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.

$$\sum_{s \in V} \lambda_sf(s) \geq f(\sum_{s \in V} s\lambda_s)$$

$V$ is a set of points and corresponding to each point in $V$, there is a $\lambda_s$ which is a scalar so $\sum_{s \in V} s\lambda_s$ is basically another point in the space, which is a linear combination of these points.

$s \in \mathbb{R}^n$ i.e. each s is a point in n-dimensional hyperspace.
$f$ is a convex function.

In my case it is also true that $V$ is actually the set of all vertices of the unit hypercube.

The property has been used in a proof in context of submodular functions and has been attributed to the convexity of the function f.

Edit :

It is also true that :

$$\sum_{s \in V} \lambda_s=1 \; \text{ and } \lambda_s\ge 0 \; , \; \forall s \; .$$

share|improve this question
1  
If $\lambda_s=0$ then this says that $0\ge f(0)$, which need not be true. –  George Lowther Jun 26 '11 at 17:50
4  
You probably forgot to put in a condition that $\sum_{s \in V} \lambda_s=1$. Which is a convex combination, not just a linear combination. –  Raskolnikov Jun 26 '11 at 18:11

1 Answer 1

up vote 6 down vote accepted

This, if properly formulated, is called Jensen's inequality.

By properly formulated, I mean that only convex combinations are allowed, not any linear combination. Thus, it is required that

$$\sum_{s \in V} \lambda_s=1 \; \text{ and } \lambda_s\ge 0 \; , \; \forall s \; .$$

share|improve this answer
1  
You also need $\lambda_s\ge0$, right? –  George Lowther Jun 26 '11 at 18:33
1  
Indeed, I'll add that. –  Raskolnikov Jun 26 '11 at 18:55
    
Thankyou very much. –  AnkurVijay Jun 27 '11 at 6:08

Your Answer

 
discard

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.