x)-\overline{f} \rangle \\
&=& \phi(\overline{f})
\end{eqnarray}
(Note: If $\phi$ is differentiable at $\overline{f}$, then you can take $\xi = \nabla \phi(\overline{f})$, otherwise you can use the Hahn Banach theorem to show that there is a hyperplane 'separating' $(\overline{f}, \phi(\overline{f}))$ from $\text{epi } \phi$, and let $\xi$ be the 'domain' component of the hyperplane.)
