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Let $a_1,a_2,\ldots,a_n\geq 1$, and let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$. Consider the sum

$$S(f)=\sum_{i}f(a_i)-\sum_{i<j}f(a_i+a_j)+\sum_{i<j<k}f(a_i+a_j+a_k)-\cdots+(-1)^{n-1}f(a_1+\cdots+a_n).$$

This question shows that if $f(x)=\frac1x$, then $S(f)>0$ for all $a_1,\ldots,a_n$. If we perturb $f$ a tiny bit, say $f(x)=\frac{1}{x}-\frac{1}{100x^{100}}$, I would imagine that $S(f)>0$ still always holds. But the proof method for $f(x)=\frac1x$ is hard to generalize to other functions. Can we prove it in some other way?

More generally, is there a theorem out there stating sufficient conditions under which $S(f)>0$ always holds?

Denis Serre
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nan
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    The proof for $1/x$ may be directly generalized to Laplace transforms of non-negative measures supported on non-negative reals: $f(a)=\int e^{-at} d\mu(t)$ – Fedor Petrov Jun 30 '15 at 12:22
  • Is this inspired by the recent Popoviciu question http://mathoverflow.net/questions/210350 ? I remember seeing some criteria involving $n$-th derivatives, but I'm not sure if I still can find them. – darij grinberg Jun 30 '15 at 15:16
  • I think your condition is what is called Condition $\left(C_{n,n,n-1}\right)$ in Corollary 6.12 of Pecaric, Proschan, Tong, Convex Functions, Partial Orderings, and Statistical Applications ( https://books.google.de/books?id=rCAOFpic7AkC&pg=PA177&lpg=PA177&dq=superadditive+of+n-th+order&source=bl&ots=5CN5K6Tc5w&sig=xb_5THO0cR-yNE7G2PeaLcll5Hs&hl=en&sa=X&ei=JrOSVavLOKaaygO72ZKIBw&redir_esc=y#v=onepage&q=superadditive%20of%20n-th%20order&f=false ). I am not fully sure, though, since I might be misreading the $\cdots$ in the formula. – darij grinberg Jun 30 '15 at 15:21
  • I expect as Darij that it could be inspired, and so, thanks a lot nan. I had no idea that this actually holds for $x \mapsto \frac{1}{x}$. Actually the reason why I've asked about Popoviciu generalization, is because I need to prove the following inequality http://mathoverflow.net/questions/210236/inequality-in-a-shape-of-inclusion-exclusion-formula and its generalization for arbitrary number of elements. And since the function I use is basically a fraction, then the above fact seems to be most likely applicable to my problem. – Marek Adamczyk Jul 03 '15 at 20:55
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    This paper: http://unix.cc.wmich.edu/~ledyaev/Spring2013/sendov.pdf studies such functions (and the CM case as noted by Fedor is discussed, as are additional properties such as convexity, harmonic convexity, etc.) – Suvrit Apr 02 '17 at 03:31

2 Answers2

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If $f$ is a polynomial with $\deg f< n$, then $S(f)=f(0)$.

ADDED. More generally, for an analytic function $f(x)$, $$S(f) = f(0) - h(x),$$ where $h(x)$ is the Hadamard product of $f(x)$ and the function $$g(x) := \int_0^{\infty} e^{-t} (1-e^{a_1tx})\cdots(1-e^{a_ntx})\,\mathrm{d}t.$$ It can be seen that $[x^k]\ g(x)=0$ for $k<n$, implying the above result for polynomials.

Max Alekseyev
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The inclusion-exclusion principle of probability/volume implies that $S(f)$ will be positive if for each $a_1, \ldots, a_n$, there can be found subsets $K_1, \ldots, K_n$ (of some measure space) such that $f(a_1 + \cdots + a_k)$ is the measure of $K_1 \cap \cdots \cap K_k$ for each $k \leq n$.

In particular, $S(f) > 0$ if $f$ maps sums into products, and $\int_{\mathbb{R}^+} f(x)dx = 1$, e.g. $f(x) = e^{-x}$.

pinaki
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