In these lecture notes on Statistical Learning Theory we find the following definitions for covering numbers:
Definition. Let $(\mathcal{W}, d)$ be a metric space and $\mathcal{F} \subset \mathcal{W}$. For every $\epsilon>0,$ denote by $N(\epsilon, \mathcal{F}, d)$ the minimal number of open balls (with respect to metric $d$ ) needed to cover $\mathcal{F}$. That is, $N(\epsilon, \mathcal{F}, d)$ is the minimal cardinality of the set $\left\{f_{1}, \ldots, f_{m}\right\} \subset \mathcal{W}$ with the property that for every $f \in \mathcal{F}$ there is some $f_{i}$ such that $d\left(f, f_{i}\right)<\epsilon .$ The set $\left\{f_{1}, \ldots, f_{m}\right\}$ is called an $\epsilon$ -cover of $\mathcal{F}$.
and for uniform covering numbers:
We will be interested in metrics induced by samples. For every sample $\left\{x_{1}, \ldots, x_{n}\right\}$ let $\mu_{n}$ be the empirical measure of the sample. For $1 \leq p \leq \infty$ and a function $f,$ put $$ \|f\|_{L_{p}\left(\mu_{n}\right)}=\left(\frac{1}{n} \sum_{i=1}^{n}\left|f\left(x_{i}\right)\right|^{p}\right)^{1 / p} $$ and in particular, we have $\|f\|_{L \infty\left(\mu_{n}\right)}=\max _{1 \leq i \leq n}\left|f\left(x_{i}\right)\right| .$ Let $N\left(\epsilon, \mathcal{F}, L_{p}\left(\mu_{n}\right)\right)$ be the covering number of $\mathcal{F}$ at scale $\epsilon$ with respect to the norm $L_{p}\left(\mu_{n}\right)$ Theorem 9 - $1 .$ For any class $\mathcal{F}$ of real-valued functions, any sample $S=\left\{x_{1}, \ldots, x_{n}\right\}$ and $\epsilon>0$ $$ N\left(\epsilon, \mathcal{F}, L_{1}\left(\mu_{n}\right)\right) \leq N\left(\epsilon, \mathcal{F}, L_{2}\left(\mu_{n}\right)\right) \leq N\left(\epsilon, \mathcal{F}, L_{\infty}\left(\mu_{n}\right)\right) $$ Definition. For $\epsilon>0,$ define the uniform covering number $$ N_{p}(\epsilon, \mathcal{F}, n)=\sup _{\mu_{n}} N\left(\epsilon, \mathcal{F}, L_{p}\left(\mu_{n}\right)\right) $$
Now assume $x_1, ..., x_n$ are samples of some random variable $x$ and that we addionally observe some random variable $z$ with samples $z_1, ..., z_n$. Let $z$ and $x$ be (jointly) distributed according to $\mu$, and let $\mu_n$ be the corresponding empirical measure. Consider the linear projection $T[f](z_i)=\mathbb{E}[f(x)|z=z_i]$ and more generally the class $T[\mathcal{F}]=\{T[f]=\mathbb{E}[f(x)|z=\cdot]: f\in\mathcal{F}\}$.
Question: Can I bound $\mathcal{N}_2(\epsilon, T[\mathcal{F}], n)\leq \mathcal{N}_\infty(\epsilon, \mathcal{F}, n)$?
It's easy to see that that $\mathcal{N}(\epsilon, T[\mathcal{F}], L_{p}\left(\mu\right))\leq \mathcal{N}(\epsilon, \mathcal{F}, L_{p}\left(\mu\right))$, where $$ \|f\|_{L_{p}\left(\mu\right)}=\left(\mathbb{E}\left|f\left(x\right)\right|^{p}\right)^{1 / p} $$ since we can easily compare $\|T[f]\|_{L_{p}\left(\mu\right)}$ and $\|f\|_{L_{p}\left(\mu\right)}$ via Jensen's inequality. However, $\mathcal{N}(\epsilon, T[\mathcal{F}], L_{p}\left(\mu_n\right))\leq \mathcal{N}(\epsilon, \mathcal{F}, L_{p}\left(\mu_n\right))$ should not be true in general, since $$ \|T[f]\|_{L_{p}\left(\mu_{n}\right)}=\left(\frac{1}{n} \sum_{i=1}^{n}\left|\mathbb{E}[f\left(x\right)|z_i]\right|^{p}\right)^{1 / p} $$ cannot be directly compared to $\|f\|_{L_{p}\left(\mu_{n}\right)}$. Now I wonder whether the supremum in the definition of the uniform covering numbers is enough to make this upper bound work.
