0
$\begingroup$

I have the following lemma that I would like to find a source to cite for. Let $L$ be a subset of $\mathbb Z^d_{>0}$. I would like to claim that the limit $$\lim_{z \to (1,\ldots,1)^-} (\sum_{v \in L} z^v) \prod_{i=1}^d(1-z_i)$$ exists. Here $z^v$ mean $z_1^{v_1}z_2^{v_2}\ldots z_d^{v_d}$, with $v=(v_1,\ldots, v_d)$ a vector in $\mathbb Z^d$.

For the 1-dimensional case (d=1), an argument using Cesaro sum (http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation) will show that the limit exists, and I believe it can be extended to higher dimension as well, though it can be very messy, and I would like to avoid that. Does anyone know a cleaner method (or even better, a source to cite directly from) to solve this problem? Thanks in advance.

$\endgroup$
  • $\begingroup$ What does $z^v$ mean, when $z$ and $v$ are vectors? $\endgroup$ – Gerry Myerson Apr 24 '14 at 5:33
  • $\begingroup$ Here, $v$ is a vector and $z$ is a polynomial in multivariable. So $z^v$ mean $z_1^{v_1}z_2^{v_2}\ldots z_d^{v_d}$, with $v=(v_1,\ldots, v_d)$ a vector in $\mathbb Z^d$. I have changed the post to add this. Thanks for the comment. $\endgroup$ – sweehong Apr 24 '14 at 5:44
  • 2
    $\begingroup$ Are you sure that the limit exists? Even in case $d=1$, I have doubts. Your limit seems to be related to the asymptotic density of the subset $L$. If for example $L$ is the union of the sets $\{2\cdot 2^n,\dots,3\cdot 2^n-1\}$ ($n\geq 0$), then $L$ does not have a well defined density and some numerical computations I did suggest that your limit does not exist. $\endgroup$ – Andreas Maurischat Apr 24 '14 at 14:18
  • $\begingroup$ Thanks for your reply. I agree with your comment, it may not even be clear for $d=1$, and I find a flaw in my argument using Cesaro sum. It will be interesting to see if the counterexample you gave is indeed a counterexample (I believe it is). $\endgroup$ – sweehong Apr 25 '14 at 16:04
2
$\begingroup$

After my negative comment, at least a positive answer in case d=1:

If the asymptotic density of $L\subset \mathbb{N}$ exists, then the considered limit exists and equals this density. This can be seen as follows: For $n\in\mathbb{N}$, let $L_{\leq n}:=L\cap \{1,\dots, n\}$. Then $$D:=\lim_{n\to \infty} \frac{\#L_{\leq n}}{n}$$ is the asymptotic density of $L$ (assuming it exists).

If $\frac{a}{b}\in\mathbb{Q}$ is a rational number bigger than $D$, then for $n\gg 0$, $\#L_{\leq n}< \frac{an}{b}$. In particular for $n=kb$, there are less than $ak$ numbers in $L$ smaller or equal to $kb$. Therefore, fix a $k\in\mathbb{N}$ big enough, then for all $l\geq k$ and $r\in \{0,\dots, a\}$, the $(al+r)$-th number in $L$, call it $n_{al+r}$, is bigger than $bl+r$.

Hence for all $0<z<1$: $$ \sum_{l=k}^\infty \sum_{r=0}^{a-1} z^{n_{al+r}} \leq \sum_{l=k}^\infty \sum_{r=0}^{a-1} z^{bl+r} = \left(\sum_{l=k}^\infty z^{bl}\right)\left(\sum_{r=0}^{a-1} z^r \right) =\frac{z^{bk}}{1-z^b}\frac{1-z^a}{1-z} $$ and so $$ (1-z)\sum_{n\in L}z^n\leq (1-z)\sum_{j<ak}z^{n_{ak}}+ z^{bk}\frac{1-z^a}{1-z^b} $$ As the last sum tends to $\frac{a}{b}$ as $z$ tends to $1$, one gets $$\limsup_{z\to 1^-} \ (1-z)\sum_{n\in L}z^n\leq \frac{a}{b} $$ for all $\frac{a}{b}>D$. Similarly, for any rational number $\frac{c}{d}<D$, one gets the estimate $$\liminf_{z\to 1^-} \ (1-z)\sum_{n\in L}z^n\geq \frac{c}{d}$$ Hence, the limit exists and is equal to $D$.

$\endgroup$
  • $\begingroup$ I think we also have the converse to be true for $d=1$ i.e. If the asymptotic density of $L$ does not exist, then $\lim_{z\to 1^-} \sum_{v \in L} z^v (1-z) $ does not exist. I am thinking that by using Borel summation (en.wikipedia.org/wiki/Borel_summation), the sum $ \sum_{v \in L} z^v (1-z)= \sum_{k=1}^{\infty} a_kz^k$ is equivalent to $\int_0^{\infty} e^{-t} BA(tz) dt$ for $|z|<1$, with $BA(t)=\sum_{k=1}^{\infty} \frac{a_k}{k!}t^k$. The integral is a continuous function of $z$, and if we choose our series to be nasty enough, we can probably show that the integral diverge for $z=1$. $\endgroup$ – sweehong Apr 25 '14 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.