I need to prove the following statement :
Assume that $\mathrm{SL}_{2}(\mathbb{R})$ acts on $(X,d)$ by isometries. Assume that $x\in X$ and $a_{n}\in A=\{\begin{pmatrix} e^{t/2}&0 \\ 0&e^{-t/2} \end{pmatrix}, t \in \mathbb{R}\backslash\{0\}\}$ such that $a_{n}\to\infty$ and $a_{n}\cdot x\to x$. Then $x$ is a fixed point.
So I need to show that $\forall g \in SL_{2}(\mathbb{R}) : g.x=x$.
We know that since $\mathrm{SL}_{2}(\mathbb{R})$ acts on $(X,d)$ by isometries : $\forall g \in SL_{2}(\mathbb{R}), \forall x,y \in X : d(g.x,g.y)=d(x,y)$
Since $a_{n}\cdot x\to x$ : $d(a_{n}.x,x)\longrightarrow 0$ as $n\rightarrow \infty$ and since $a_{n}\to \infty$ we have a sequence $(t_{n})$ in $\mathbb{R}\backslash\{0\}$ with $t_{n}\longrightarrow \infty$
KAK decomposition : For any $g\in SL_{2}(R)$ there exists $k,l \in SO(2)$ and $a \in A$ so that $g=kal$.
I thought that maybe I could use the KAK decomposition to reduce my problem to the case were $a.x=x$ ($*$) for $a \in A$ but I am not sure if I can do this, and if I can I don't know how to prove that ($*$) would hold.
I really don't know where to start and I would appreciate any kind of hints !
Thank you beforehand for your help