= Solution
Identify the <circle> with $\mathbb T=\mathbb R/(2\pi\mathbb Z)$ and use normalized <Fourier coefficients> and <convolution>:
$$
\widehat f(n)=\frac1{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}\,dt,\qquad (f*g)(t)=\frac1{2\pi}\int_{-\pi}^{\pi}f(t-s)g(s)\,ds.
$$
For $0\leq r<1$, summing two <geometric series> gives the <Poisson kernel on the circle>:
$$
1+\sum_{n\geq1}r^n(e^{int}+e^{-int})=\frac1{1-re^{it}}+\frac1{1-re^{-it}}-1=\frac{1-r^2}{1-2r\cos t+r^2}=P_r(t).
$$
This <Fourier series> converges absolutely and uniformly for each fixed $r<1$. Consequently we can integrate it term by term in the <convolution>; the substitution $u=t-s$ gives
$$
\boxed{(f*P_r)(t)=\sum_{n\in\mathbb Z}\widehat f(n)r^{|n|}e^{int}.}
$$
The resulting series is itself absolutely and uniformly convergent, since $|\widehat f(n)|\leq\|f\|_\infty$ and $\sum_n r^{|n|}<\infty$.
The <Poisson kernel on the circle> is nonnegative and has normalized integral one, as its constant <Fourier coefficient> is one. Its mass concentrates near zero. Indeed, for $r\geq1/2$ and $\delta\leq|s|\leq\pi$,
$$
0\leq P_r(s)=\frac{1-r^2}{(1-r)^2+2r(1-\cos s)}\leq\frac{1-r^2}{1-\cos\delta}.
$$
Hence its normalized integral outside $(-\delta,\delta)$ tends to zero as $r\uparrow1$. Given $\eta>0$, <uniform continuity> of $f$ gives a $\delta>0$ such that $|f(t-s)-f(t)|<\eta$ whenever the circular distance of $s$ from zero is less than $\delta$. Splitting the <convolution> error into this arc and its complement yields
$$
\sup_t|(f*P_r)(t)-f(t)|\leq\eta+2\|f\|_\infty\frac1{2\pi}\int_{\delta\leq|s|\leq\pi}P_r(s)\,ds.
$$
The second term tends to zero, and $\eta$ is arbitrary. Thus \b[the Abel sums converge uniformly to $f$], or $\boxed{\|f*P_r-f\|_\infty\to0}$. This is <uniform Poisson summability of continuous circle functions>: positivity, unit mass and concentration supply the required <approximate identity> argument.
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