Mollifying a zero extension leaks across the boundary

Given: the Axiom of Choice; the constant $u\equiv 1$ on $(0,1)$; its zero extension ${\chi }_{(0,1)}$; a nonnegative even unit-mass $\rho \in C_c^{\mathrm{\infty }}(\mathbb{R})$ supported in $[-1,1]$; a scale $0<\epsilon <1\mathrm{/}2$ and its rescaling ${\rho }_{\epsilon }$; the restriction $v_{\epsilon }=({\rho }_{\epsilon }\ast {\chi }_{(0,1)}){\mathrm{\mid }}_{(0,1)}$; and $1\le p\le \mathrm{\infty }$.

Under Countable Choice (implied by the assumed Axiom of Choice), $W_0^{1,p}(0,1)$ is the closure of $C_c^{\mathrm{\infty }}(0,1)$ in the $W^{1,p}(0,1)$ norm: $w\in W_0^{1,p}(0,1)$ if and only if for every $\delta >0$ there is a test function on $(0,1)$ within $\delta$ of $w$ (Zero-boundary Sobolev space as a norm closure).

Under the assumed Axiom of Choice, every $w\in W^{1,p}(I)$ on a nonempty open interval has exactly one continuous representative $w^{\ast }$, which is locally absolutely continuous; if $I=(a,b)$ has finite endpoints then $w^{\mathrm{\prime }}\in L^1(a,b)$, $w^{\ast }$ extends uniquely to an absolutely continuous function on $[a,b]$, and $w^{\ast }(x)=w^{\ast }(a)+{\int }_a^x{w}^{\mathrm{\prime }}$ for $x\in [a,b]$ (One-dimensional $W^{1,p}$ functions have unique absolutely continuous representatives).

Mean-value and endpoint estimate: for the representative of [F2] on $I=(0,1)$ there is $a\in (1\mathrm{/}4,3\mathrm{/}4)$ with $\mathrm{\mid }{w}^{\ast }(a)\mathrm{\mid }\le 2{\int }_0^1\mathrm{\mid }w\mathrm{\mid }$, and likewise at the endpoint $1$, hence $\mathrm{\mid }{w}^{\ast }(0)\mathrm{\mid }\le 2{\int }_0^1\mathrm{\mid }w\mathrm{\mid }+{\int }_0^1\mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }$; by Hölder on the finite interval this is at most $C_p\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,p}(0,1)}$ with a constant $C_p$ depending only on $p$ (Holder's inequality for integrals, including the endpoint cases, [F2]).

If $\phi \in C_c^{\mathrm{\infty }}(0,1)$, then the continuous representative of $[F2]$ is $\phi$ itself and $\phi (0)=\phi (1)=0$ by compact support in the open interval. [F2, given]

For ${\chi }_{(0,1)}\in L^1(\mathbb{R})$, the convolution ${\rho }_{\epsilon }\ast {\chi }_{(0,1)}$ is smooth on $\mathbb{R}$ and equals ${\int }_{\mathbb{R}}{\chi }_{(0,1)}(y){\rho }_{\epsilon }(x-y)dy$; it is the classical convolution of Convolution with a mollifier is smooth, and derivatives pass under the integral sign. By the support hypothesis on $\rho$ and its rescaling in The mollifier family generated by a unit-mass smooth bump, ${\rho }_{\epsilon }$ is supported in $[-\epsilon ,\epsilon ]$.

The zero extension ${\chi }_{(0,1)}$ does not belong to $W^{1,p}(\mathbb{R})$ for any $1\le p\le \mathrm{\infty }$ (A nonzero boundary value creates a zero-extension jump).

Choice use. The Axiom of Choice licenses the representative interface [F2], including its fundamental-theorem prerequisites. Countable Choice is inherited by the closure and convolution interfaces [F1] and [F5] and selects the sequence of test approximants in step 2.1.