Integration by parts and the endpoint-vanishing space on intervals

[Bounded integration by parts] Let $u,v\in H^1(0,L)$. By [F2], the products $u^{\ast }\overline{v^{\ast }}$ and the derivative classes $u^{\mathrm{\prime }}\bar{v}$ and $u\overline{v^{\mathrm{\prime }}}$ are integrable; by [F1] in its complex form applied to $F=u^{\ast }$ and $G=v^{\ast }$ (whose real and imaginary parts are absolutely continuous with $L^1$ derivatives), ${\int }_0^L(u^{\mathrm{\prime }}\bar{v}+u\overline{v^{\mathrm{\prime }}})=u^{\ast }(L)\overline{v^{\ast }(L)}-u^{\ast }(0)\overline{v^{\ast }(0)},$ which is exactly (a).

[Zero traces lie in $H_0^1$] Let $u\in H^1(0,L)$ satisfy $u^{\ast }(0)=u^{\ast }(L)=0$. For $0<\epsilon <L\mathrm{/}4$ put ${\chi }_{\epsilon }(x):=\sigma (2x\mathrm{/}\epsilon -1)\sigma (2(L-x)\mathrm{/}\epsilon -1)$ with $\sigma$ as in [F3]. Then ${\chi }_{\epsilon }$ is smooth, $0\le {\chi }_{\epsilon }\le 1$, equals one on $[\epsilon ,L-\epsilon ]$, vanishes on $[0,\epsilon \mathrm{/}2]\cup [L-\epsilon \mathrm{/}2,L]$, and has $\mathrm{\mid }{\chi }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le 4C_{\sigma }\mathrm{/}\epsilon$, supported in $S_{\epsilon }:=[0,\epsilon ]\cup [L-\epsilon ,L]$. The AC product rule [F1] gives $u_{\epsilon }:={\chi }_{\epsilon }u\in H^1(0,L)$ with weak derivative ${\chi }_{\epsilon }^{\mathrm{\prime }}u+{\chi }_{\epsilon }{u}^{\mathrm{\prime }}$; its support lies in $[\epsilon \mathrm{/}2,L-\epsilon \mathrm{/}2]$. Extending by zero to $\mathbb{R}$ preserves this weak derivative because the product vanishes near both endpoints. Choose a finite $R\ge 1$ with $\mathrm{supp}\rho \subseteq [-R,R]$. For $0<\delta <\epsilon \mathrm{/}(4R)$ the convolution $u_{\epsilon ,\delta }:={\rho }_{\delta }\ast u_{\epsilon }$ is smooth by [F4] and its support lies in $[\epsilon \mathrm{/}2-R\delta ,L-\epsilon \mathrm{/}2+R\delta ]\subset (0,L)$, because the integrand vanishes outside the sum of the two supports. The zero extension of $u_{\epsilon }^{\ast }$ is absolutely continuous on every bounded interval, with compact support and derivative $u_{\epsilon }^{\mathrm{\prime }}$. For fixed $x$, apply [F1] on an interval containing that support to $u_{\epsilon }^{\ast }(t)$ and ${\rho }_{\delta }(x-t)$; the boundary term vanishes and $\frac{d}{dt}{\rho }_{\delta }(x-t)=-{\rho }_{\delta }^{\mathrm{\prime }}(x-t)$, so $({\rho }_{\delta }\ast u_{\epsilon }{)}^{\mathrm{\prime }}={\rho }_{\delta }^{\mathrm{\prime }}\ast u_{\epsilon }={\rho }_{\delta }\ast u_{\epsilon }^{\mathrm{\prime }}$. Applying [F4] separately to the function and derivative proves $u_{\epsilon ,\delta }\to u_{\epsilon }$ in $H^1$ as $\delta \to 0$, with every such approximant in $C_c^{\mathrm{\infty }}(0,L)$.

[The cutoff converges to $u$] By [F5], $\mathrm{\parallel }({\chi }_{\epsilon }-1)u{\mathrm{\parallel }}_2\to 0$ and $\mathrm{\parallel }({\chi }_{\epsilon }-1)u^{\mathrm{\prime }}{\mathrm{\parallel }}_2\to 0$. The zero traces and the integral representative in [F2] give, for $0\le x\le \epsilon$, $\mathrm{\mid }{u}^{\ast }(x){\mathrm{\mid }}^2\le x{\int }_0^{\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2$, and for $L-\epsilon \le x\le L$, $\mathrm{\mid }{u}^{\ast }(x){\mathrm{\mid }}^2\le (L-x){\int }_{L-\epsilon }^L\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2$. Thus the derivative term of step 1.2 satisfies $\mathrm{\parallel }{\chi }_{\epsilon }^{\mathrm{\prime }}u{\mathrm{\parallel }}_2^2\le (4C_{\sigma }\mathrm{/}\epsilon {)}^2({\epsilon }^2\mathrm{/}2){\int }_{S_{\epsilon }}\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2=8C_{\sigma }^2{\int }_{S_{\epsilon }}\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2\to 0$ by [F5], since $\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^2$ is integrable and the strip indicators tend to zero almost everywhere. Hence $u_{\epsilon }\to u$ in $H^1$; each $u_{\epsilon }$ lies in the $H^1$ closure of $C_c^{\mathrm{\infty }}(0,L)$ by step 1.2, and that closure is closed, so $u\in H_0^1(0,L)$.

[Second-order identity] Let $u,v\in H^2(0,L)$. Applying (a) to the pair $(u^{\mathrm{\prime }},v)$ and to the pair $(u,v^{\mathrm{\prime }})$, both in $H^1(0,L)$, gives ${\int }_0^L{u}^{\mathrm{\prime }\mathrm{\prime }}\bar{v}=[u^{\mathrm{\prime }\ast }\overline{v^{\ast }}{]}_0^L-{\int }_0^L{u}^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}},{\int }_0^{L}u\overline{v^{\mathrm{\prime }\mathrm{\prime }}}=[u^{\ast }\overline{v^{\mathrm{\prime }\ast }}{]}_0^L-{\int }_0^L{u}^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}}\mathrm{.}$ Subtracting the second identity from the first gives (c).

[Traces of $H_0^1$ classes vanish] Conversely let $u\in H_0^1(0,L)$; by definition there are $u_n\in C_c^{\mathrm{\infty }}(0,L)$ with $u_n\to u$ in $H^1$. For each $x\in [0,L]$ the trace bound of [F2] gives $\mathrm{\mid }{u}_n^{\ast }(x)-u^{\ast }(x)\mathrm{\mid }\le L^{-1\mathrm{/}2}\mathrm{\parallel }{u}_n-u{\mathrm{\parallel }}_{L^2}+L^{1\mathrm{/}2}\mathrm{\parallel }{u}_n^{\mathrm{\prime }}-u^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2},$ so $u_n^{\ast }\to u^{\ast }$ uniformly on $[0,L]$. Since $u_n^{\ast }(0)=0=u_n^{\ast }(L)$, it follows that $u^{\ast }(0)=0=u^{\ast }(L)$. Together with step 2.1 this proves (b).

[Half-line and line] Let $u,v\in H^1(0,\mathrm{\infty })$. Step 1.1 on $(0,R)$ gives ${\int }_0^{R}u\overline{v^{\mathrm{\prime }}}=u^{\ast }(R)\overline{v^{\ast }(R)}-u^{\ast }(0)\overline{v^{\ast }(0)}-{\int }_0^R{u}^{\mathrm{\prime }}\bar{v}$. The boundary term at $R$ tends to zero by [F2], while both integrals converge absolutely by Cauchy–Schwarz (or $\mathrm{\mid }u\overline{v^{\mathrm{\prime }}}\mathrm{\mid }\le (\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^2)\mathrm{/}2$), so ${\int }_0^{\mathrm{\infty }}u\overline{v^{\mathrm{\prime }}}=-u^{\ast }(0)\overline{v^{\ast }(0)}-{\int }_0^{\mathrm{\infty }}{u}^{\mathrm{\prime }}\bar{v}$. For $u,v\in H^1(\mathbb{R})$, apply this identity to their positive-half-line restrictions and to the reflected negative-half-line restrictions $u(-x),v(-x)$. Reflection reverses the derivative sign, so the negative-half-line identity has the opposite origin term $+u^{\ast }(0)\overline{v^{\ast }(0)}$. The representatives on the two sides have the same trace at zero by [F2]; adding cancels the origin terms and gives the claimed whole-line identity.