Functions of uniform-mixing processes

Question

Let $\{X_t\}_{t=1}^\infty$ be a stationary process which is uniform mixing with coefficients $\phi_r$ such that $\sum_{r=}^\infty\phi_r<\infty$. Consider the periodogram for $\omega\in[-\pi,\pi]$ $$I(\{X_t\}_{t=1}^T,\omega) = \frac{1}{2\pi T}\left|\sum_{t=1}^TX_te^{-i\omega t}\right|^2.$$ For fixed $T$ and fixed $\omega$, I can see that the periodogram is still $\phi$-mixing because the periodogram is a measurable function $f$ of $(X_1,\ldots,X_T)$, so $$\sigma\left(I\left(\{X_t\}_{t=1}^T\right)\right)\equiv \sigma\left(f\left(X_1,\ldots, X_T\right)\right) \subseteq \sigma\left(X_1,\ldots, X_T\right)$$ and $$\alpha\left(I\left(\{X_t\}_{t=1}^T\right), T\right) \leq \alpha\left(\left\{X_t\right\}_{t = 1}^T, T\right).$$ My questions are:

1. Is what I wrote above correct?
2. Does it hold even if $T\to\infty$?
3. Do I know if $I$ is $\phi-$mixing as a function of $\omega$?

Thanks for your help!

Answer 1

1. Unfortunately, not quite: we talk about the $\alpha$-mixing coefficient either between two $\sigma$-algebras or of a sequence (the mixing coefficient between the past and future), but not of a finite sequence.