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Identify the circle with $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and use normalized Fourier coefficients and convolution:For $0\le r<1$, summing two geometric series gives the Poisson kernel on the circle: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 $u=t-s$ givesThe resulting series is itself absolutely and uniformly convergent, since $|\hat{f}(n)|\le \Vert f{\Vert }_{\infty }$ 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 $r\ge 1/2$ and $\delta \le |s|\le \pi$,Hence its normalized integral outside $(-\delta ,\delta )$ tends to zero as $r\uparrow 1$. 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 yieldsThe second term tends to zero, and $\eta$ is arbitrary. Thus the Abel sums converge uniformly to $f$, or $\boxed{\Vert f\ast P_r-f{\Vert }_{\infty }\to 0}$. This is uniform Poisson summability of continuous circle functions: positivity, unit mass and concentration supply the required approximate identity argument.