If $u\in H_0^1(\Lambda,\mathbb R^2)\cap H^2(\Lambda,\mathbb R^2)$, is $\nabla^\perp\cdot\Delta u\in L^2(\Lambda)$? Let $\Lambda\subseteq\mathbb R^2$ be open (and sufficiently regular for the subsequent consideration), $u\in H_0^1(\Lambda,\mathbb R^2)\cap H^2(\Lambda,\mathbb R^2)$ and $$w:=\nabla^\perp\cdot\Delta u,$$ where $$\nabla^\perp:=\left(-\frac\partial{\partial x_2},\frac\partial{\partial x_1}\right).$$ It's been a while, since I've worked in the context of distribution theory, but if I remember correctly, $v$ should, by definition of this space, belong to $H^{-1}(\Lambda)$. On the other hand, $H^{-1}(\Lambda)$ is isometrically isomorphic to $H_0^1(\Lambda)^\ast$, which in turn (since $H_0^1(\Lambda)$ is a Hilbert space) is isometrically isomorphic to $H_0^1(\Lambda)$. But I guess there I made a mistake in my reasoning at some point, since this would mean that $w\in H_0^1(\Lambda)$.

So, where is my mistake? And beyond that: Is there a suitable assumption (other than $u\in H^3(\Lambda)$, of course), which would yield that $w$ at least belongs to $L^2(\Lambda)$?

Remark: To give some motivation for the question: In this paper, at the end of page 9, the authors consider the $L^2(\Lambda)$-norm of a $w$ arising in such a way, which doesn't make sense to me (without a suitable identification), if $w$ is only a distribution.
 A: About the main question: It seems that you are confusing the different representations of the space $H^{-1}$ (as it is quite common) that use different duality products. When we describe $H^{-1}$ as derivatives of $L^2$ functions, then we are using the distribution product but when we are using Riesz representation theorem (isometry between a Hilbert space and its dual) then you have to use the scalar product as duality product, in this case
$$ (f,  g)_{H_0^1} =  \int\limits_\Lambda \nabla f \cdot \nabla g \, dx $$
Other way to say it is that just because there is an isomorphism (lets call it $T$) it doesn't mean they are equal, so the correct implication would be
$$ w\in H^{-1} \Rightarrow T(w) \in H_0^1 $$
and in general $T(w) \not= w$.
To clarify ideas lets consider a simple one dimensional example:
As the functions in $H^1(-1,1)$ are continuous, de delta distribution is in $H^{-1}(-1,1)$. Does this mean that $\delta \in H^{1}(-1,1)$? of course not, as is not even in $L^2(-1,1)$, but it means that there is a function $f\in H_0^1(-1,1)$ such that
$$ g(0) = \langle \delta,g \rangle = (f, g)_{H_0^1(-1,1)} = \int_{-1}^1 f'g' \, dx \quad \forall g \in H_0^1(-1,1)$$
and in fact such $f$ exists
$$f(x) = \frac{1 - |x|}{2}$$
(you can find it by noticing that the problem is just the weak formulation of $-f'' = \delta$)
So it is this function and not $\delta$ that belongs to $H_0^1(-1,1)$.
About your second question: The condition $\nabla^\perp f\in L^2$ with $f\in L^2$ is equivalent to $f \in H^1$ so in your case the condition is equivalent to $\Delta u \in H^1$, so if your domain is regular enough and you have appropriate boundary conditions, this implies $u\in H^3$ by standard regularity results for Poisson equation.
That being said, I cannot find in the article that you link where the $L^2$ norm of such quantity is consider. In page 9 $w$ is defined as $\nabla^\perp u$ which is of course in $L^2$. On the next page $\nabla^\perp Au$ is mentioned but I don't see that they imply that it belongs to $L^2$. Could you give the specific equation number that you mentioned?
