Continuous piecewise-affine functions with zero endpoints lie in H1-zero

Let $g$ equal the slope of $v$ on the interior of each partition interval, with arbitrary values at the finitely many breakpoints. Both $v$ and $g$ are bounded, hence square integrable on $(0,1)$. Integration by parts on each partition interval gives ${\int }_0^{1}v{\phi }^{\mathrm{\prime }}=-{\int }_0^{1}g\phi$ for every $\phi \in C_c^{\mathrm{\infty }}(0,1)$: the interior endpoint terms cancel because $v$ is continuous, and the outer endpoint terms vanish because $\phi$ is compactly supported. Thus [F2] identifies $g$ as the weak derivative, and [F1] gives $v\in W^{1,2}(0,1)$.

Put $L={\max }_i\mathrm{\mid }(v{\mathrm{\mid }}_{[x_{i-1},x_i]}{)}^{\mathrm{\prime }}\mathrm{\mid }$. For $0<\delta <1\mathrm{/}4$, let ${\chi }_{\delta }$ be zero on $[0,\delta ]\cup [1-\delta ,1]$, linear from zero to one on $[\delta ,2\delta ]$, equal to one on $[2\delta ,1-2\delta ]$, and linear from one to zero on $[1-2\delta ,1-\delta ]$. Then $v_{\delta }={\chi }_{\delta }v$ is continuous and piecewise quadratic, with compact support in $[\delta ,1-\delta ]$. Its piecewise derivative is bounded; integration by parts on its finitely many pieces cancels interior boundary terms by continuity, as in step 1.1, so $v_{\delta }\in W^{1,2}(0,1)$. Since $\mathrm{\mid }v(x)\mathrm{\mid }\le L\min (x,1-x)$, its difference from $v$ is supported on strips of total length at most $4\delta$; there $\mathrm{\mid }v\mathrm{\mid }\le 2L\delta$, $\mathrm{\mid }{v}^{\mathrm{\prime }}\mathrm{\mid }\le L$, and $\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le 1\mathrm{/}\delta$. The product rule therefore gives $\mathrm{\parallel }v-v_{\delta }{\mathrm{\parallel }}_{L^2}^2\le 16L^2{\delta }^3$ and $\mathrm{\parallel }(v-v_{\delta }{)}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}^2\le 36L^2\delta$. The $W^{1,2}$ formula in [F1] then gives $v_{\delta }\to v$ in $W^{1,2}(0,1)$ as $\delta \to 0^{+}$.

Extend $v_{\delta }$ by zero to $V_{\delta }\in W^{1,2}(\mathbb{R})$ using [F5]. Choose the [F6] bump with $r=1\mathrm{/}2$ and $R=1$. Its integral $c$ is finite because it is bounded and supported in $(-1,1)$, and $c\ge 1$ because it equals $1$ on an interval of length $1$. Normalize it by $c$ and take its mollifier family from [F7]. The convolution $V_{\delta }\ast {\rho }_{\epsilon }$ is smooth by [F10]. Differentiating under the integral and integrating by parts on the finitely many polynomial pieces of $V_{\delta }$ gives $(V_{\delta }\ast {\rho }_{\epsilon }{)}^{\mathrm{\prime }}=V_{\delta }^{\mathrm{\prime }}\ast {\rho }_{\epsilon }$: interior boundary terms cancel by continuity and outer terms vanish by compact support. By [F8]–[F9], the function and its derivative converge in $L^2(\mathbb{R})$ to $V_{\delta }$ and $V_{\delta }^{\mathrm{\prime }}$ as $\epsilon \to 0^{+}$. The normalized bump has support in $[-1,1]$, so $0<\epsilon <\delta$ places the convolution support inside $(0,1)$ and it belongs to $C_c^{\mathrm{\infty }}(0,1)$.

Given $\eta >0$, choose $\delta$ so that step 2.1 gives $\mathrm{\parallel }v-v_{\delta }{\mathrm{\parallel }}_{W^{1,2}}<\eta \mathrm{/}2$, and then choose sufficiently small $\epsilon$ in step 3.1 so that $\mathrm{\parallel }{v}_{\delta }-(V_{\delta }\ast {\rho }_{\epsilon }){\mathrm{\mid }}_{(0,1)}{\mathrm{\parallel }}_{W^{1,2}}<\eta \mathrm{/}2$. The triangle inequality [F4] gives a $C_c^{\mathrm{\infty }}(0,1)$ approximation to $v$ within $\eta$. Since $\eta$ was arbitrary, [F3] yields $v\in H_0^1(0,1)$.