6
$\begingroup$

Suppose a function $g(x)$ has a convergent power series expansion with real (not necessarily integer or rational) exponents on $[0,1]$ of the form $$ g(x)=\frac{1}{x^\delta}(1+\sum_{i=1}^\infty c_i x^{\Delta_i}) $$ where $\delta>0$, $0<\Delta_1<\Delta_2<\ldots$ and, crucially, $c_i>0$. Suppose further that $$ g(x)=g(1-x) $$ What is known about such functions? Do they occur anywhere in mathematics? (They do occur in theoretical physics, namely in the study of conformal field theories. In fact this condition is a toy model for operator product expansion associativity.)

I can use the Hardy-Littlewood tauberian theorem to show that $c_i$'s should have a certain asymptotic behavior. But this does not use the full strength of the condition $g(x)=g(1-x)$. Can I learn more about the asymptotics, e.g. about the subleading terms, via a smart use of this condition?

I'm also interested in questions which are not asymptotic. E.g. I am interested how large $\Delta_1$ can be for a fixed $\delta$. Expanding around $x=1/2$ to third order I can easily show that $\Delta_1\le 2\delta+3$. I know that this bound is not best possible, since expanding to higher order and doing some numerics I can somewhat improve it. I'm interested in the exact bound on $\Delta_1$ as a function of $\delta$ and which functions $g(x)$ saturate it.

I can also study the question for $\delta$ integer under an extra assumption that all $\Delta_i$'s are integers. E.g. for $\delta=1$ I can find a function for which $\Delta_1= 3$: $$ g(x)=\frac{P(x)+P(1-x)}{2x(1-x)}=x^{-1}(1+2x^3+x^4+x^5+\ldots) $$ where $P(x)=3x^2-x^4$. I can also show that $\Delta_1= 4$ is impossible. But I'm most interested in the case when $\delta$ and $\Delta_i$'s are real.

$\endgroup$
2
  • $\begingroup$ Would you care to elaborate on the comment on OPE associativity? I ask cause in my article arxiv.org/abs/1105.4280 we find exponentials of the dilogarithm as OPE of certain fields from a CFT perspective and they satisfy the identity g(x)=g(1-x). $\endgroup$ Sep 9, 2015 at 10:21
  • $\begingroup$ The full condition for OPE associativity involves conformal blocks of the form $x^\Delta{}_2F_1(\Delta,\Delta,2\Delta,x)$ rather than simple powers $x^\Delta$. $\endgroup$ Sep 12, 2015 at 19:46

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.