Define $$W^{1,2}[0,\infty) := \{ f \in L^2[0,\infty): f' \in L^2[0,\infty)\}.$$ It is a Hilbert space with the inner product $$ \langle f,g \rangle = \langle f,g \rangle_{L^2} + \langle f',g' \rangle_{L^2}.$$ It turns out to be a conmutative Banach algebra without unity with the pointwise multiplication. It can be show that if $f \in W^{1,2}[0,\infty)$ then $f$ is absolutely continuous in every bounded suvinterval of $[0,\infty)$.
Let $f \in W^{1,2}[0,\infty)$, and denote $E = f^{-1}(\{0\})$ the zero set of $f$. I'm trying to construct a sequence $(f_n)$ in $W^{1,2}[0,\infty)$ such that $f_n \rightarrow f$ in $W^{1,2}[0,\infty)$ with the following property: Let $E_n = f_n^{-1}(\{0\})$ the zero sets of each $f_n$. Then, $E_n$ contains an open set $U_n$ of $E$ with compact complement.
My attempt: define $$\phi_n(t) = \min \{1, n \cdot \text{d}(x, (E\cup[n,\infty))+ [-1/n,1/n])\}.$$ Then, and put $f_n = \phi_n\cdot f.$ This sequence is in $W^{1,2}[0,\infty)$ and fulfill the properties, but I can't show that $f_n \rightarrow f$. It's easy to see that $f_n \rightarrow f$ in $L^2$, but I can't get a proof of the convergence of $f_n'$ to $f'$ in $L^2$.
Can anyone help me to prove this convergence? Has anyone an idea to construct an alternative sequence? Thank you very much!