Pointwise convergent functions need not converge as distributions without local control

F2 supplies the bump by taking a nonzero nonnegative test with the required support and dividing by its positive finite integral. For $j\ge 1$, the support of $f_j$ lies in $(3\mathrm{/}(2j),5\mathrm{/}(2j))$. Thus $f_j(x)=0$ for every $j\in \mathbb{N}$ when $x\le 0$, and for each fixed $x>0$ it is zero once $j>5\mathrm{/}(2x)$. In particular the pointwise limit is zero even at the origin. Each $f_j$, including $f_0$, is smooth and compactly supported, so F1 applies.

For every test $\phi$ and every $j\ge 1$, the substitution $t=jx-2$ from F3 gives $u_{f_j}(\phi )=\int \rho (t)\phi ((t+2)\mathrm{/}j)dt$. Since $\mathrm{\mid }t+2\mathrm{\mid }\le 5\mathrm{/}2$ on the bump support, [step 1.1, F1, F3] $\mathrm{\mid }{u}_{f_j}(\phi )-\phi (0)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }s\mathrm{\mid }\le 5\mathrm{/}(2j)}\mathrm{\mid }\phi (s)-\phi (0)\mathrm{\mid }⟶0.$ Here positivity and unit mass give the inequality, and continuity at zero gives the limit. Thus the weak limit is ${\delta }_0$. Choose a cutoff test equal to one near zero by F2; its pairings are eventually one, whereas the zero regular distribution pairs to zero. This is the failed conclusion for the explicit witness sequence. Its mass is one for every $j$, concentrated in a shrinking interval; pointwise convergence alone does not control these pairings. $\mathrm{\square }$