= Solution
Convergence $\phi_j\to\phi$ in the <space of test functions> means that all <supports> lie in one <compact set> $K\subset\mathbb R$ and $\|\phi_j^{(r)}-\phi^{(r)}\|_\infty\to0$ for every integer $r\geq0$. An eventual common <compact support> is equivalent, since finitely many initial <supports> can be added to $K$.
Write the standard <mollifier> as $\rho_\varepsilon(x)=\varepsilon^{-1}\rho(x/\varepsilon)$, where $\rho\in\mathcal D(\mathbb R)$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\varepsilon\leq1$, the <convolution> $\rho_\varepsilon*\phi$ has <support> in the fixed <compact set> $\operatorname{supp}\phi+[-1,1]$. The given <convolution derivative identity> yields
$$
(\rho_\varepsilon*\phi)^{(r)}(x)-\phi^{(r)}(x)
=\int_{-1}^{1}\rho(y)\bigl[\phi^{(r)}(x-\varepsilon y)-\phi^{(r)}(x)\bigr],dy.
$$
Each $\phi^{(r)}$ is uniformly continuous; more explicitly, the <mean value theorem> bounds the absolute value by $\varepsilon\|\phi^{(r+1)}\|_\infty\int|y|\rho(y)\,dy$, uniformly in $x$. Thus
$$
\boxed{\rho_\varepsilon*\phi\longrightarrow\phi\text{ in }\mathcal D(\mathbb R).}
$$
If $\phi_j\to0$ in the <space of test functions> and $K$ is a common <compact support>, then $\|\phi_j\|_{L^1}\leq|K|\|\phi_j\|_\infty\to0$, where $|K|$ is its <Lebesgue measure>.
The <mollifiers> themselves have no limit in the <L1 space>. If $\rho_\varepsilon\to g$ in that <norm>, then for each $\delta>0$ the integral of $|g|$ on $\{|x|>\delta\}$ is zero, because $\rho_\varepsilon$ vanishes there for $\varepsilon<\delta$. Hence $g=0$ <almost everywhere>. This contradicts $\|\rho_\varepsilon\|_{L^1}=1$. Therefore $\boxed{\rho_\varepsilon\text{ does not converge in }L^1(\mathbb R)}$.
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