What can be the condition I want to prove that when $u_n\rightarrow u$ when $n$ tends to $\infty$ $$J(u_n)=\frac12||u_n||^2-\int_0^1 (A(su_n),u_n) ds\rightarrow J(u)$$
I have that $A$ is continuous so $(A(su_n),u_n)\rightarrow (A(su),u)$ but what is the condition on $A$ to obtain that $\int_0^1(A(su_n),u_n) ds\rightarrow \int_0^1 (A(su),u) ds$ ?
Thank you .
 A: You don't need any condition.
I interpret your question as follows: $H$ is a Hilbert space (inner product space will do), $u_n, u \in H$ with $u_n \to u$ and $A :H \to H$ is a continuous operator (not necessary linear).
It is easy to see (e.g. using the definition) that
$$
K := \{u_n \mid n \in \Bbb{N}\} \cup \{u\}
$$
is compact. Furthermore, $I := [0,1]$ is also compact and hence so is $[0,1] \times K$.
But the multiplication map
$$
\Phi : \Bbb{R} \times H \to H, (\alpha, x) \mapsto \alpha x
$$
is continuous, so that $L := \Phi([0,1] \times K) \subset H$ is compact.
Since $A$ is continuous, it is bounded on $L$, say by $C>0$. All in all, we get
$$
\left| (A(s u_n) , u_n\right| \leq \Vert A(s u_n) \Vert \cdot \Vert u_n \Vert \leq C \cdot \Vert u_n \Vert \leq C',
$$
since $u_n \to u$.
All in all, we have shown that the sequence of integrands is uniformly bounded and pointwise convergent. Since the domain of integration has finite measure, the dominated convergence theorem completes the proof.
