Projection operator: why squared norm of the sum of them is equal (or smaller) than the sum of the squared norms?

Question

I am working through the proof of Lemma 2 in this paper (page 25, need it for my own research) and I am stuck at the very first step. Here, I will formulate a bit simplified version of this step.

Suppose we have a causal and stationary error process $\{\mathbf{X}_t\}_{t}$ which can be represented as $ \mathbf{X}_{t} = \mathbf{H}(\ldots,u_{t-1},u_{t})$ with $u_{t}$ being i.i.d. random variables and $\mathbf{H} := (H_{1}, H_{2}, \ldots, H_{d})^\top: \mathbb{R}^\mathbb{Z} \rightarrow \mathbb{R}^d$ being a measurable function such that $\mathbf{H}(\ldots,u_{t-1},u_{t})$ is well defined. Denote $\mathcal{I}_{t} = (\ldots,u_{t-2},u_{t-1}, u_{t})$.

Define a projection operator as \begin{align*} &\mathcal{P}_{t}(\cdot) := \mathbb{E}[\cdot|\mathcal{I}_{t}] -\mathbb{E}[\cdot|\mathcal{I}_{t-1}], \end{align*}

We are interested in the following quantity:

\begin{align*} \kappa_{s}^{\mathcal{P}} := \frac{1}{T}\sum_{t=1}^T \mathcal{P}_{t-s}( \mathbf{X}_{t}), \end{align*} or, more precisely, in $||\kappa_{s}^{\mathcal{P}}||^2$ where $||\cdot||$ denotes a usual $L^2$ norm.

The author states that \begin{align} ||\kappa_{s}^{\mathcal{P}}||^2 &= \Big|\Big| \frac{1}{T}\sum_{t=1}^T \mathcal{P}_{t-s}( \mathbf{X}_{t}) \Big|\Big|^2 \\ &\leq \frac{1}{T^2} \sum_{t=1}^T \Big|\Big| \mathbb{E} \big(\mathbf{X}_{t}|\mathcal{I}_{t-s}\big) -\mathbb{E} \big(\mathbf{X}_{t}|\mathcal{I}_{t-s-1}\big) \Big|\Big|^2, \tag{1}\label{eq1} \end{align} and I just can't get my head around it. It is not neither a triangular inequality, nor Jensen's inequality, and I just do not know how to get there.

Apparently, this type of proof strategy is quite common in the literature, but the only explanation I could find is something like the one here (the end of the proof of Lemma 3, page 528), where they state that $\mathcal{P}_{t-s}(\Delta_t) $, $1 \leq t \leq T$, are martingale differences, hence, the inequality I am interested in is actually an equality. But as I am not an expert in stochastic processes, I also feel a little lost with this reasoning...

To summarise, my questions are as follows:

1. Can somebody explain me how the author got \eqref{eq1}?
2. If it is the consequence of the series of projection being a martingale difference sequence, are there any restrictions on the process $\mathbf{X}_t$ such that $\mathcal{P}_{t-s}(\mathbf{X}_t)$ is one?
3. This one is probably very simple, but still. Suppose $\mathbf{X}_t$ satisfies some property (from 2.) such that $\mathcal{P}_{t-s}(\mathbf{X}_t)$ is a martingale difference sequence. Then how can we get $\big|\big| \frac{1}{T}\sum_{t=1}^T \mathcal{P}_{t-s}( \mathbf{X}_{t}) \big|\big|^2 = \frac{1}{T^2} \sum_{t=1}^T\big|\big| \mathcal{P}_{t-s}( \mathbf{X}_{t}) \big|\big|^2$? Which of the properties for mds we should use in order to obtain this equality?

Answer 1

Will reference the book Stochastic Limit Theory (SLT), 2021, Davidson.

Assuming $E[|X_t|]<\infty$ then $\mathcal{P}_{t-s}(X_t)$ is a martingale difference sequence in $t$ (pg 316 SLT).

This means that by Theorem 16.6 (SLT) that $$\text{Cov}(\mathcal{P}_{t-s}(X_t),\mathcal{P}_{u-s}(X_u))=0 \;\;\; \text{for $u\neq t$}.$$

Also $E[\mathcal{P}_{t-s}(X_t)]=E\left[E[X_t|\mathcal{I}_t]-E[X_t|\mathcal{I}_{t-1}] \right]=E[X_t]-E[X_t]=0$.

Then we can just write $$\begin{align} ||\sum_{t=1}^T\mathcal{P}_{t-s}(X_t)||^2&=\text{Cov}\left(\sum_{t=1}^T\mathcal{P}_{t-s}(X_t),\sum_{t=1}^T\mathcal{P}_{t-s}(X_t)\right)\\ &= \sum_{t=1}^T \text{Cov}\bigg(\mathcal{P}_{t-s}(X_t),\mathcal{P}_{t-s}(X_t)\bigg)\\&=\sum_{t=1}^T ||E[X_t|\mathcal{I}_{t-s}]-E[X_t|\mathcal{I}_{t-s-1}]||^2\end{align}$$