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

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.