= Solution
The <weak convergence of distributions> $u_j\to u$ means
$$
\langle u_j,\phi\rangle\longrightarrow\langle u,\phi\rangle
\quad\text{for every }\phi\in\mathcal D(\mathbb R).
$$
For $\theta\in\mathcal D(\mathbb R)$, its <convolution> with a <distribution> is the <smooth function>
$$
(\theta*u)(x)=\langle u_y,\theta(x-y)\rangle.
$$
The subscript records the variable on which the <distribution> acts. With $\check\theta(x)=\theta(-x)$, its <regular distribution> satisfies the useful pairing identity
$$
\langle\theta*u,\phi\rangle=\langle u,\check\theta*\phi\rangle.
$$
Indeed integrating $\theta(x-y)\phi(x)$ in $x$ gives $(\check\theta*\phi)(y)$; continuity of the <distribution> justifies moving the pairing through that integral, by approximation with sums of <test functions> in one common <compact support>.
Part (i), applied to the reflected <mollifier>, gives $\check\rho_\varepsilon*\phi\to\phi$ in the <space of test functions>. Continuity of $u$ therefore proves
$$
\boxed{\langle\rho_\varepsilon*u,\phi\rangle
=\langle u,\check\rho_\varepsilon*\phi\rangle\longrightarrow\langle u,\phi\rangle,
\qquad\rho_\varepsilon*u\to u\text{ in }\mathcal D'.}
$$
Reflection is harmless even if the chosen standard <mollifier> is not even.
For the <mollifiers> themselves, a change of variable gives
$$
\langle\rho_\varepsilon,\phi\rangle=\int\rho(y)\phi(\varepsilon y)\,dy\longrightarrow\phi(0)\int\rho(y)\,dy=\phi(0).
$$
Thus $\boxed{\rho_\varepsilon\to\delta_0\text{ in }\mathcal D'(\mathbb R)}$, with limit the <Dirac delta distribution>, despite the failure of convergence in the <L1 space>.
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