Let $\Omega \subset \mathbb{R}^{n}$ be a bounded domain. Let $a \in B C\left(\mathbb{R}, \mathbb{R}_{s y m}^{n \times n}\right)$ with $$ a(u) \xi \cdot \xi \geq \underline{\alpha}|\xi|^{2}, \quad \xi \in \mathbb{R}^{n}, \quad u \in \mathbb{R} $$ for some $\underline{\alpha}>0$. Then, the operator $$ A: \stackrel{\circ}{H}^{1}(\Omega) \rightarrow H^{-1}(\Omega):=\left(\stackrel{\circ}{H}^{1}(\Omega)\right)^{\prime} $$ given by $$ \langle A u, \phi\rangle_{\dot{H}^{1}(\Omega)}:=\int_{\Omega} a(u) \nabla u \cdot \nabla \phi \mathrm{d} x, \quad u, \phi \in \stackrel{\circ}{H}^{1}(\Omega), $$ is hemicontinuous .
I know there may be different definitions of hemicontinuous. The one I am using is: a mapping is hemicontinuous if the mapping $t\to \langle A+tv, \phi\rangle$ is a $C[0,1]$ mapping. Here $u, v$ are also elements of $\stackrel{\circ}{H}^{1}(\Omega)$.
So I guess we just use the definition and show sequential continuity. Assume $t_n \to t$. then we want $$ \langle A (u+t_nv), \phi\rangle_{\dot{H}^{1}(\Omega)}:=\int_{\Omega} a(u+t_nv) \nabla u \cdot \nabla \phi \mathrm{d} x $$ to converge to $$ \langle A (u+tv), \phi\rangle_{\dot{H}^{1}(\Omega)}:=\int_{\Omega} a(u+tv) \nabla u \cdot \nabla \phi \mathrm{d} x $$
Well, I'm not sure how we deal with the right hand side now...