I have a densely defined linear operator on $H^{s}(\mathbb{R})$ for all $s<1/2$. I would like to prove it is closed. All the term of the operators are closed except that one $$ {\mathcal{L}}v\equiv {\rm{sgn}}{(x)}\,e^{-|x|}\left(v_0^++v_0^-\right), $$ where $v_0^\pm\equiv \lim_{x\rightarrow 0^\pm} v(x)$. I thus need to prove that if a sequence $v_n\rightarrow v$ in $H^s(\mathbb{R})$ for all $s<1/2$, and ${\mathcal{L}}v_n$ converges, then $v$ is in the domain of $\mathcal{L}$ and ${\mathcal{L}}v_n$ converges to ${\mathcal{L}}v$. In other words, I need to prove that the graph of ${\mathcal{L}}$ is closed.
In my case, that means that if $v_n\rightarrow v$ in $H^s(\mathbb{R})$ and $(v_{n0}^++v_{n0}^-)$ converges, then $(v_{n0}^++v_{n0}^-)$ converges to $(v_0^++v_0^-)$.
If that helps, there is the following condition on the domain of my operator, which the equation above only represent one part of: $v$ is such that $(1-e^x)v \in H^s(\mathbb{R})$ for all $s<3/2$.
I made up examples of sequences, which all turn out to be consistent with that result. So, I have hopes it is true. I just cannot prove it in general. My main obstacle is that I do not know how to use the norm on $H^s$ properly to prove my statement.