Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 5 / 4

Let $\varphi$ be a test function supported in $[-R,R]$. Symmetric cancellation givesThe integrand is bounded by $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, so the limit exists andThis is a continuous linear functional on every fixed-support test space, hence a distribution. It also specifies its local order as at most one.

For the boundary value, split the ordinary complex denominator:The real part paired with $\varphi$ isIt converges to the principal value by dominated convergence theorem, using the same bound $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. In the second part set $x=\epsilon t$:Here $\Vert \varphi {\Vert }_{\infty }/(1+t^2)$ is an integrable majorant. The correct Sokhotski–Plemelj formula is thereforeThe resulting functional is continuous, with bound $2R\Vert {\varphi }^{\prime }{\Vert }_{\infty }+\pi \Vert \varphi {\Vert }_{\infty }$, so it too is a distribution.

The printed PDF has the wrong sign on its delta term. For a real even nonnegative test function with $\varphi (0)>0$, the principal value is zero while the imaginary part of every regularized integral is negative. Its limit is $-\pi \varphi (0)$, disproving the printed plus sign directly. A plus sign instead belongs to the boundary value $(x-i0{)}^{-1}$.

For every compactly supported smooth test function $\varphi$, integration by parts in the definition of the weak derivative givesBoth $\varphi$ and ${\partial }_{\ell }\varphi$ belong to $L^2({\mathbb{R}}^n)$, so the assumed weak limits implyThis is exactly the distributional characterization of ${\partial }_{\ell }g=h_{\ell }$. Since $g,h_1,\dots ,h_n$ are all in $L^2$, the defining condition of the H1 space is satisfied. ThusFor complex functions the same argument uses conjugated test functions in the $L^2$ pairing. No prior identification of an $H^1$ weak limit is needed.