We know that we may decompose $L^2(\mathbb{R})$ as $L^2(\mathbb{R^-})\oplus L^2(\mathbb{R^+})$. Can we write $L^2(\mathbb{R})=L^2(\mathbb{R_-^*})\oplus L^2(\mathbb{R_+^*})$?
Now, let $H^1(\mathbb{R})$ be the Sobolev space. Can we write $H^1(\mathbb{R})=H^1(\mathbb{R^-})\oplus H^1(\mathbb{R^+})$. What about the subspace $X:=\{u\in H^1(\mathbb{R}): u(0)=0\}$, can we write $X=\{u\in H^1(\mathbb{R_-^*}): u(0)=0\}\oplus \{u\in H^1(\mathbb{R_+^*}): u(0)=0\}$?
Many thanks in advance.
Math.
For a Lebesgue space, functions are defined almost everywhere, so given $f \in L^2((-\infty,0)) \oplus L^2((0,\infty))$, there is a unique class in $L^2(\mathbb{R})$ which restricts to $f$ on $(-\infty,0)\cup(0,\infty)$. In other words, the value at the single point $0$ does not matter.
In contrast, the value at a single point does matter for Sobolev spaces. For example, the function $$ x \mapsto x/|x| $$ is not in $H^1((-1,1))$ but its restrictions are in $H^1((-1,0))$ and $H^1((0,1))$. (Can you generalize this example to the infinite rays $(-\infty,0)$ and $(0,\infty)$?)
The question about $H^1_0((-\infty,0)) + H^1_0((0,\infty)) \subseteq H^1(\mathbb{R})$ is more subtle., and I do not have a solution at the moment.
I learned Sobolev spaces out of Brezis, in which one only defines $H^1(\Omega)$ when $\Omega$ is an open subset of some $\mathbb{R}^N$. So, I'm not sure what you might mean by $H^1((-\infty,0])$ except to mean $H^1((-\infty,0))$.