# Weak version of Karamata's Tauberian theorem

I first posted this on mathematics. However, I got no answer there and it seems adapted here too. Also, it seems to be harder than I first thought.

Karamata's Tauberian theorem states the following. Let $$A(z)=\sum a_nz^n$$ be a power series with non-negative coefficients $$a_n$$ and radius of convergence 1. Let $$\beta>0$$. Then, as the real variable $$s\in [0,1]$$ tends to 1, $$\sum_{n\geq 0}a_ns^n\underset{s\to 1}{\sim} c/(1-s)^\beta$$ if and only if $$\sum_{k=0}^na_k\underset{n\to \infty}{\sim}c'n^{\beta}$$, where $$c$$ and $$c'$$ are determine each other. Moreover, if $$a_n$$ is non-increasing, then one can replace $$\sum_{k=0}^na_k$$ with $$a_n$$, that is, we have $$a_n\sim c''n^{\beta-1}$$. See for example Corollary 1.7.3 in Bingham, Goldie and Teugel's book Regular variation.

I'm wondering if the following is true. For two functions $$f$$ and $$g$$, write $$f\asymp g$$ if there exists $$C\geq 0$$ such that $$f\leq Cg$$ and $$g\leq Cf$$. Let $$A(z)=\sum a_nz^n$$ be a power series with non-negative coefficients $$a_n$$ and radius of convergence 1. Then, $$\sum_{n\geq 0}a_ns^n \asymp 1/(1-s)^\beta$$ for $$s\in [0,1]$$ if and only if $$\sum_{k=0}^na_k\asymp n^{\beta}$$ if and only if $$a_n\asymp n^{\beta-1}$$. The implicit constants are asked not to depend on $$s$$ and $$n$$ respectively.

• Surely there are typos in the statements of Karamata's theorem: you mean $A(s) \sim c/(1-s)^{\beta}$ (twice)? – David Handelman Mar 1 '20 at 23:53
• @DavidHandelman Sure, thanks for noticing ! I edited accordingly – M. Dus Mar 2 '20 at 8:52
• The asymptotics $a_n \approx n^{\beta}$ and $\sum_{k=0}^n a_k \approx n^{\beta}$ sound inconsistent; $a_n$ should behave like $n^{\beta-1}$. Also, if $(a_n)$ is decreasing, then $\beta$ can be at most 1. – Giorgio Metafune Mar 2 '20 at 9:29
• I do not have access to Bingham–Goldie–Teugels book right now, but did you check in the chapter on "$O$-regular variation" (Chapter 2, I believe)? At least in the continuous case (the Laplace transform instead of the power series), both Karamata's Tauberian theorem and monotone density theorem have their couterparts for the "$O$-regular variation" (that is, roughly speaking, with "$\sim$" replaced by "$\asymp$"). – Mateusz Kwaśnicki Mar 2 '20 at 20:07
• @GiorgioMetafune Of course you're right, thank you very much. I'm particuarly interested in the case where $\beta <1$. – M. Dus Mar 3 '20 at 15:10

This seems to follow easily from de Haan–Stadtmüller Theorem; see Theorem 2.10.2 in the Bingham–Goldie-Teugels book:

Theorem: Let $$U$$ be non-decreasing, and vanish on $$(-\infty, 0)$$. The following are equivalent:

(i) $$U \in OR$$;

(ii) $$\hat{U}(1/\cdot) \in OR$$;

(iii) $$\hat{U}(1/t) \asymp U(t)$$ ($$t \to \infty$$).

Here $$\hat{U}(\lambda) = \int_{[0, \infty)} e^{-\lambda x} U(dx)$$ is the Laplace–Stieltjes transform of the measure $$U(dx)$$, and $$OR$$ stands for the class of $$O$$-regularly varying functions at infinity, that is, positive functions $$f$$ such that $$\limsup_{x \to \infty} \frac{f(\lambda x)}{f(x)} < \infty$$ for every $$\lambda \in (0, \infty)$$.

Now suppose that $$A(z) = \sum_{n = 0}^\infty a_n z^n$$ with $$a_n \geqslant 0$$. If we let $$U(x) = \sum_{k = 0}^n a_k$$ for $$x \in [n, n+1)$$, then $$\hat{U}(\lambda) = \sum_{n = 0}^\infty a_n e^{-\lambda n} = A(e^{-\lambda}) .$$ Thus the above theorem reads as follows:

Corollary: With the above notation, the following are equivalent:

(i) $$\sum_{k = 0}^{\lfloor \cdot \rfloor} a_k \in OR$$;

(ii) $$A(e^{-1/\cdot}) \in OR$$;

(iii) $$A(e^{-1/t}) \asymp \sum_{k = 0}^{\lfloor t \rfloor} a_k$$ ($$t \to \infty$$).

Clearly, if $$\sum_{k = 0}^n a_k \asymp n^\beta,$$ then (i) holds, and by (iii) we have $$A(e^{-1/t}) \asymp t^\beta.$$ On the other hand, if $$A(s) \asymp (1 - s)^{-\beta}$$ as $$s \to 1^-$$, then (ii) is satisfied, and by (iii) we have $$\sum_{k = 0}^n a_k \asymp (1 - e^{-1/n})^{-\beta} \asymp n^\beta.$$

• Well thank you very much. I was not aware of this result and I should have looked more carefully at Bingham–Goldie-Teugels book first. – M. Dus Mar 3 '20 at 15:19
• You are welcome. The Regular variation book is very useful, but I always have problems finding the result I need there. – Mateusz Kwaśnicki Mar 3 '20 at 20:18

This is too long for a comment but I did not check all the details. I think that the solution can be found following the arguments in the book of Titchmarsh "The theory of functions" pp. 224 and following. First of all, the series expansion of $$(1-x)^{-\beta}$$ has positive coefficients $$b_n \approx n^{\beta-1}$$. Therefore, if $$a_n \approx n^{\beta-1}$$, then $$F(x) \approx (1-x)^{-\beta}$$. More generally, the same holds if $$s_n:=\sum_{k=0}^n a_k \approx n^\beta$$, using $$F(x)=(1-x)\sum_{n=1}^\infty s_n x^{n-1}$$. The converse, that is going from $$F$$ to $$(s_n)$$, should follow from the preceeding consideration by adapting the explanations in 7.52, where the author shows how to obatin weaker forms of Karamata's theorem with $$\beta=1$$, using more elementary tools .

• Nice answer. Thank you very much. – M. Dus Mar 3 '20 at 15:21