 Solution

Because white-noise observations are not themselves in ${\ell }^2$, define the least-squares estimator as a minimizer of the Gaussian least-squares contrast, equivalently a maximizer over $\theta \in \Theta$ ofwhere $W$ is the isonormal Gaussian process on ${\ell }^2$. The entropy assumption makes $W$ sample-continuous on compact $\Theta$, so a maximizer exists.

Put $\Delta =\hat{\theta }-{\theta }_0$. Comparison with ${\theta }_0$ gives the basic inequalityForthe entropy assumption and the Dudley entropy integral giveThe Borell-TIS inequality further gives

Set $r_n=c{n}^{-4/9}$. This is the balanceOn the shell $2^j{r}_n\le \Vert \theta -{\theta }_0{\Vert }_2<2^{j+1}{r}_n$, the basic inequality would requireFor $c$ sufficiently large, the expectation bound is at most half this threshold for every $j$. Borell concentration then bounds the shell probability bySumming the geometric sequence of shell bounds gives a quantity tending to zero, uniformly in ${\theta }_0\in \Theta$. Therefore