Let $V\subset H\subset V^*$ be an evolution triple and suppose that $u\in W^{1,2}(0,T;V,H)$, where $$W^{1,2}(0,T;V,H)=\{f\in L^2(0,T,V)\,|\,f'\in L^2(0,T,V^*)\}.$$
Now, let $\{w_1, w_2,...\}$ be a basis of $V$ (H-space). Define $$u_n=\sum_{k=1}^{n}c_{kn}(t)w_{k}.$$ What must be the regularity of $c_{kn}$ in order to have $u_n\in W^{1,2}(0,T;V,H)$ or $u_n\in W^{1,2}(0,T;V_n , H)$, where $V_n$ is an n-dimensional subspace of $V$? I was thinking of $c_{kn}\in W^{1,2}(0,T;\mathbb{R})$. But then, how to prove that $u_n'\in L^2(0,T; V^*), L^2(0,T; V_{n}^*)$?
Denote $L^2=L^2(0,T,V)$ and let $c_{kn}\in W^{1,2}(0,T;\mathbb{R})$. Observe that $$\|u_n\|^2_{L^2}\le\sum_{k=1}^{n}\int_{0}^{T}|c_{kn}'(t)|^2\,dt\cdot\|w_k\|^2_{V}\\ \le \sum_{k=1}^{n}\int_{0}^{T}c\cdot\|c_{kn}\|_{W^{1,2}(0,T,V)}^2\,dt\cdot\|w_k\|^2_{V} \le C\cdot T\sum_{k=1}^{n}\|w_k\|^2_{V}<\infty.$$ This implies that $u_n\in L^2(0,T;V)$. Now, with the framework of evolution triple, we may indentify $u'_{n}(t)$ with an element of the dual, i.e., $V^*$. Moreover, the following holds $$\langle u'_{n}(t),v\rangle_{V^*\times V}=\langle u_n'(t),v\rangle_{H}.\qquad (*)$$ Since $\|u_{n}'\|^2_{L^2(0,T; V^*)}=\int_0^{T}\|u'_{n}(t)\|^2_{V^*}\, dt$ and $\|u'_{n}\|_{V^*}=\sup_{\|v\|\le 1}|\langle u_n'(t),v\rangle_{V^*\times V}$ it is easy to deduce that $\|u_{n}'\|_{L^2(0,T; V^*)}<\infty$ (look at $(*)$). This means that $u_n'\in L^2(0,T;V^*)$.