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In a previous post, the following inequality has been proven $${\left( {\sum\limits_{i = 1}^n {{W_i}} } \right)^a} \le \sum\limits_{i = 1}^n {{W_i}^a}$$ where $W_i\gt 0$, $0\lt a\lt 1$. I guess it is more correct to say that this is always greater, and it is valid for $a\gt 0$ not just $0\lt a \lt 1$.

I am trying to see if one can generalize it to something like $$f\left( \sum\limits_{i = 1}^n W_i \right) \le \sum\limits_{i = 1}^n {f({W_i})}$$ where ${W_i}\gt 0$ ?

Under what circumstances and functions can this be true ?

It has been proven for the power function $f(x)=kx^a$ where $a,k\gt 0$.

Are there any other cases? Do you think there is any basic inequality to prove this?

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2 Answers 2

The condition $$f\left( {\sum\limits_{i = 1}^n {{W_i}} } \right) \le \sum\limits_{i = 1}^n {f({W_i})} $$ where ${W_i}$>0 is equivalent to $$f\left( {\sum\limits_{i = 1}^2 {{W_i}} } \right) \le \sum\limits_{i = 1}^2 {f({W_i})} $$ where ${W_i}$>0. These are subadditive functions. See Subadditivity

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I know von Neumann entropy is subadditive, but that's a function of a different kind, defined on a different kind of space. But I think the Shannon entropy has the same property. –  Raskolnikov Nov 26 '10 at 12:07

Many inequalities similar to this one come from Jensen's Inequality.

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Jensen's inequality cannot prove a nontrivial result of this form directly. Jensen's inequality only applies if f is homogeneous of degree 1 and concave, but the only such functions are the linear functions f(x) = ax, and then the inequality is an equality. –  Qiaochu Yuan Nov 26 '10 at 1:20
    
You're right; I expressed myself badly, sorry. I've tried to adjust my answer. –  Daniele A Nov 26 '10 at 11:43
    
Where is it needed that the function be homogenuous in Jensen's inequality? You must have a particular version in mind that I don't know. I thought only convexity (concavity) is needed? Besides, for linear functions, the inequality usually is just an equality. Or are you just refering to how it would apply in the specific case proposed by the initiator of the thread? –  Raskolnikov Nov 26 '10 at 11:49
    
@Raskolnikov: I am referring to how it would apply in the specific case proposed by the OP. –  Qiaochu Yuan Nov 26 '10 at 11:53
    
OK, thanks, my bad. –  Raskolnikov Nov 26 '10 at 12:04

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