Covering numbers for products of functions from two spaces?

Question

Exercise (HW1): Let $\mathcal{F}$ and $\mathcal{G}$ be classes of measurable function. Then for any probability measure $Q$ and any $1 \leq r \leq \infty$, (i) $N_{[]}\left(2 \epsilon, \mathcal{F}+\mathcal{G}, L_{r}(Q)\right) \leq N_{[]}\left(\epsilon, \mathcal{F}, L_{r}(Q)\right) N_{[]}\left(\epsilon, \mathcal{G}, L_{r}(Q)\right)$; (ii) provided $\mathcal{F}$ and $\mathcal{G}$ are bounded by 1 , $$ N_{[]}\left(2 \epsilon, \mathcal{F} \cdot \mathcal{G}, L_{r}(Q)\right) \leq N_{[]}\left(\epsilon, \mathcal{F}, L_{r}(Q)\right) N_{[]}\left(\epsilon, \mathcal{G}, L_{r}(Q)\right) $$

This is the question. I need to proof part (ii) for my work.Assume those Ns are coverning numbers and not bracketing numbers as he considers

Reference. Question is taken from page 21,2nd HW1 part (ii)

I am getting no clue. Can anyone help me in this?

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$2.2$ Bracketing numbers Let $(\mathcal{F},\|\cdot\|)$ be a subset of a normed space of real functions $f: \mathcal{X} \rightarrow \mathbb{R}$ on some set $\mathcal{X}$. We are mostly thinking of $L_{r}(Q)$-spaces for probability measures $Q$. We shall write $N\left(\varepsilon, \mathcal{F}, L_{r}(Q)\right)$ for covering numbers relative to the $L_{r}(Q)$-norm $\|f\|_{Q, r}=\left(\int|f|^{r} d Q\right)^{1 / r}$. Definition $2.9$ ( $\varepsilon$-bracket). Given two functions $l(\cdot)$ and $u(\cdot)$, the bracket $[l, u]$ is the set of all functions $f \in \mathcal{F}$ with $l(x) \leq f(x) \leq u(x)$, for all $x \in \mathcal{X}$. An $\varepsilon$-bracket is a bracket $[l, u]$ with $\|l-u\|<\varepsilon$.

Definition 2.10 (Bracketing numbers). The bracketing number $N_{[]}(\varepsilon, \mathcal{F},\|\cdot\|)$ is the minimum number of $\varepsilon$-brackets needed to cover $\mathcal{F}$.

Definition $2.11$ (Entropy with bracketing). The entropy with bracketing is the logarithm of the bracketing number.

In the definition of the bracketing number, the upper and lower bounds $u$ and $l$ of the brackets need not belong to $\mathcal{F}$ themselves but are assumed to have finite norms.

Answer 1

Given two functions $l(\cdot)$ and $u(\cdot)$, the bracket $[l, u]$ is the set of all functions $f \in \mathcal{F}$ with $l(x) \leq f(x) \leq u(x)$, for all $x \in \mathcal{X}$. An $\varepsilon$-bracket is a bracket $[l, u]$ with $\|l-u\|<\varepsilon$.

If $[l,u]$ is a $2\epsilon$-bracket, then it is contained in the $\|\cdot\|$-ball of radius $\epsilon$ centered at $(l+u)/2$, since $l \le f \le u$ implies $$\|f - (l+u)/2\| \le \frac{1}{2} \|f-l\| + \frac{1}{2} \|f - u\| \le \|u-l\| = \epsilon.$$

Hence $N(\epsilon,\cal F,||\cdot||)\leq N_{[]}(2\epsilon,\cal F,||\cdot||). $

This means a set of $2\epsilon$-brackets covers $\cal F$, then this set is also a set of balls of radius $\epsilon$ that can cover $\cal F$.

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For any $f\in \mathcal{F}$ and $g\in\mathcal{G}$, we can find $\epsilon-$bracket $[l_1, u_1]$ and $[l_2,u_2]$ contain $f$ and $g$ respectively.

And the largest distance is \begin{align*} ||u_1u_2 - l_1l_2||_{Q,r} & = ||u_1u_2 - l_1u_2 + l_1u_2 - l_1l_2||_{Q,r}\\ & \le ||u_1u_2 - l_1u_2||_{Q,r} + ||l_1u_2 - l_1l_2||_{Q,r}\\ &\le 2\epsilon \end{align*}

So, $$ N_{[]}(2\epsilon,\mathcal{F}\cdot\mathcal{G},L_{r}(Q))\leq N_{[]}(\epsilon,\mathcal{F},L_{r}(Q))N_{[]}(\epsilon,\mathcal{G},L_{r}(Q)). $$

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Hence we can write,

So, $$ N(\epsilon,\mathcal{F}\cdot\mathcal{G},||\cdot||)\leq N(\frac{\epsilon}{2},\mathcal{F},||\cdot||)N(\frac{\epsilon}{2},\mathcal{G},||\cdot||). $$

N.B:-

Sincere thanks to these answers

1. Bracketing numbers for products of functions from two spaces

2. bracketing-number-vs-covering-number

[I am new to this field, requesting experts to check the proof and give the feedback,TIA]