I' struggling with an argument in van der Vaart proof of theorem 5.21 at p.52:
I don't understand when he says
[...] $\mathbb{G}_n \psi_{\hat{\theta}_n}- \mathbb{G}_n \psi_{\theta_0}\rightarrow_p 0$
For a nonrandom sequence $\hat{\theta}_n$ this is immediate from the fact that the means of these variables are zero, while the variances are bounded by $P ||\psi_{\hat{\theta}_n}- \psi_{\theta_0}||^2$.
Suppose parameters are uni-dimensional, hence $||\psi_{\hat{\theta}_n}- \psi_{\theta_0}||^2=|\psi_{\hat{\theta}_n}- \psi_{\theta_0}|^2=(\psi_{\hat{\theta}_n}- \psi_{\theta_0})^2$.
I understand that $\mathbb{E}_P(\mathbb{G}_n \psi_{\hat{\theta}_n}- \mathbb{G}_n \psi_{\theta_0})=0$
I don't understand why the variance is bounded by the expression above. In fact,
$$ Var(\mathbb{G}_n \psi_{\hat{\theta}_n}- \mathbb{G}_n \psi_{\theta_0})= Var(\sqrt{n} (\mathbb{P}_n\psi_{\hat{\theta}_n}-\mathbb{E}_P(\psi_{\hat{\theta}_n}(X_i)) - \mathbb{P}_n \psi_{\theta_0}))= nVar(\mathbb{P}_n\psi_{\hat{\theta}_n}- \mathbb{P_n \psi_{\theta_0}})=n\mathbb{E}_P(\frac{1}{n}\sum_{i=1}^n (\psi_{\hat{\theta}_n}(X_i)- \psi_{\theta_0}(X_i)))^2=\frac{1}{n}\mathbb{E}_P(\sum_{i=1}^n (\psi_{\hat{\theta}_n}(X_i)- \psi_{\theta_0}(X_i)))^2 $$
How can this be less than $\mathbb{E}_P(\psi_{\hat{\theta}_n}(X_i)- \psi_{\theta_0}(X_i))^2$