Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1 / b

For $u\in L_{\mathrm{loc}}^1(\mathbb{R})$, a function $g\in L_{\mathrm{loc}}^1(\mathbb{R})$ is its weak derivative whenfor every test function $\phi \in C_c^{\infty }(\mathbb{R})$. Equivalently, the distributional derivative of $u$ is represented by the locally integrable function $g$.

For $u(x)=\sin |x|$, the ordinary derivative away from zero isThis bounded function is locally integrable. For a test function $\phi$, integration by parts on $(-\pi ,0)$ and $(0,\pi )$ produces boundary terms at zero which cancel because $u$ is continuous there and $u(0)=0$. HenceTherefore $u$ has the weak derivativeThe jump in the ordinary derivative creates no Dirac delta function; such a term would arise from a jump in the function itself.