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Suppose that we have $q$ positive integers $a_1, \ldots, a_q$ satisfying $a_1 \leq \cdots \leq a_q \leq q$. I'm interested in the possible types of behaviors for the function given by $$f(x) = (a_1^{x-1} + \cdots + a_q^{x-1})^{1/x},$$ where $x \in [2,\infty)$. In particular, I'm interested in the behavior at integer values of $x$ in that range, but I think that the continuous version of $f$ might be easier to handle.

It isn't hard to see that $\lim_{x \to \infty} f(x) = a_q$. I can show that if there is an $x_0 \in (2,\infty)$ with $f(x_0) > a_q$, then $f$ is decreasing at $x_0$, and furthermore I can show that $f$ is either eventually decreasing, increasing, or constant (this almost entirely comes from the number of $i$ with $a_i = a_q$).

Each example $(a_1,\ldots,a_q)$ that I've used also has the nice property that it is either always decreasing, always constant (this in fact only occurs if $q=a_q$ and $a_i = a_q$ for each $i$), or decreasing on $[2,x_0)$ and increasing on $(x_0,\infty)$ for some $x_0$ which depends on the $a_i$. Does anyone know of either a proof or counterexample of this property? Is any other behavior possible? Simply computing the derivative of $f$ doesn't seem to be too helpful, but perhaps I'm missing the correct viewpoint on this derivative.

Edit: added the restriction $a_q \leq q$. Without this restriction, $f(x)$ can be strictly increasing as well, e.g. $a_1=3,a_2=4$. An example of the decreasing/increasing is $a_1=1, a_2=2$.

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I don't have time to think about this properly now, but it seems very likely that it would help to think about power means (a.k.a. generalized means). See e.g. Wikipedia or Hardy, Littlewood and Pólya's text Inequalities. The basic theorem is that the power mean of order $x$ increases with $x$, strictly so unless all the numbers you're taking the mean of are equal. In your case you might want to relate $f(x)$ to the $x$th order power mean of $a_1, \ldots, a_q$, weighted by $a_1^{-1}, \ldots, a_q^{-1}$. – Tom Leinster May 16 '12 at 16:09
I don't have time to think about this carefully either, but one way to prove that a function only changes direction once is to prove that it is log convex. Googling "log convex" and "power mean" turned up , which looks like it might have the tools you need. – David Speyer May 16 '12 at 19:14
Thanks for the suggestions --- I've tried using a H\"{o}lder/Jensen/etc inequality, but haven't found the right thing to use them on yet. @Tom My initial use of the power mean theorem was hurt by the sum of the weights, which also get hit by the $x$th root; I shall keep poking around along these lines. @David Unless I'm missing something here, I don't think that the log convex trick will work, since $f$ isn't necessarily convex (e.g. $a_1=1, a_2=2$). – John Engbers May 17 '12 at 15:41
My suggestion is to show that $\log f$ is convex, not that $f$ is. A plot of $\log f$ looks pretty good^%28x-1%29%29%2Fx&t=mfftb‌​01 – David Speyer May 17 '12 at 17:08
up vote 1 down vote accepted

As @David points out, the function is log-convex -- this is a general fact about $L^p$ norms (as pointed out by @Tom, this is a special case of an $L^p$ norms where the measure of a point $x_i$ is proportional to $1/a_i.$) Not one, not two, but three proofs of this fact is given in Terry Tao's excellent as usual blog post. Since a convex function has at most one minimum, that seems to resolve your conjecture.

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I seem to be missing something here. Lemma 2 that gives the result, but doesn't this state that the function from $1/x$ to $f = (a_1^{x-1} + \cdots + a_q^{x-1})^{1/x}$ is log convex, and so the function $f$ might not be log convex? For example, plotting $g(x) = \log(1+2^{x-1})/x$ from 1.5 to 20 gives a non-convex function (while plotting $g(1/x)$ does appear convex, which is what the lemma tells us).^%28x-1%29%29%2Fx+for+x%3D1‌​.5+to+20 – John Engbers May 17 '12 at 18:44
True, but since $1/x$ is a monotonic function of $x$ and you only care about increase and decrease of your function, that's just as good. – Igor Rivin May 17 '12 at 19:08
Thank you Igor - I've got it now. – John Engbers May 17 '12 at 19:24

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