= Uniqueness of the Fourier transform on integrable functions
{title2=$\widehat f=\widehat g\Longrightarrow f=g\text{ almost everywhere}$}
For <integrable functions> on the real line, equality of their <Fourier transforms> implies equality almost everywhere. To prove it, subtract and assume $\widehat h=0$. Convolve $h$ with a <Gaussian heat kernel> $q_t$. The <convolution theorem> gives $\widehat{h*q_t}=0$, which is integrable, so the <Fourier inversion theorem> gives $h*q_t=0$ almost everywhere. Finally $q_t$ is an <approximate identity>: $\|h*q_t-h\|_1\leq\int q_t(y)\|h(\cdot-y)-h\|_1dy\to0$. This follows from <translation continuity in Lp> with exponent one and the vanishing Gaussian tails. Hence $h=0$ in $L^1$. If the two original functions are continuous, equality almost everywhere strengthens to equality everywhere, since a nonzero continuous difference has magnitude bounded below by a positive constant on some interval.
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