Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question
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

1 Answer 1

up vote 0 down vote accepted

(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.)

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.