On Gowers' approach of Green-Tao Theorem ($\mathcal{D}f$s span $L^q(\mathbb{Z}_N)$). I am trying to understand the Gowers' approach to the Green-Tao Theorem, and so far I am doing well.
Although, there is one point that I am not understanding.
Here comes:
Let $f:\mathbb{Z}_N\rightarrow\mathbb{R}$ be an function, and define its dual pointwise as
$$\mathcal{D}f(x) := \mathbb{E}_{x\in\mathbb{Z}_N}\left(\prod_{\begin{matrix}\omega\in\{0,1\}^{k-1}\\\omega\neq(0,0,\ldots,0)\end{matrix}}f(x+\omega\cdot h) : h\in\mathbb{Z}_N^{k-1}\right),$$
where $\omega \cdot h = \omega_1\cdot h_1 + \ldots + \omega_{k-1}\cdot h_{k-1}$.
At some point, it is defined a "basic norm" for $g$ given by
$$||g||_B = \inf\left\{\sum_{j=1}^n |\lambda_j| : g = \sum_{j=1}^n\lambda_j\mathcal{D}f_j\right\}.$$
See, for this to be well-defined, we need the fact that $\mathcal{D}f_j$s span completely the space of functions, and that is the part I am not getting it.
The Gowers' article Decompositions, approximate structure, transference, and the Hahn-Banach theorem (W. T. Gowers) does emphasize this issue at page 33, but only assumes the spanning property for convenience.
Maybe one more idea is worth telling: on the context of the Green-Tao Theorem, the set $X$ from which we take the $f$s and apply the $\mathcal{D}$ operator is $X = \{f: 0\leq f\leq\nu, \nu \text{ is a pseudorandom measure}\}$. (The $\leq$ relation is taken pointwise, for all elements on the domain)
 A: Some time ago I made a proof and I think oportune to post here:
Let $d\geq2$ be an integer, and let us define $f_j = \chi_j :\mathbb{Z}_N\rightarrow\mathbb{R}$, where $\chi_j$ is the characteristic function of the element $j$.
Clearly, these functions are linearly independent.
Now, we are going to analyze the terms of the expected value:
First, we notice that for $h = (0,0,\ldots,0)$, the correspondent term of the dual function is $\prod_{\omega\in\{0,1\}^d, \text{ } \omega\neq 0}f_j(x)$, which results in 0 for any $x\neq j$, since it is the multiplication of at least two characteristic functions. In the case of $x=j$, this multiplication results in $1$, and we don't need to worry about the other terms because they don't affect the positivity of the sum.
So, we can assume $h \neq (0,0,\ldots,0)$ or, equivalently, that there is a $k$ such that if $h = (h_1, h_2,...,h_k,...,h_d)$, then $h_k\neq0$.
Let's consider $\omega = (1,1,...,1)$ and $\tilde{\omega}=(1,1,...,1,0_k,1,...,1)$ (the purpose of the subscript is only to indicate that the $0$ is in position $k$).
The calculations yields $\omega\cdot h=1h_1+1h_2+...+1h_d$ and $\tilde{\omega}\cdot h = 1h_1+1h_2+...+1h_d - h_k = \omega\cdot h - h_k$, and given that $h_k\neq0$, those have different values, so we can conclude that at least one of the values $f_j(x+\omega\cdot h)$ and $f_j(x+\tilde{\omega}\cdot h)$ is equal to $0$, hence $\prod_{\omega\in\{0,1\}^d, \omega\neq 0}f_j(x + \omega\cdot h) = 0$.
We came to the conclusion that $\mathcal{D}f_j(x) > 0$, if $x=j$, and $\mathcal{D}f_j(x) = 0$, if $x\neq j$.
More explicity, $\mathcal{D}f_j(x) = \lambda(x)\chi(x)$, where $\lambda(x)$ is a strictly positive function.
Then, the set $\{\mathcal{D}f_1(x), \mathcal{D}f_2(x), \ldots, \mathcal{D}f_N(x)\}$ is linearly independent, and we are done. $\square$
