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Hello.

I have a research note coming out soon, and I'm stuck showing that a weird kind of function is continuous. I need to it show a new method of bounding exponential growth factors in combinatorial classes.

The function in question is $f: [0,1] \rightarrow (0,\infty)$, which sends a parameter $l \in [0,1]$ to the $z$ value of the unique positive critical point ($P''>0$) of a function $P_{S,l}(z) = \sum_{(i,j) \in S} z^{j\cdot l + i\cdot(1-l)}$, where $S \subset \{ 0,1,-1 \}^2$.

For several different sets $S$, I have numerical experiments supporting the claim that this function is continuous. I've considered trying the $L^2$ norm, but I don't get very far before I'm swamped with unmanageable amounts of output. I'm looking for a shortcut that will give continuity, I'm not really concerned with how refined the bounds are. Any help or references are greatly appreciated.

Cheers, Sam

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    $\begingroup$ I'm missing something. The $\delta_{ij}$ means you are only summing on those $(i,j)\in S$ with $i=j$, in which case the exponent of $z$ is $i$, not depending on $l$. $\endgroup$
    – Pietro Majer
    Commented Dec 11, 2012 at 20:09
  • $\begingroup$ Sorry Pietro, I mean that $\delta_{ij} = 1$ if and only if $(i,j) \in S$, on second look it's actually redundant since I'm only summing over the vectors in $S$. Thanks for helping me catch it :) Sam $\endgroup$
    – Sam
    Commented Dec 11, 2012 at 21:17
  • $\begingroup$ What is the domain of $P_{S,l}$? (Positive reals, I guess) $\endgroup$
    – Pietro Majer
    Commented Dec 12, 2012 at 19:11
  • $\begingroup$ Pietro: Yes, the domain of $P_{S,l}(z)$ is the positive reals. Cheers, Sam $\endgroup$
    – Sam
    Commented Dec 13, 2012 at 17:48

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If you know that P''>0, then the implicit function theorem should be applicable to give you continuity.

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  • $\begingroup$ Oooh, I'll try this. Thanks Michael. $\endgroup$
    – Sam
    Commented Dec 11, 2012 at 21:19
  • $\begingroup$ This would work if I had $P''>0$ on my domain. Unfortunately, I was hasty in asserting this above. I started with a function, for example, $P(z) = z^a + z^b + z^{-a} + z^{-b}$, where $a,b \in \mathbb{Z}$, which does have $P''>0$. But I wanted to be able to vary a parameter, and vary the exponents in the function above with it. This led to the parameterisation above, and I assumed that convexity would be carried with it, but I'm finding it hard to show. I've made plots for different sets $S$ and they all seem like convex functions. Back to the drawing board. $\endgroup$
    – Sam
    Commented Dec 17, 2012 at 22:49

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