Given a vector $ v $ with $ {v}_{n} \in \mathbb{C} $, $ n \in \mathbb{Z} $ and denoting $ {v}_{n}^{\left( k \right)} = {v}_{n - k} $, namely, a shifted vector by $ k $ elements (Mind the vector is infinitely long, Basically, discrete sampled function for that matter).
How could one prove that there exist $ \alpha > 0 $ such that:
$$ \sum_{k = -\infty}^{\infty} {2}^{- \left| k \right| } \left \langle {v}^{\left( 0 \right)}, {v}^{\left( k \right)} \right \rangle \geq \alpha {\left\| v \right\|}^{2} > 0 $$
One could see it a weighted sum of the Auto Correlation Function of the vector $ v $.
The above could be written as:
$$ \sum_{k = -\infty}^{\infty} {2}^{- \left| k \right| } \sum_{n = -\infty}^{\infty} {v}_{n} {v}_{n - k} \geq \alpha {\left\| v \right\|}^{2} $$
Defining $ a \left[ k \right] = {2}^{- \left| k \right|} $.
Moreover, the Auto Correlation function of $ v $ defined as $ {r}_{vv} \left[ k \right] = \left \langle {v}^{\left( 0 \right)}, {v}^{\left( k \right)} \right \rangle = \sum_{n = -\infty}^{\infty} {v}_{n} {v}_{n - k} $.
Pay attention that Auto Correlation is Hermitian Function.
Using the definition of Convolution one could write:
$$ \left( {r}_{vv} \ast a \right) \left[ 0 \right] = \sum_{k = -\infty}^{\infty} {2}^{- \left| k \right| } \left \langle {v}^{\left( 0 \right)}, {v}^{\left( k \right)} \right \rangle $$
Using the Convolution Theorem one could write that:
$$ \left( {r}_{vv} \ast a \right) \left[ 0 \right] = \int_{- \pi}^{\pi} {R}_{vv} \left( \omega \right) A \left( \omega \right) d \omega $$
Where $ R \left( \omega \right) $ and $ {R}_{vv} \left( \omega \right) $ are the DTFT of $ {r}_{vv} \left[ k \right] $ and $ a \left[ k \right] $ respectively.
One should notice the DTFT of $ a \left[ k \right] $ is defined only one sided. Yet since its symmetrical it can well calculated:
$$ \begin{align*} A \left( \omega \right) & = DTFT \left\{ a \left[ k \right] \right\} = \sum_{k = -\infty}^{\infty} a \left[ k \right] {e}^{-j \omega k} = \sum_{k = 0}^{\infty} {2}^{-k} {e}^{-j \omega k} + \sum_{k = 0}^{\infty} {2}^{-k} {e}^{j \omega k} - 1 \\ & = \frac{1}{1 - 0.5 {e}^{-j \omega}} + \frac{1}{1 - 0.5 {e}^{j \omega}} - 1 = \frac{1 - {c}^{2}}{1 - 2 c \cos \left( \omega \right) + {c}^{2}} = \alpha \left( \omega \right) > 0 \quad \forall 0 < c < 1 \end{align*} $$
In the above $ c = {2}^{-1} = 0.5 $ yet actually this will hold for any $ c < 1 $.
Analyzing $ \alpha \left( \omega \right) $ by looking for its minimum value (Maximizing the denominator) over $ \omega \in \left[ -\pi, \pi \right] $:
$$ \alpha \left( \omega \right) = \frac{1 - {c}^{2}}{1 - 2 c \cos \left( \omega \right) + {c}^{2}} \geq \frac{1 - {c}^{2}}{1 + 2 c + {c}^{2}} = \frac{\left( 1 + c \right) \left( 1 - c \right)}{\left( 1 + c \right) \left( 1 + c \right)} = \frac{1 - c}{1 + c} = \alpha $$
So the integral is given by:
$$ \begin{align*} \int_{- \pi}^{\pi} {R}_{vv} \left( \omega \right) A \left( \omega \right) d \omega & = \int_{- \pi}^{\pi} {R}_{vv} \left( \omega \right) \frac{1 - {c}^{2}}{1 - 2 \alpha \cos \left( \omega \right) + {c}^{2}} d \omega \\ & \geq \alpha \int_{- \pi}^{\pi} {R}_{vv} \left( \omega \right) d \omega = \alpha {\left\| v \right\|}^{2} \end{align*} $$
As requested.
By the way the value of $ \left( {r}_{vv} \ast a \right) \left[ 0 \right] $ must be real since $ a \left[ k \right] $ is symmetric and $ {r}_{vv} $ is hermitian function and hence the sum is real.