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.

  • 1
    $\begingroup$ There is one obvious problem with defining $h(x) = i \int_0^xg(y) dy$, Take, for example, $g(y) = (e^{-y^2})'$. If you integrate from 0, you will have $\lim_{x \to\pm\infty} h(x) = -1$, and so $h(x) \not\in L^2$. $\endgroup$ – Willie Wong Oct 29 '13 at 14:36
  • $\begingroup$ Now, the "obvious fix" is to try to integrate back from infinity. But there you run into a problem (why is there a problem?) Can you see how to use this problem to construct a counterexample? $\endgroup$ – Willie Wong Oct 29 '13 at 14:50

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.

  • $\begingroup$ Thanks. I can now figure it out on my own. $\endgroup$ – AlexE Feb 2 '15 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.