I need to show that $\Gamma-\lim F_k=F$ and $\Gamma-\lim G_k=G$ doesn't always imply $\Gamma-\lim(F_k + G_k)=F+G$.
Our definition of $\Gamma-$convergence is the following: $F_k\;$ $\Gamma$-converges to $F$ if 1) For all sequences $(u_k)$ with $u_k \rightarrow u \in X$: $F(u) \leq \liminf F_k(u_k)$ and 2)$\;\forall u \in X: \exists$ a sequence $u_n \rightarrow u$ with $F(u)=\lim F_k(u_k)$.
We've seen in our lecture that for $F_k: \mathbb{R} \rightarrow \mathbb{R},\; F_k(x)=\sin(kx)$ follows $\Gamma-\lim \sin(kx)=-1$. This is true because $F(x)=-1 \leq \liminf\, F_k(x_k)$ and for all $k \in \mathbb{Z}$ we have $F_k(\frac{3+4k}{2n}\pi)=-1$ and every $x\in \mathbb{R}$ can be approximated by $\frac{3+4k}{2n}\pi$ and so we obtain $F(x)=-1= \lim F_k(x_k)$.
I've thought that I could use this to create the counterexample I need. My idea was to show an analog to the example from the lecture that $\Gamma-\lim\sin^2(kx)=0$ and $\Gamma-\lim\cos^2(kx)=0$ but $\Gamma-\lim(\sin^2(kx)+\cos^2(kx))=\Gamma-\lim1=1 \neq 0$. I'd like to ask if my reasoning is correct or am I somewhere mistaken?
I found out this unanswered question (even if the comment by Federico actually provides a feedback). For readers that, like me, come across this question, I would like to add my feedback as well in saying that the reasoning is perfectly fine and the counterexample is really nice. Note indeed that $\sin$ and $\cos$ are the typical examples to compute $\Gamma-$limits easily just by looking for the infima of the subsequential limits and a particular recovery sequence. The fundamental $\sin$ and $\cos$ trigonometric identity then does the final job in showing the additivity does not hold.
Here is another: $X= \mathbb{R}$, $F_h(x)= \sin(hx)$ and $G_h(x)= -\sin(hx)$. They they both $\Gamma-$converge to $-1$, but their sum $\Gamma-$ converge to $0$.
Note, though, that $\Gamma-$limit is stable under continuous perturbations, that is:
If $F_{n} \stackrel{\Gamma}{\rightarrow} F$ and $G$ is continuous, then $$ F_{n}+G \stackrel{\Gamma}{\rightarrow} F+G $$