Gaussian Fourier summability at Lebesgue points

By F1 and Gaussian integrability, the defining integral for $S_{t}f(x)$ is absolutely convergent. Inserting the definition of $\widehat{f}$, the product integrand has modulus $\mathrm{\mid }f(y)\mathrm{\mid }{e}^{-\pi t\mathrm{\mid }\xi {\mathrm{\mid }}^2}$, whose double integral is $\mathrm{\parallel }f{\mathrm{\parallel }}_1{t}^{-n\mathrm{/}2}<\mathrm{\infty }$. Use a Borel representative of f as supplied in F5's convolution construction; the resulting integrand is product measurable. F4 therefore gives $S_{t}f(x)=\int f(y)[\int e^{-\pi t\mathrm{\mid }\xi {\mathrm{\mid }}^2}{e}^{2\pi i(x-y)\cdot \xi }d\xi ]dy=\int f(y)k_t(x-y)dy$ by F2. This last integral exists at every x since k_t is bounded; substitution $y\mapsto x-y$ gives the stated convolution convention.

F2 and F5 now give $\mathrm{\parallel }{S}_{t}f-f{\mathrm{\parallel }}_1\to 0$. For pointwise recovery, take $K(y)=e^{-\pi \mathrm{\mid }y{\mathrm{\mid }}^2}$, $\mathrm{\Phi }(r)=e^{-\pi r^2}$ and $\epsilon =\sqrt{t}$ in F3. This profile is bounded, nonnegative and decreasing, its radial integral is one, and the dilated kernel is exactly k_t. Thus at every specified Lebesgue value a, $S_{t}f(x)\to a$. The two assertions use different estimates; norm convergence alone has not been used to infer pointwise convergence.