A number sequence problem involving binomial transform

Question

Let $\{b_n\}_{n\geq0}$ be a sequence such that $b_nb_{n+1}=0$ and define $$a_n:=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}b_k.$$ If $\lim_{n\to\infty}a_n=0$, can we conclude that $b_n=0$ for all $n$?

More generally, if $\{b_n\}_{n\geq0}$ is a sequence with infinitely many zeros and $\lim_{n\to\infty}a_n=0$, can we still conclude that $b_n=0$ for all $n$?

Finally, what if we only assume $\{b_n\}_{n\geq0}$ contains at least ONE zero?

Till now, only the original question remains unsolved.

Remark: This question arised from the computation of the $K_0$ groups of the smooth noncommutative $\mathbb{R}^{2n}$ which come from noncommutative field theory, see $K_0$ groups of noncommutative $\mathbb{R}^{2n}$. I showed that the $K_0$ groups of original noncommutative $\mathbb{R}^{2n}$ are all $\mathbb{Z}$. Then mimicking the smooth noncommutative tori, I construct the smooth noncommutative $\mathbb{R}^{2n}$ and I want to show that the $K_0$ groups of the smooth cases are still all $\mathbb{Z}$. But this brings many new problems. First I consider a special class of the projectors of smooth noncommutative $\mathbb{R}^{2}$, and show that if this conjecture is true, then this class of projectors are 0 or 1. The other two questions I asked here are also related to the characterization of projectors of smooth noncommutative $\mathbb{R}^{2n}$.

Answer 1

Unfortunately the argument that I originally posted contained a gap at the end. The gap is explained at the end of the proof, where I also state 3 partial results.

Let $$f(z)=\sum_{n=0}^\infty b_nz_n/n!,\quad g(z)=\sum_{n=0}^\infty a_nz^n/n!.$$ Then your relation between $a_n$ and $b_n$ means $$g(z)=e^{-z}f(z).$$ Suppose that $a_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform $$F(z)=\sum_{n=0}^\infty b_n z^{-n-1}$$ is analytic outside of this disk $D$. Thus the power series $$F(1/z)=\sum_{n=0}^\infty b_nz^{n+1}$$ has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since your condition that $b_nb_{n+1}=0$ implies that the density of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have a singularity on every arc of the circle of convergence which is bigger than semi-circle.

This argument does not work when $F(1/z)$ has infinite radius of convergence.

1. To exclude this case, one may assume that $|a_n|$ tends to zero with geometric speed, that is $|a_n|=O(\delta^n)$ for some $\delta\in(0,1)$. Then $0\not\in K$, and $F(1/z)$ is not entire.

2. And of course, if the condition $b_nb_{n+1}=0$ is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.

3. The argument above also shows that under the stated conditions one can conclude that $b_n=O(\epsilon^n),\forall \epsilon>0$.

Thus my argument needs either a stronger assumption (1) or (2), or leads to a weaker conclusion (3).

Reference.

L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2 (Chap. 1 also contains all other facts used in this argument).

Answer 2

Perhaps it is useful to start with the final question and work backwards - the following only answers the final question, but I could well imagine the idea could be extended to the next-to-final question. In the case of the final question, the answer is, no, we can't conclude $b_k =0$ for all $k$. A counterexample can be readily constructed:

Consider $b_k $ of the form $b_k = st^k $. Then, we just have a binomial sum, $a_n = s(-1+t)^n \rightarrow 0 $ for $n\rightarrow \infty $ if $0<t<2$ (and $t\neq 1$ in order to avoid $0^0 $). Now, we can simply combine two such cases such that there is a cancellation of the coefficients for some specific $k$. Consider $$ b_k = st^k - s^{\prime } t^{\prime \, k} $$ in which case $$ a_n = s(-1+t)^n - s^{\prime } (-1+t^{\prime } )^n $$ with $a_n \rightarrow 0 $ for $n\rightarrow \infty $ as long as we respect $0<t,t^{\prime } <2$ (and $t\neq 1$, $t^{\prime } \neq 1$). Now it is simple to arrange $s,t,s^{\prime } , t^{\prime } $ such that $b_k =0 $ for some $k$. For example, if we want $b_2 =0$, we can choose $s=9$, $t=1/3$ and $s^{\prime } =4$, $t^{\prime } =1/2$, i.e., $b_k = 3^{2-k}-2^{2-k} $.

The generalization to a finite set of vanishing $b_k $ is immediate; for example, to have $b_2 =b_3 =0$, use $$ b_k = \left( 3^{2-k} - 2^{2-k} \right) - \frac{10}{3} \left( 5^{2-k} - 4^{2-k} \right) $$ It remains to be investigated whether this type of scheme can be extended to an infinite set of vanishing $b_k $ while retaining a nontrivial, finite structure of $b_k $ overall.