Exercise 14.3: Sharper Rates via Entropy Integrals
Note that $f \mapsto \frac1n \sum_{i=1}^n \e f(x_i)$ is a sub-Gaussian process with respect to the metric $d(f, g) = \norm{f - g}_n$. Therefore, \begin{equation} \E\sbrac{ \sup_{f \in \F}\,\frac1n \sum_{i=1}^n \e f(x_i) } \lesssim \frac1{\sqrt{n}}\int_0^D \sqrt{\log N(u, \F, \norm{\vardot}_n)} \isd u \end{equation} where \begin{equation} \F = \set{f_\theta(x) = \lra{\phi(x), \theta}:\theta \in \R^3,\, \norm{f}_\infty \le 1} \end{equation} and $D = \sup_{f,g \in \F} = \norm{f - g}_n \le 2$. Note that $\theta \mapsto \phi(X) \theta$ is an isomorphism between $\R^3$ and $\col(\phi(X))$ with the norm $\norm{\phi(X) \theta}_n = \norm{f}_n$. Hence \begin{align} N(u, \F, \norm{\vardot}_n) &\le N(u, \set{\theta \in \R^3 : \norm{\lra{\phi(\vardot),\theta}}_\infty \le 1}, \norm{\phi(X)\vardot}_n) \newline &\le N(u, \set{\theta \in \R^3 : \norm{\phi(X)\theta}_\infty \le 1}, \norm{\phi(X)\vardot}_n) \newline &\le N(u, \set{\theta \in \R^3 : \norm{\phi(X)\theta}_n \le 1}, \norm{\phi(X)\vardot}_n) \newline &\le 3 \log(1 + 2 / u), \end{align} so the entropy integral is finite and independent of $n$.