Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 4

Weak convergence of distributions means $⟨T_n,\phi ⟩\to ⟨T,\phi ⟩$ for every test function $\phi \in C_c^{\infty }(\Omega )$. The distributional derivative is defined byIt is linear and continuous: differentiation preserves compact support and bounds each test-function seminorm by one of higher derivative order, so composing it with the continuous functional $T$ again gives a distribution. Finally, for every fixed test function,Thus distributional differentiation preserves the stated convergence.

The function $U(x)=1/|x|$ is locally integrable in ${\mathbb{R}}^3$, so it defines a distribution. The claimed distributional identity means that for every $\phi \in C_c^{\infty }({\mathbb{R}}^3)$,Choose a ball containing the support and remove $B_{\epsilon }(0)$. On the punctured region $\Delta U=0$. Green's second identity givesThe outer boundary contributes zero. On the inner boundary, the outward normal for the punctured region is $-\hat{r}$, so ${\partial }_{n}U=+1/{\epsilon }^2$, fixing the sign.

The first term is $O(\epsilon )$; the second equals $-{\int }_{S^2}\phi (\epsilon \omega )d\omega \to -4\pi \phi (0)$. Local integrability allows the left side to converge to its full-space integral. Therefore $\boxed{-\Delta (1/|x|)=4\pi {\delta }_0}$. In this sign convention, $1/(4\pi |x|)$ is the fundamental solution of the Laplace equation for the operator $-\Delta$.

The functions are nonnegative and normalized to have integral one. On the disk $|x|\le (2n{)}^{-1/4}$, Bernoulli inequality gives $(1-|x{|}^4{)}^n\ge 1-n|x{|}^4\ge 1/2$. ConsequentlyFor any fixed $0<a<1$, the mass outside radius $a$ is at mostThis is an approximate identity. For any test function,Let $n\to \infty$ and then $a\downarrow 0$ to conclude $\boxed{g_n\to {\delta }_0\text{ in }{\mathcal{D}}^{\prime }({\mathbb{R}}^2)}$.

Polar coordinates also give the exact normalization through the beta function: