Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For which functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ does the inequality

$f^{-1}\left(\sum\limits_{k=1}^n f(x_k+y_k)\right) \leq f^{-1}\left(\sum\limits_{k=1}^n f(x_k)\right) + f^{-1}\left(\sum\limits_{k=1}^n f(y_k)\right)$

hold for any nonnegative real numbers $x_1, \dots, x_n, y_1, \dots, y_n$?

Note that for $f(x)=x^p$, $p \geq 1$, this is the classical Minkowski inequality.

We can assume that $f$ is strictly inreasing to guarantee that $f^{-1}$ exists. Any other natural conditions on $f$ are acceptable. For example: does the above inequality hold for any stricly increasing convex function with $f(0)=0$?

share|improve this question
    
Recall that the Minkowski inequality turns into equality when $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ are proportional. Thus, if you perturb the function $x^p$ a bit in a neighborhood of some point $a$ (keeping monotonicity, convexity, whatnot) then you may set $a=x_n$ or $a=x_n+y_n$ refuting your inequality for the perturbed function. –  Ilya Bogdanov Mar 3 at 19:43

1 Answer 1

The question that you are asking was asked in "On Generalizations of Minkowski's Inequality in the Form of a Triangle Inequality" by F. Mulholland (1949). In that paper, Mulholland established a sufficient condition on $f$, namely that it should satisfy $f(0)=0$, be increasing on $x \ge 0$ and be g-convex, i.e., $\log f(e^x)$ is convex on the reals.

However, Mulholland's condition is not necessary, which seems to have been very recently shown in On functions that solve Mulholland inequality and on compositions of such functions by M. Petrík (2013)---but given the elementary nature of the question, perhaps the non-necessity of the above g-convexity hypothesis of Mulholland has also previously been observed.

share|improve this answer
    
Thank you very much! –  Bogdan Mar 4 at 13:33
    
You're welcome! –  Suvrit Mar 4 at 15:22

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.