Let:
$\Omega$ a bounded domain of $\mathbb{R}^d$ with a Lipshitz boundary $\partial\Omega =\Gamma_1 \cup\Gamma_2\cup\Gamma_3.$
$S_{d}$ is the space of second order symetric tensors on $\mathbb{R}^d \left(d=2,3\right)$
$\sigma.\tau=\sigma_{ij}.\tau_{ij} \forall \sigma,\tau\in S_{d} $
$W=\{\sigma=\sigma_{ij}; \sigma_{ij}=\sigma_{ji}\in L^2\left(\Omega\right)/1\leq i\leq d\}=L^2\left(\Omega; S_{d}\right)$
$H=\{u=\left(u_{i}\right)\ u_{i}\in L^2\left(\Omega\right)\leq i\leq d \}=L^2\left(\Omega\right)^{d}$
$H_{1}=\{u=\left(u_{i}\right)\ u_{i} \in H^1\left(\Omega\right)\ 1\leq i\leq d\}=H^1\left(\Omega\right)^d$
$W_{1}=\{\sigma \in W ;Div \sigma \in H\}$
$ V=\{ v\in H_{1}\left(\Omega\right) ;v=0$ in $\Gamma_{1}\}$
$\left(\sigma,\tau\right)_W=\int_{\Omega}\tau_{ij}\sigma_{ij}dx \forall \sigma,\tau \in W$
$\left(u,v\right)_H=\int_{\Omega}u_iv_idx \forall u,v \in H$
$\left(u,v\right)_{H_{1}}=\left(u,v\right)_H+\left(\epsilon\left(u\right),\epsilon\left(v\right)\right)_W \forall u,v \in H_{1}$
$\left(\sigma,\tau\right)_{W_{1}}=\left(\tau,\sigma \right)_W+\left(Div \tau ,Div \sigma \right)_H$
$\left(u,v\right)_V=\left(\epsilon\left(u\right),\epsilon\left(v\right)\right)_W \forall u,v \in V$
$\epsilon\left(u\right)=\epsilon_{ij}\left(u\right)=\frac{1}{2}\left(\frac{\partial u_{i}}{\partial x_{j}}+\frac{\partial u_{i}}{\partial x_{j}} \right)$
There exists $C$ such that $\lvert \epsilon\left(v\right)\rvert_{W}\geq C\lvert v \rvert_{H_1\left(\Omega\right)}$
The non linear operator of viscosity$ A:\Omega \times S^{d}\rightarrow S^{d} $satisfy:
1-There exists a constant $L_{A}$ such that: $\lvert A\left(x,\epsilon_1\right)-A\left(x,\epsilon_2\right)\rvert\leq L_{A} \lvert \epsilon_{1}-\epsilon_{2} \rvert$
2-There exists $m_A\geq 0$ such that: $\left(A\left(x,\epsilon_1\right)-A\left(x,\epsilon_2\right).\left( \epsilon_{1}-\epsilon_{2}\right)\right)\geq m_{A}\lvert\epsilon_{1}-\epsilon_{2}\rvert^2 \forall \epsilon_1,\epsilon_2 \in S^d, a.e x \in \Omega $
3-$x\rightarrow \left(A\left(x,\epsilon\right)\right)$ is lebesgue measurable in $\Omega$.
4-The mapping $x\rightarrow \left(A\left(x,0\right)\right)$ belongs to $W$.
If $ u^{'}_n$ is a bounded sequence in $L^2\left(0,T;V\right)$ such that $u_n^{'}$ converges weakly to $u^{'}$ in $L^2\left(0,T;V\right)$ , how can we show that $\left(A\epsilon\left(u^{'} _n\right),\epsilon\left(v\right)\right)_{W}$ converges weakly to $\left(A\epsilon\left(u^{'}\right),\epsilon\left(v\right)\right)_{W}$ in $L^2\left(0,T\right)$?
I tried to do it but I have found a difficulty because of the nonlinearity of the operator $A$.
NB: $u$ depend of $x$ and $t$ with $x\in \Omega$ and $t \in \left[0,T\right]$and $u^{'}$ is the derivative of $u$ with respect to $t$.
I don't think that this will be true. Let us ignore the time-dependence and replace $\varepsilon(u_n)$ by $f_n$.
Then, you essentially ask whether the weak convergence $f_n \rightharpoonup f$ implies $$A(f_n) \rightharpoonup A(f)$$ for some non-linear, pointwise defined mapping $A$. This is, however, not true.
Let us take $$A(f) = \begin{cases} f & \text{for } f < 0 \\ 2 \, f & \text{for } f \ge 0.\end{cases}$$ This $A$ satisfies your assumption.
Now, take a standard oscillating sequence $f_n$ which takes only the values $-1$ and $1$ with $f_n \rightharpoonup 0$. Then, $A(f_n)$ takes the values $-1$ and $2$ and will converge weakly to $(-1+2)/2 = 1/2 \ne 0 = A(0)$.