of $N$ and $i$.

Denote $$A_{s,t}(\beta^{H,N}_i):=\Big( \frac{|\beta_{i}^{H,N}(t)-\beta_{i}^{H,N}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H,N}(r)-\beta_{i}^{H,N}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big).$$

Then, we have $$\begin{aligned}
\label{Ast}
    A_{s,t}(\beta^{H,N}_i)& \leq & \Big( \frac{|\beta_{i}^{H,N}(t)-\beta_{i}^{H,N}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H,N}(r)-\beta_{i}^{H,N}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{s \leq t \leq \tau_N\}}\cr
    &&+\Big( \frac{|\beta_{i}^{H,N}(t)-\beta_{i}^{H,N}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H,N}(r)-\beta_{i}^{H,N}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{\tau_N \leq s \leq t\}}\cr
    &&+\Big( \frac{|\beta_{i}^{H,N}(t)-\beta_{i}^{H,N}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H,N}(r)-\beta_{i}^{H,N}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{s \leq \tau_N \leq t\}}\cr
    & \leq &
    \Big( \frac{|\beta_{i}^{H}(t)-\beta_{i}^{H}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H}(r)-\beta_{i}^{H}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{s \leq t \leq \tau_N\}}\cr
    &&+\Big( \frac{|\beta_{i}^{H}(\tau_N)-\beta_{i}^{H}(\tau_N)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H}(\tau_N)-\beta_{i}^{H}(\tau_N)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{\tau_N \leq s \leq t\}}\cr
    &&+  \Big(\frac{|\beta_{i}^{H}(\tau_N)-\beta_{i}^{H}(s)|}{(\tau_N-s)^{1-\alpha}}\frac{(\tau_N-s)^{1-\alpha}}{(t-s)^{1-\alpha}}\Big)\mathbf{1}_{\{s \leq \tau_N \leq t\}}\cr
    &&+\Big(\int_s^t\frac{|\beta_{i}^{H}(r \wedge \tau_N)-\beta_{i}^{H}(s)|}{(r\wedge \tau_N-s)^{2-\alpha}} \frac{(r\wedge \tau_N  -s)^{2-\alpha}}{(r -s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{s \leq \tau_N \leq t\}} \cr
    &\leq & 
        C \Big( \frac{|\beta_{i}^{H}(t)-\beta_{i}^{H}(s)|}{(t-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H}(r)-\beta_{i}^{H}(s)|}{(r-s)^{2-\alpha}}\mathrm{d}r\Big)\mathbf{1}_{\{s \leq t \leq \tau_N\}}\cr
    &&+ C   \Big(\frac{|\beta_{i}^{H}(\tau_N)-\beta_{i}^{H}(s)|}{(\tau_N-s)^{1-\alpha}}+\int_s^t\frac{|\beta_{i}^{H}(r \wedge \tau_N)-\beta_{i}^{H}(s)|}{(r\wedge \tau_N-s)^{2-\alpha}} \mathrm{d}r\Big)\mathbf{1}_{\{s \leq \tau_N \leq t\}},\end{aligned}$$ where $C$ is a constant which is independent of $N$ and $i$.

Thus, by (\[Ast\]), we have $$\begin{aligned}
    \|\beta_{i}^{H,N}\|_{\alpha,0,T}= \sup_{0\le s<t\le T} A_{s,t}(\beta^{H,N}_i) \leq C \|\beta^H_i\|_{\alpha,0,\tau_N},\end{aligned}$$ almost surely. This completes the proof of Lemma \[betaHn1\].

To proceed, by Lemma \[smooth\] and Lemma \[betaHn1\], for any $i\in \mathbb{N}$ and $\varpi \in(1-\alpha, H)$, we have $$\begin{aligned}
\label{betaNn-N}
\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|_{\alpha,0,T} &\leq& CK_{\varpi}(\beta_{i}^{H,N})(\frac{1}{n})^{\varpi+\alpha-1} \cr
&\leq& C\|\beta_{i}^{H,N}\|_{1-\varpi,0,T}(\frac{1}{n})^{\varpi+\alpha-1}\cr
&\leq& C\|\beta_{i}^{H}\|_{1-\varpi,0,\tau_N}(\frac{1}{n})^{\varpi+\alpha-1},\end{aligned}$$ almost surely. Moreover, since $\varpi \in(1-\alpha, H)$, by Lemma \[betaHn1\] and (\[betaNn-N\]), we have $$\begin{aligned}
\label{betaNn-Nn}
\Lambda_{\alpha,B^{H,N, n}}^{0,T}
&\leq& \frac{ \sum_{i=1}^\infty\sqrt{\lambda_i}\big(\|\beta_{i}^{H,N,n}-\beta_{i}^{H,N}\|_{\alpha,0,T}+\|\beta_{i}^{H,N}\|_{\alpha,0,T}\big)}{\Gamma(1-\alpha)\Gamma(\alpha)}\cr
&\leq& \frac{C  \sum_{i=1}^\infty\sqrt{\lambda_i}(\|\beta_{i}^{H}\|_{\alpha,0,\tau_N}+\|\beta_{i}^{H}\|_{1-\varpi,0,\tau_N})}{\Gamma(1-\alpha)\Gamma(\alpha)}\cr
&\leq& CN.\end{aligned}$$

Now, let us consider $$\begin{aligned}
\label{map1}
u_t^{N,n}&=& S_t u_{0}+\int_{0}^{t} S_{t-s} f(u^{N,n}_s) \mathrm{d} s+\int_{0}^{t} S_{t-s} \sigma(u^{N,n}_s) \mathrm{d} W_s\cr
&&+\sum_{i=1}^{\infty} \sqrt{\lambda_{i}}\int_{0}^{t} S_{t-s} g(u^{N,n}_s)e_i  \mathrm{d} \beta_{i}^{H,N,n}(s),\end{aligned}$$ or 