Hence, $\mathcal{X}{1}(t, b) \in L{2, \mathcal{F}}\left(C\left([0, T] ; \mathbf{M}{\infty, \varpi}\right)\right)$ and we can define $r{2}\left(t, \mathcal{X}{1}(b), \omega, q\right)$ as the unique solution of, driven by the input process $\mathcal{X}{1}(t, b)$. Suppose we have already defined $r_{m}\left(t, \mathcal{X}{m-1}(b), \omega, q\right)$ as (unique) solutions of with empirical measure processes $\mathcal{X}{m-1}(t, b) \in L_{2, \mathcal{F}}\left(C\left([0, T] ; \mathbf{M}{\infty, \varepsilon}\right)\right)$ for $m=1, \ldots n$. As before, the solution $r{n}\left(t, \mathcal{X}{n-1}(b), \omega, q\right.$ satisfies the assumptions of Proposition and we define the empirical measure process of $r{n}\left(t, \mathcal{X}{n-1}(b), \omega, q\right)$ by $$ \mathcal{X}{n}(t, b, \omega):=\int \delta_{r_{n}\left(t, \mathcal{X}{n-1, b}, \omega, q\right)} \mathcal{X}{0, b}(\omega, \mathrm{d} q)
$$
Again implies
$$
E \sup {0 \leq t \leq T} \gamma{ए}^{2}\left(\mathcal{X}{n}(t, b)\right) \leq c{T} E \gamma_{\varpi}^{2}\left(\mathcal{X}_{0, b}\right)
$$
We show by induction that the right-hand side of equals 0 . First of all, note that
$$
\mathcal{X}{0, b{2}} 1_{\Omega_{b_{1}}}=\mathcal{X}{0, b{1}} 1_{\Omega_{b_{1}}},
$$
because both sides equal $\mathcal{X}{0} 1{\Omega_{b_{1}}}$. If
$$
\mathcal{X}{n-1}\left(\cdot, b{2}, \omega\right) 1_{\Omega_{b_{1}}} \equiv \mathcal{X}{n-1}\left(\cdot, b{1}, \omega\right) 1_{\Omega_{b_{1}}},
$$
MATH4078 COURSE NOTES ：
For $\sigma \in \mathcal{O}(d)$ we define the rotation operator $S{\sigma}$ on $\mathbf{H}{0, \varpi}$ by $$ \left(S{\sigma} f\right)(r):=f_{\sigma}(r):=f(\sigma r) .
$$
Similarly, we define the shift operator $U_{h}$ on $\mathbf{H}{0, \varpi}$, where $h \in \mathbf{R}^{d}$, by $$ \left(U{h} f\right)(r):=f_{h}(r):=f(r+h) .
$$
Moreover, we use $f_{\sigma h}(r)$ to denote the image of the composition $S_{a} U_{h}$ :
$$
\left(S_{a} U_{h} f\right)(r):=f_{\sigma h}(r):=f(\sigma(r+h)) .
$$