# Continuous embedding of magnetic Sobolev space into regular Sobolev space

I am working on Aharonov-Bohm operators for my thesis, and one of the articles given to me as a reference (here it is) tells me (intequality (2.1) on p. 6 in the link) that the magnetic Sobolev space defined with the AB operator is continuously embedded into $H_0^1$, that is:

$$\|u\|_{H_0^1(\Omega)}\leq C\|(i\nabla+A_a)u\|_{L^2(\Omega)}.$$

(Notes: $A_a$ is $\alpha(-\frac{x_2-a_2}{|x-a|^2},\frac{x_1-a_1}{|x-a|^2})$, $\Omega$ is a bounded simply connected domain containing $a$, $u$ is a generic function in the magnetic Sobolev space, and the RHS of the above inequality is taken as the norm w.r.t. which we complete $\mathcal C_0^\infty(\Omega\smallsetminus\{a\})$ to get the magnetic Sobolev space, denoted $\mathcal D^{1,2}_{A_a}(\Omega)$.)

It seems relatively easy to prove that there is a continuous embedding into $L^2$, seen as, being $\Omega$ bounded, so is $|x-a|$, so that $\frac{1}{|x-a|}$ is bounded away from zero, and we have inequality (2.2) (which holds also for $D=\Omega$, provided $u$ has zero trace, and holds for the $D$'s stated in the article for general $u$) which allows us to complete the following inequality chain:

$$\|u\|_{L^2}^2\leq\sup|x-a|^2\left\|\frac{u}{|x-a|}\right\|_{L^2}^2\leq C^2\sup|x-a|^2\|u\|_{\mathcal D^{1,2}_{A_a}(\Omega)}^2.$$

However, the Sobolev norm is either:

$$\|u\|_{H_0^1}=\sqrt{\|u\|_{L^2}^2+\|\nabla u\|_{L^2}^2},$$

or, equivalently (since the trace is zero and by Poincaré inequality):

$$\|u\|_{H_0^1}=\|\nabla u\|_{L^2}.$$

So I have the gradient to deal with. What I need to prove is therefore that for some $K$ we have:

$$\|\nabla u\|_{L^2(\Omega)}\leq K\|u\|_{\mathcal D^{1,2}_{A_a}(\Omega)}.$$

I tried working on the RHS without a constant and making lower estimates, but I only managed to show non-negativity, which is pretty useless. Here is what I did:

\begin{align*} \|(i\nabla+A_a)u\|_{L^2}^2={}&\int_\Omega|i\nabla u+A_au|^2dx=\int_\Omega(|\nabla u|^2+|A_au|^2+i(\nabla u;A_a)\overline u-i(A_a;\nabla u)u)dx={} \\ {}={}&\int_\Omega(|\nabla u|^2+|A_au|^2+2\operatorname{Im}[(A_a;\nabla u)u])dx, \end{align*}

where:

$$(v;w)=v^T\overline w,$$

the Euclidean inner product (with conjugation on the second entry rather than the first one). So I would have to estimate that imaginary part, hopefully to eliminate the $|A_au|^2$ and only partially cancel out the $|\nabla u|^2$. The only estimate I was able to come up with for this is Cauchy-Schwarz plus Young, giving me:

$$\left|2\operatorname{Im}[(A_a;\nabla u)u]\right|\leq2|A_a||\nabla u||u|\leq|\nabla u|^2+|A_a|^2|u|^2=|\nabla u|^2+|A_au|^2,$$

but to use that I would have to say twice that imaginary part is bigger than the opposite of that right-hand side, which happens to cancel out everything and leave me with:

$$\|(i\nabla+A_a)u|_{L^2}^2\geq0,$$

which I already knew to be true. So what can I do to prove my desired inequality? Is there a useful estimate for that imaginary part? Did I make any mistake in the above?

Note

Thinking about it, I realized that, if $u$ is real-valued (as I guess the article assumes), the imaginary part in question is zero, so I'm already done. So is there a way to generalize this estimate to complex-valued $u$s? Is it even true in that generality?

where the $C$ is the constant of an inequality proved here. The passage from $A_a$ to $\frac{1}{|x-a|}$ is because: