# 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 .

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Are you asking how to prove Jensen's inequality, or how it relates to your formula? – copper.hat Jan 8 '13 at 19:54
@copper.hat : i would like to know how to prove the above relation . – Theorem Jan 8 '13 at 19:55

(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.)

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I am sure i will take some time to go through the solution . I am dealing with subdifferential for the first time . I will come back with lots of questions i think :D – Theorem Jan 8 '13 at 20:47
OK. The subdifferential is the main thing here. It may be better to assume $\phi$ is differentiable first, then the subdifferential inequality is straightforward to prove. – copper.hat Jan 8 '13 at 20:49
In the third step , how can u move $\phi (\bar f)$ outside the integral ? Does it not depend on $x$ ? and i didn't quite get how the integral of the measure on $\Omega$ is $1$ . – Theorem Jan 9 '13 at 20:09
$\phi(\overline{f})$ is a constant, so it can be moved outside the integral. I defined the measure $\mu$ as a normalized version of $L^N$, the Lebesgue $N$-dimensional measure ($\Omega$ is bounded, and hence has finite measure, and must have non-zero measure for the question to make sense.) – copper.hat Jan 9 '13 at 22:58