Characterisation of second Sobolev space through Laplacian Let $u \in H^1(\Omega)$ with $\Omega \subset \mathbb{R}^n$. We say that $\Delta u \in L^2(\Omega)$ iff there exists $f \in L^2(\Omega)$ such that
\begin{equation}
\int_\Omega u \Delta \varphi \operatorname{dx} = \int_\Omega f\varphi \operatorname{dx} \quad (\varphi \in C_c^\infty(\Omega)).
\end{equation}
In this case we write $\Delta u =f$. My question is whether $H^2(\Omega)= \{u \in H^1(\Omega) : \Delta u \in L^2(\Omega)\}$? Maybe there are conditions on $\Omega$ such that the equality holds?
 A: Some additional assumptions on $\Omega$ and $u$ are needed. A typical result would be (section 6.3.2 in Partial Differential Equations by Evans): if $\partial \Omega$ is $C^2$ smooth, $u\in H^1_0(\Omega)$ (notice the zero boundary condition), and $\Delta u\in L^2(\Omega)$, then $u\in H^2(\Omega)$. A more general form, Theorem 8.12 in Elliptic Partial Differential Equations by Gilbarg and Trudinger, does not require zero boundary condition but makes the assumption that the boundary trace of $u$ is compatible with some $H^2(\Omega)$ function, which is essentially the same thing (just subtract that function).
I'll give some reasons for why we need such assumptions
Geometry of $\Omega$
In $\mathbb R^2$ consider $u(x_1,x_2)=\log(x_1^2+x_2^2)$. This is a harmonic function on $\mathbb R^2\setminus\{0\}$, so $\Delta f = 0$. Its gradient satisfies $\|\nabla u(x)\| \sim \|x\|^{-1}$ and the Hessian matrix is of size $\|x\|^{-2}$. Neither is square integrable in a neighborhood of $0$, but the Hessian is "more nonintegrable". So, it $\Omega$ has an outward cusp touching $0$, say $\Omega = \{x : 0\le x_2\le x_1^2 \le 1 \}$, then 
$$
\int_\Omega \|\nabla u\|^2 \approx \int_0^1 \frac{1}{t^2}t^2\,dt <\infty
$$ 
but 
$$
\int_\Omega \|D^2 u\|^2 \approx \int_0^1 \frac{1}{t^4}t^2\,dt  = \infty
$$ 
Boundary conditions
Let $\Omega$ be the unit disk in $\mathbb R^2$. Any integrable function on $\partial \Omega$ extends to a harmonic function on $\Omega$. A discontinuity, or some other poor behavior on the boundary is reflected in the growth of derivatives toward that point. Derivatives of higher order are affected more strongly, so it is possible to make the boundary values bad enough so that $u\notin H^2(\Omega)$ yet not bad enough to lose $u\in H^1(\Omega)$. 
