Define $g$ by the inverse <integral>. Its absolute convergence gives $|g(x)|\leq(2\pi)^{-d}\|\widehat f\|_1$. If $x_n\to x$, the exponential factors converge pointwise and their differences are bounded by $2|\widehat f|$. The <dominated convergence theorem> proves $g(x_n)\to g(x)$. Therefore \b[$g$ is bounded and <continuous> and equals $f$ almost everywhere].

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 24J / d / Solution

 Solution (source code)