Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 6 / d

Both kernels are integrable functions, so the convolution theorem givesThe uniqueness of the Fourier transform on integrable functions gives equality almost everywhere. Both sides are continuous, so it holds at every point:For completeness, continuity of the convolution follows by dominated convergence: $q_t$ is bounded and continuous, so for $x_j\to x$ the integrands $q_t(x_j-y)q_s(y)$ converge pointwise and are dominated by $\Vert q_t{\Vert }_{\infty }{q}_s(y)$.

The identity is the heat semigroup property: evolving for time $s$, then for time $t$, is the same as evolving once for time $s+t$.