Let $\Omega$ have the segment property. Define $$ E(\Omega)=\{u\in L^2(\Omega)^n\mid \nabla\cdot u\in L^2(\Omega)\}, $$ where $L^2(\Omega)^n=L^2(\Omega;\mathbb{R}^n)$ and the derivatives taken in the sense of distributions in $\Omega$. $E(\Omega)$ is a Hilbert space with the scalar product $$ [u,v]_{E(\Omega)}:=(u,v)_{L^2(\Omega;\mathbb{R}^n)}+\int_\Omega (\nabla\cdot u)(\nabla\cdot v)\ dx $$ where $(u,v)_{L^2(\Omega; \mathbb{R}^n)}$ is the inner product in $L^2(\Omega)^n$: $$ (u,v)_{L^2(\Omega;\mathbb{R}^n)}=\int_\Omega u(x)\cdot v(x)\ dx $$ with $u(x)\cdot v(x)$ denoting the scalar product in $\mathbb{R}^n$. The space $E(\Omega)$ appears in the study of Navier-Stokes equations.
Would anybody give an example showing the difference between $E(\Omega)$ and $H^1(\Omega)^n:=H^1(\Omega;\mathbb{R}^n)$?
The space $E(\Omega)$ does not require the gradient to be square integrable: it's enough for the divergence to be square integrable. So, $H^1\subset E$ and to construct a vector field $u\in E\setminus H^1$, we should look for something with large derivatives, which either do not contribute to divergence or cancel out.
For example, in two dimensions let $\Omega$ be the square $(-1,1)^2$ and $$u(x_1,x_2)=\left(x_1+(1-x_2^2)^{-1/3},\ x_2\right)$$ Then $u\in L^2(\Omega)$ and $\nabla \cdot u\equiv 2$. However, $u$ is not in $H^1$ because $$\frac{\partial}{\partial x_2} \left(x_1+(1-x_2^2)^{-1/3}\right) =\frac{2x_2}{3}(1-x_2^2)^{-4/3} $$ is not in $L^2(\Omega)$.
In terms of mechanics, the field $u$ exhibits strong shear flow, which is something that divergence is not concerned with.