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Let $(a_n)$ be a bi-infinite sequence of complex numbers, and suppose we have the equation

$$a_n = C (a_{n+1} + a_{n-1})$$

for some constant $C$ and each $n\in \mathbb Z$. I am trying to prove that $(a_n)$ is not square summable, i.e. $\displaystyle \sum_{n\in\mathbb Z} |a_n|^2 = \infty$. Does anyone have any suggestions?

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    $\begingroup$ The sequence $a_n = 0$ satisfies the recurrence relation but is square summable, but I guess this is a very minor remark. $\endgroup$ – i like xkcd Jan 25 '13 at 23:20
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It doesn't really matter that the sequence is bi-infinite. You can still solve the recursion explicitly. Try to find non-trivial solutions $a_n = r^n$. Inserted into the equation, we get $$ r^n = C(r^{n+1} + r^{n-1}) \Leftrightarrow r = Cr^2 + C. $$

If $C=1$, the quadratic will have a double root $r=1$ and the recursion is solved by $a_n = A + Bn$. For other values of $C$, there will be two distinct roots $r_1$ and $r_2$, and the solution is $a_n = Ar_1^n + Br_2^n$. Note that $r_1r_2 = 1$, so either one of the $r$:s will have modulus bigger than $1$ or both will be of modulus $1$.

In either case you can check that all (non-trivial) solutions fail to to be square integrable, since either $\lim_{n\to\infty} a_n \neq 0$ or $\lim_{n\to-\infty} a_n \neq 0$.

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