# When is alternating sum $\sum_{i}f(a_i)-\sum_{i<j}f(a_i+a_j)+\ldots+(-1)^{n-1}f(a_1+\ldots+a_n)$ always positive?

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

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?

• 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 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 ( books.google.de/… ). 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 mathoverflow.net/questions/210236/… 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 3 '15 at 20:55
• This paper: 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 2 '17 at 3:31

## 2 Answers

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, implying the above result for polynomials.

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. $$f(x) = e^{-x}$$.