Let $F(z)=\displaystyle \sum_{k=0}^\infty a_kz^k,\;|z|<R $ and $F(R)=\displaystyle \sum_{k=0}^\infty a_kR^k$ (the series converges).

Assume that $F(\alpha_j)=0,\;j=1,2,\dots ,m$, where all $|\alpha_j|<R$, Then $$F(z)=(z-\alpha_1)\dots (z-\alpha_m)\cdot \displaystyle \sum_{k=0}^\infty b_kz^k,\;|z|<R. $$ This is obvious, because $F(z)/[(z-\alpha_1)\dots (z-\alpha_m)]$ is analytic in $|z|<R.$

My question. Does $\sum_{k=0}^\infty b_kR^k$ converge?

The question is related to the research on the $q$-Bernstein polynomials. The relevant information is contained, for example, in S. Ostrovska, The $q$-Versions of the Bernstein Operator: From Mere Analogies to Further Developments, Results in Mathematics, 69(3), 275-295, Section 4.2.


We can normalise $R=1\ $. By induction the question boils down to this: Let $$ F(z)=(z-a)\sum_{n=0}^\infty b_nz^n=-ab_0+\sum_{n=1}^\infty (b_{n-1}-ab_n)z^n. $$ If this series converges for $z=1$, we have to show that $\sum_{n=0}^\infty b_n$ converges.

Let $S_N=\sum_{n=1}^Nb_{n-1}$ and let $T_N=S_N-aS_{N+1}\ $. Then we assume that $T_N$ converges. As $|a|<1$, the series $\displaystyle\sum_{j=0}^\infty a^j$ converges and hence the series $\displaystyle\sum_{j=0}^\infty a^jT_{N+j}\ $ converges. Now $\displaystyle\sum_{j=0}^Ma^jT_{N+j}=S_N-a^{M+1}\ S_{N+M+1}\ \ $. This means that $a^jS_{N+j}\ $ converges for $j\to\infty$. Let $A_N$ denote the limit, then $\displaystyle F_N=\sum_{j=0}^\infty a^jT_{N+j}=S_N-A_N$. Now $$ A_N=\lim_ja^{j+k}\ S_{N+k+j}\ =a^k\lim_ja^j\ S_{N+k+j}\ =a^kA_{N+k}\ \ , $$ so that $A_N=a^{-N}A$ for some $A\in\mathbb C$. We show that $A=0$. For this write $\displaystyle S_N(z)=\sum_{j=1}^Nb_{n-1}\ z^{n-1}$. Then $S_N(a)$ converges to $\displaystyle S(a)=\sum_{n=1}^\infty n_{n-1}\ a^{n-1}$. We have $A=\lim_N a^NS_N$ and $$ a^jS_j-aS_j(a)=\sum_{k=1}^{j-1}b_{n-1}\ (a^j-a^k)=a^jS_{j-1}-aS_{j-1}\ (a). $$ The left hand side converges to $A-aS(a)$ and the right hand side to $aA-aS(a)$. We conclude that $A=aA$ and hence $A=0$.

Therefore $F_N=S_N$. We claim that $F_N$ is a Cauchy sequence. For this consider $$ F_{N+k}-F_N=\sum_{j=0}^\infty a^j\left(T_{N+j}-T_{N+k+j}\ \ \right). $$ As $T_N$ is a Cauchy-sequence, the right hand side becomes arbitrarily small as $N$ increases. Therefore $F_N$ is Cauchy, hence convergent and so is $S_N$.

  • $\begingroup$ I think there is a small misprint: it should be $S_N-a^{M+1}S_{N+M+1} $ in the formula for the sum $\sum_{j=0}^M a^jT_{N+j}.$ $\endgroup$ – Deepti Aug 4 '16 at 6:29
  • $\begingroup$ @Anton: Where does your proof use that $F(a)=0$ ? It is clear that the statement does not hold without this assumption. $\endgroup$ – Alexandre Eremenko Aug 4 '16 at 7:41
  • $\begingroup$ @Alexandre Eremenko: In the definition of $F(z)$. $\endgroup$ – user1688 Aug 4 '16 at 7:56
  • $\begingroup$ @Anton: Does it matter that the limit $A$ depends on $N$? $\endgroup$ – Deepti Aug 4 '16 at 10:00
  • $\begingroup$ @Deepti: it does, however it can be repaired and I did so. $\endgroup$ – user1688 Aug 4 '16 at 12:38

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.