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If $$\alpha_k = \sum_l a_l \ \ g((k-l)T-l\Delta T)$$ $$s_k = \sum_l \alpha_l \ \ q((k-l)T+k\Delta T)$$

where $a_l \in \pm1$ and $g(t) = \frac {\sin(\pi t/T)}{\pi t/T}$ and $q(t) = \frac {\sin(\pi t/(T+\Delta T))}{\pi t/(T+\Delta T)}$,

can we prove $\forall k, s_k =a_k $?

My attempt:

$$s_k = \sum_l \sum_i a_i g((l-i)T-i\Delta T) \ \ q((k-l)T+k\Delta T)$$ $$ \ \ \ =\sum_i a_i \sum_l g((l-i)T-i\Delta T) \ \ q((k-l)T+k\Delta T)$$ If $k=i$, $$=a_k \sum_l g((l-k)T-k\Delta T) \ \ q((l-k)T-k\Delta T)$$ Now, is $\sum_l g((l-k)T-k\Delta T) \ \ q((l-k)T-k\Delta T)=1$? How to proceed for $k\not=i$?

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