what is the meaning of $H_0^1(\mathbb{R})$ space?

Question

So, I was looking at the definition for the $H_0^1(\Omega)$ space, and I was wondering that if $\Omega = \mathbb{R}$.

This is really to embed the boundary conditions of my operator, $\mathcal{L}$ which is posed on the real line, but also has conditions such that for $\frac{du}{dt} = \mathcal{L}u$, $u \longrightarrow 0$ as $x\longrightarrow \pm \infty$.

For the space $H_0^1(\Omega)$, the norm is given by $||\frac{du}{dx} ||_{L^2}$, but this norm is motivated using Poincare's inequality that I am not sure can be used on the entire real line? On wikipedia for Poincare's inequality, it says that it must be bounded in at least one direction.

Answer 1

Results that $H^1_0(\Omega)=\overline{C^{\infty}_0(\Omega)}^{\|\cdot\|_1}$, where $\|\cdot\|^2_1 :=\|\cdot \|^2_{L^2}+\|\nabla \cdot \|^2_{L^2}$. But, $C^{\infty}_0(\mathbb{R}^n)$ is dense over $H^1(\mathbb{R}^n)$ (it requires convolution theory) so in this limit case $H^1(\mathbb{R}^n)=H_0^1(\mathbb{R}^n)$. See H.Brezis Functional Analysis, Sobolev Spaces and Partial Differential Equations, chapter 9 pp 287.

Answer 2

Yes, for Poincaré's equation you need that $\Omega$ is bounded. You can construct a counterexample by looking at $u_r(t) := u(rt)$ for any $u \in C_0^1(\mathbb{R})$. Just play with $r \rightarrow 0$ and $r \rightarrow \infty$. The only similar inequalities that hold on $\mathbb{R}$ spaces are Gagliardo-Nirenberg-type equations.

If we define $H_0^1(\Omega)$ as the closure of $C_0^\infty(\Omega)$ w.r.t. the norm $\lVert u \rVert := \lVert u \rVert_{L^2(\Omega)} + \lVert u' \rVert_{L^2(\Omega)}$, we can make sense out of $H_0^1(\mathbb{R})$. We have $H_0^1(\mathbb{R}) = H^1(\mathbb{R})$. This is because $C^\infty(\mathbb{R})$-functions can be approximated by $C_0^\infty(\mathbb{R})$-functions using a sequence of functions whose support grows.

Of course, those functions tend to $0$ as $x \rightarrow \pm\infty$, since they are $L^2$.