Lebesgue integration by parts in Sobolev space $W^{1,2}(\mathbb{R})$

Question

Let $\phi, \psi \in W^{1,2}(\mathbb{R}) \subset L^2(\mathbb{R})$ and we want to integrate by parts the following piece: $$\int_{\mathbb{R}}\phi(x)\psi'(x)dx$$ Supposedly, it should look like this: $$\int_{\mathbb{R}}\phi(x)\psi'(x)dx = \phi(x)\psi(x)\bigg\rvert_{\mathbb{R}}-\int_{\mathbb{R}}\phi'(x)\psi(x)dx$$ The quastion is what exactly is the first item? Is it $\phi(x)\psi(x)\bigg\rvert_{-\infty}^{+\infty}$? But limits $\lim_{x\to+\infty}\phi(x), \lim_{x\to-\infty}\phi(x), \lim_{x\to+\infty}\psi(x), \lim_{x\to-\infty}\psi(x)$ in a usual sense may not exist. So how then?

Answer 1

$C_c^\infty$, the space of $C^\infty$ functions with compact support is dense in $W^{1,2}$. Let $\{\phi_n\}$, $\{\psi_n\}$ be sequences in $C_c^\infty$ converging to $\phi$ and $\psi$ respectively in $W^{1,2}$. We have $$ \int_{\mathbb{R}}\phi_n\,\psi'_n=-\int_{\mathbb{R}}\phi'_n\,\psi_n. $$ Taking limits as $n\to\infty$ we get $$ \int_{\mathbb{R}}\phi\,\psi'=-\int_{\mathbb{R}}\phi'\,\psi. $$ since $\phi_n\to\phi$, $\phi'_n\to\phi'$, $\psi_n\to\psi$ and $\psi'_n\to\psi'$ in $L^2$.