Weak convergence of distributions

A sequence of distributions $u_j$ converges weakly to $u$ when $⟨u_j,\phi ⟩\to ⟨u,\phi ⟩$ for every test function $\phi$.

If $f\in L^1(0,1)$ has mean zero, $j$ is a positive integer and $\varphi \in \mathcal{D}(0,1)$, then for every integer $N\ge 1$,Here the overline denotes the periodic extension, not complex conjugation. Let $h_N=P^{N}f$ use periodic primitives with zero mean. Repeated integration by parts gives the signed identityThere are no boundary terms, and integration over exactly $j$ periods gives $\Vert {\overline{h}}_N(j\cdot ){\Vert }_1=\Vert h_N{\Vert }_1\le \Vert f{\Vert }_1$. This proves the estimate. If zero-mean $g_j$ satisfy $\Vert g_j{\Vert }_1=O(j^m)$ for fixed real $m\ge 0$, choose an integer $N>m$ to obtain ${\overline{g}}_j(j\cdot )\to 0$ in weak convergence of distributions.

A sequence can converge as distributions off one point while retaining a nonconvergent multiple of a Dirac delta function at that point. If $u_j=v+c_j{\delta }_p+o_{{\mathcal{D}}^{\prime }}(1)$ with $c_j$ nonconvergent, restriction to the punctured domain tends to $v$, but a test nonzero at $p$ detects the obstruction to global convergence.

For $a>0$, polar-coordinate reduction and integration by parts give $m\sin (m||x{|}^2-a|)=2\delta (|x{|}^2-a)-\pi \cos (ma){\delta }_0+o_{{\mathcal{D}}^{\prime }}(1)$ in two dimensions. The circle term comes from the cusp in the folded phase, while the origin term is a radial endpoint contribution. A punctured-domain limit therefore need not extend across the quadratic critical point.