Poisson kernel transform and Abel summability on the line

Put $q_a(x)=e^{-2\pi a\mathrm{\mid }x\mathrm{\mid }}$. It is integrable with integral $1\mathrm{/}(\pi a)$, by F1 on finite half-intervals and F2 for the positive exponential tails. Integration on $[0,R]$ gives $[1-e^{-2\pi (a+i\xi )R}]\mathrm{/}[2\pi (a+i\xi )]$. The omitted absolute tail is $e^{-2\pi aR}\mathrm{/}(2\pi a)$, so the half-line transform is $1\mathrm{/}[2\pi (a+i\xi )]$. Reflecting the negative half gives $1\mathrm{/}[2\pi (a-i\xi )]$. Their sum is $a\mathrm{/}[\pi (a^2+{\xi }^2)]=P_a(\xi )$.

The rational P_a is bounded on a compact core and bounded by $a\mathrm{/}(\pi x^2)$ for $\mathrm{\mid }x\mathrm{\mid }\ge 1$, hence integrable by the elementary convergent inverse-square improper tail and F2. F3 applied to q_a gives ${\widehat{P}}_a(-x)=q_a(x)$ almost everywhere. Both sides are continuous, so equality holds everywhere (a nonzero continuous difference cannot vanish a.e. on an interval). Evenness then gives the stated transform and, at zero, $\int P_a=1$.

In the Abel integral, inserting $\widehat{f}$ gives a double absolute bound $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }{q}_a{\mathrm{\parallel }}_1<\mathrm{\infty }$. F4 exchanges the integrals and step 1.1 identifies the inner inverse integral with $P_a(x-y)$, since P_a is even. Thus the mean is $f\ast P_a$, absolutely at every x since P_a is bounded. Finally $P_a(y)=a^{-1}{P}_1(y\mathrm{/}a)$ and $\mathrm{\Phi }(r)=1\mathrm{/}[\pi (1+r^2)]$ is bounded, decreasing and integrable with radial mass one. F5 gives the asserted Lebesgue-point limit. All integral and pointwise suppliers carry the stated countable choice.