I have a (potentially) infinite real sequence $\{b_j\}$, $j={0,1,2...}$, which I know is square-summable, with $b_0 \neq 0$. For a fixed positive integer K, I have a set of restrictions, $$\sum_{i=0}^\infty b_{i+k} b_i = 0,$$ for all $k\geq K$.

If the sequence were known to be finite, with a last potential non-zero term, $b_T$, I could prove that the terms $b_K = b_{K+1} = ...=b_T = 0$, by solving backwards from the last restriction, $b_{T} b_0 = 0$. I conjecture the result can be extended to the infinite case, i.e. $b_K = b_{K+1} = ... = 0$, but seek the proof.

Thank you!

  • $\begingroup$ What about the functional version ? If $f\in L^2$ is such that $\forall k, \int f(t)f(t-k) dt$ then $f=0$ ? $\endgroup$ – Gabriel Romon Aug 22 '17 at 6:41
  • $\begingroup$ I would be perfectly happy to establish this in the function space. I arrived at this question from a time-series problem in the function space (i.e. using z-transforms), but could not prove it there. I thought I could get more traction in the sequence space, but not so far. $\endgroup$ – Ryan C Aug 22 '17 at 13:11

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