- #1

Drazick

- 10

- 2

## Homework Statement

Given ##v = {\left\{ {v}_{i} \right\}}_{i = 1}^{\infty}## and defining ## {v}_{n}^{\left( k \right)} = {v}_{n - k} ## (Shifting Operator).

Prove that there exist ## \alpha > 0 ## such that

$$ \sum_{k = - \infty}^{\infty} {2}^{- \left| k \right|} \left \langle {v}^{\left ( 0 \right )}, {v}^{\left ( k \right )} \right \rangle \geq \alpha {\left\| v \right\|}^{2} $$

Basically, a Weighted Sum of the Auto Correlation Functions must be greater then its value at Zero.

Namely, the Negative Values can not be summed to match the positive values.

## Homework Equations

Probably some variant of Cauchy Schwartz inequality.

Maybe taking advantage of the Auto Correlation function being Symmetric.

## The Attempt at a Solution

I understand that behind the scenes the correlations of the shifted versions cant' have negative value which is up to half of the norm of the vector ## v ##.

Yet I couldn't use it to prove the above.

Thank You.