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SETUP:

Assume $f(\cdot)$ is continuous and strictly monotone decreasing on $[0,\infty]$, with $f(0)>0$ and $f(\infty)<0$.

Let $x_m$ be the solution of $\frac{1}{m}\sum_{i=1}^{m}f(ix)=0$, where $m$ is some positive integer. (By the above assumption, the solution always exists and is unique.)

QUESTION:

Can someone give a necessary and sufficient condition such that there exists at least one $m$, such that $mx_m\geq(m+1)x_{m+1}$?

SOME COMMENTS:

  1. This is a practical question I encounter in my work recently. It seems rather easy initially but I haven't been able to tackle it. I guess my maths is just deteriorating over years...

  2. The question title might be a bit misleading. But lacking a better summary for the problem, I just put it that way. I apologize for the potential confusion.

  3. For all "regular" functions I tried, the numerical result is always $mx_m<(m+1)x_{m+1}$. However, one can, for sure, construct weirdly shaped $f(\cdot)$ to achieve the reversed inequality. To give an example, let $f(x)$ be a piece-wise linear function that goes through points $(.37,.51)$, $(.5,.5)$, $(.74,.19)$, $(1,0)$, $(1.1,-.3)$, $(1.48,-.4)$, and $(1.5,-.5)$. It is so constructed that $x_3=.5$ and $x_4=.37$, which result in $3x_3=1.5>4x_4=1.48$. (Piece-wise linearity is not important here. One can always interpolate these points to get a sufficiently smooth, differentiable function.)

  4. If easier, $f(\cdot)$ can be assumed to be differentiable on $[0,\infty]$. (But I doubt how much that will help...)

  5. If necessary and sufficient condition is hard to derive, I am willing to see a sufficient condition that is not "too restrictive".

  6. Well, thanks all for help in advance.

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  • $\begingroup$ I believe that you assume $f$ to be continious, you should probably mention this. $\endgroup$
    – Peter Kravchuk
    Apr 24, 2013 at 16:25
  • $\begingroup$ $x_1$ is the root of $f$. If the claim holds for all $m\ge 1$, it is equivalent to $mx_m\le x_1$. This can never be achieved, because it would mean that every summand is non-negative. Where am I wrong? $\endgroup$
    – Peter Kravchuk
    Apr 24, 2013 at 16:40
  • $\begingroup$ Actually, for a continuous strictly decreasing $f$ and for positive integers $n$ and $m$ one always has $m x_m < nm x_{nm}$. $\endgroup$
    – Pietro Majer
    Apr 24, 2013 at 17:30
  • $\begingroup$ Peter: Thanks for the comments. I added the continuity. I do not require the result hold for all $m$, but only some $m$. Pietro: That is an interesting result. I will need to think about it. But I am more interested in the one-step effect, i.e. from $m$ to $m+1$. $\endgroup$
    – bzy
    Apr 24, 2013 at 20:46
  • $\begingroup$ Peter: I rephrased the question so that it now asks existence. Hope there will be no confusion any more. Thanks for the comment! $\endgroup$
    – bzy
    Apr 24, 2013 at 20:48

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