Range of the Dirac operator on the real line closed?

Question

This is quite a short question: consider the Dirac operator $-i \tfrac{d}{dx}\colon H^1(R) \to L^2(R)$, where $H^1(R)$ denotes the Sobolev space of square-integrable functions with square-integrable weak first derivative.

Is the range of the Dirac operator closed?

edit: Sure I tried to apply the definitions: given $f_n \in H^1(R)$ with $-i f_n^\prime$ convergent to some $g \in L^2(R)$, we have to show that there is some $h \in H^1(R)$ with $-i h^\prime = g$.

Now I thought about writing something like $h(x) := i\int_0^x g(y) dy$. Using a suitable generalization of the fundamental theorem of calculus that would mean that $h$ is almost everywhere differentiable with derivative $ig$ (where the derivative is defined). It remains to show that $h \in H^1(R)$. Here I think that I have to use somehow that $g$ is the limit of the $-i f_n^\prime$, but I don't see how.

Answer 1

Do you think that it is true or false ?

You can remark that your operator is injective on $H^1(\mathbb R)$. Now try to imagine what a injective and closed-range operator might yield, if you look "from the other side".

Edit : since there was no answer, I give you here more precise hints. An injective and closed-range operator has a continuous inverse. Thus, if your operator was closed-range, there would exist $c>0$ such that for any $u\in H^1(\mathbb R)$, $\|\partial_x u\|_{L^2} \geq c \|u\|_{H^1}$. Now think of a smooth bump function $\phi$, and set $\phi_n(x) = \phi(\frac{x}{n})$ and see what happens.