Is it possible to define an extension operator $E : H^1(0,1) \rightarrow H^1(\mathbb R)$ which has dense range? Namely, I wish to know if there exists $E : H^1(0,1) \rightarrow H^1(\mathbb R)$ such that
- $E$ is linear and continuous,
- For all $u \in H^1(\mathbb R)$ there holds $$ \left.\left[E(\left. u \right|_{(0,1)}) \right] \right|_{(0,1)} = \left. u \right|_{(0,1)}, $$
- $E$ has dense range.
This would be useful as for then the TVS adjoint of $E$, namely $E^* : (H^1(\mathbb R))' \rightarrow (H^1(0,1))'$, would be into hence an embedding.
Note that the condition 2 is usually replaced by $$ \forall u \in H^1(0,1),\quad \left. (Eu) \right|_{(0,1)} = u $$ which is slightly stronger. It is standard that replacing condition 2 by the one above and omitting the dense range, there do exists extension operators. However I do not know how to build such an operator satisfying 1,2 and 3.
No, such an extension operator does not exist.
To see this, let $v \in H^1(\mathbb R)$ such that $v \equiv 0$ in a neighbourhood of $[0,1]$ such that $v \neq 0$. Since the range of $E$ is dense by (3), there exists $u_n \in H^1(0,1)$ such that $Eu_n \to v$ in $H^1(\mathbb R)$. Restricting to $(0,1)$ and noting that $Eu_n \rvert_{(0,1)} = u_n$ by (2), we get $$Eu_n \rvert_{(0,1)} =u_n \to v \rvert_{(0,1)} = 0$$ in $H^1(0,1)$. Thus by continuity of $E$ using (1), we must have $Eu_n \to 0$ also.
Also unless I'm mistaken, I don't see why condition (2) is weaker than what you state below. If $u \in H^1(0,1)$, there exists some extension $\tilde u \in H^1(\mathbb R)$. Applying (2) to this $\tilde u$, noting that $\tilde u \rvert_{(0,1)} = u$, precisely gives $Eu \rvert_{(0,1)} = u$.