corresponding spaces. To prove $u_n \to u$ in ${\mathcal{H}}^{\sigma+2}_2(T)$, it suffices to show (\[sense of solution\]) holds for all $t\leq T$ (a.s.). Since the series $\sum_{k=1}^\infty \int^t_0 (g^k_n (s),\phi) dW^k_s$ converges in probability uniformly on $[0,T]$, so does $(I^{1-\gamma} \sum_{k=1}^\infty \int^\cdot_0 (g^k_n(s),\phi) dW^k_s)(t)$. By Remark \[remark 1\], considering the limit of $$(I^{1-\beta}(u_n-u_n(0)))(t), \phi)_{L_2}=\int^t_0(f_n(s, \cdot), \phi)_{L_2}ds+\sum_{k=1}^{\infty} (I^{1-\gamma} \int^{\cdot}_0 (g^k_n(s, \cdot),\phi)_{L_2}\, dW^k_s)(t)$$ for $t\leq T$, we get (\[sense of solution\]) for all $t\leq T$ (a.s.) since both sides of (\[sense of solution\]) are continuous in $t$.

\(ii) We only prove the case $\sigma=0$. The general case is covered by applying $(1-\Delta)^{\sigma/2}$ to (\[eqn 7.08.1\]). Denote $f={\mathbb{D}}u$ and $g={\mathbb{S}}u$. Notice that as an $L_2({\mathbb{R}}^d)$-valued process, $u(t)-u(0)$ satisfies $$k_{1-\beta}*(u(\cdot,x)-u(0,x))(t)=\int^t_0 f(s,x) ds+(k_{1-\gamma}* (\sum_{k=1}^\infty \int^{\cdot}_0 g^k(s,x)dW^k_s))(t) \quad \hbox{all } t\leq T \, (a.s.).$$ Taking the convolution with $k_{\beta}$ and using $$\frac{\partial}{ \partial t}(k_{\beta}*\int^{\cdot}_0 f(s,x)ds)(t)=(k_{\beta}*f)(t,x),$$ $$\frac{\partial}{ \partial t}(k_{\beta}*k_{1-\gamma}*(\sum_{k=1}^\infty \int^{\cdot}_0 g^k (s,x)dw_s)) (t)=\sum_{k=1}^\infty \frac{1}{\Gamma(1+\beta-\gamma)}\int^{t}_0 (t-s)^{\beta-\gamma}g^k(s,x)dW^k_s,$$ where the second equality is from Lemma \[derivative\], we get for all $t\leq T$ (a.s.) $$\label{eqn 10}
u(t,x)-u(0,x)=(k_{\beta}*f)(t,x)+\sum_{k=1}^\infty \frac{1}{\Gamma(1+\beta-\gamma)}\int^{t}_0 (t-s)^{\beta-\gamma}g^k(s,x)dW^k_s.$$ Hence the claim follows.

\(iii) Denote $w(t,x)=u(t,x)-u(0,x)-v(t,x)$. Then we have $\partial^{\beta}_tw(t,x)=f(t,x)$. Now we use the fact (see e.g. [@Za Lemma 2.1]) that if $\kappa$ is a positive decreasing function on $[0,T]$ then $$\kappa*\|w\|^2_{L_2} 