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?

  • 1
    $\begingroup$ 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

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 $\lim_{n\to-\infty} a_n \neq 0$.


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.