Limit of shifted indexes pairs series

Question

Suppose $\sum_\limits{n = -\infty}^\infty a_n^2 < \infty$. Prove $$\lim_{k\rightarrow \infty} \left|\sum_{n = -\infty}^\infty a_na_{n-k}\right| = 0$$

I tried some inequalities but none seem useful. Also tried to show the sum in $k$ must be finite, but had no success. Any hints?

Answer 1

Here is some intuition. Since $A:=\sum a_n^2$ is finite, both the left and right tail of the series $\sum a_n^2$ tend to zero. When you form the product $a_na_{n-k}$, you're multiplying together (pointwise) a copy of the sequence $a_n$ with a shifted version of itself. But as $k\to\infty$, the overlap between these two gets more and more sparse. This motivates breaking the sum into two parts, one piece focusing on the right tail, the other on the left tail: $$ \left|\sum a_na_{n-k}\right|\le \left|\sum_{n\ge M}a_na_{n-k}\right| +\left|\sum_{n< M}a_na_{n-k}\right| $$ for some fixed $M$ to be determined. Apply Cauchy-Schwarz to each term on the RHS to find: $$ \left|\sum_{n\ge M}a_na_{n-k}\right|^2 \le\sum_{n\ge M}a_n^2\sum_{n\ge M}a_{n-k}^2\le A\sum_{n\ge M}a_n^2\tag1 $$ and $$ \left|\sum_{n< M}a_na_{n-k}\right|^2 \le\sum_{n< M}a_n^2\sum_{n< M}a_{n-k}^2\le A\sum_{n< M}a_{n-k}^2 = A\sum_{n< M-k}a_{n}^2\tag2 $$ You see where this is going: For given $\epsilon>0$, choose $M$ to make (1) smaller than $(\epsilon/2)^2$. Having chosen this $M$, then for all $k$ large enough you can make (2) less than $(\epsilon/2)^2$.

Answer 2

Fix a positive $\varepsilon$ and let $N$ be an integer such that $\sum_{n:\left\lvert n\right\rvert\geqslant N}a_n^2\lt\varepsilon$. Then $$\left|\sum_{n = -\infty}^\infty a_na_{n-k}\right| \leqslant \left|\sum_{n = -N-1}^{N+1} a_na_{n-k}\right| +\left|\sum_{n:\left\lvert n\right\rvert\geqslant N} a_na_{n-k}\right|$$
and using Cauchy-Schwarz inequality, one can see that the second term does not exceed $\sqrt{ \varepsilon \sum_\limits{n = -\infty}^\infty a_n^2}$.

Answer 3

$b_{n}=a_{-n}$, then $(a_{n})\ast(b_{n})(k)=\displaystyle\sum_{n}a_{n}b_{k-n}=\sum_{n}a_{n}a_{n-k}$. We know that $(a_{n})\ast(b_{n})\in c_{0}$ if $(a_{n})\in l^{p}$, $(b_{n})\in l^{q}$ for $1<p,q<\infty$.