For any continuous function $f:S^1\to\mathbb{C}$ prove that $\|\sum\hat{f}(n)r^{|n|}e_n-f\|_{\infty}\to 0$ as $r\to 1$. This is Exercise 1.4.9 from "Fourier series and integrals" by Mckean.

Problem:
For any continuous function $f:S^1\to\mathbb{C}$ prove that $\|\sum\hat{f}(n)r^{|n|}e_n-f\|_{\infty}\to 0$ as $r\to 1$.

In the previous exercise, I did show that the following identity holds
$$\sum_{n=-\infty}^{\infty} \hat{f}(n)r^{|n|}e^{2\pi i n x}=\int_{S^1}\dfrac{1-r^2}{1-2r\cos(2\pi(x-y))+r^2}f(y)dy.$$
I did so by bringing the sum into the integral on the left hand side and then used some algebraic manipulations.
$\text{ }$
In this exercise, they want me to use that identity to solve 1.4.9 and there is also the following hint:
$$\int_{S^1}\dfrac{1-r^2}{1-2r\cos(2\pi(x-y))+r^2} dy=\sum\hat{1}(n)r^{|n|}e_n(0)=1.$$
"Solution":
Let $f:S^1\to\mathbb{C}$ be a continuous function. We now have that
$$\Big\|\sum\hat{f}(n)r^{|n|}e_n-f\Big\|_{\infty}=\Big\|\int_{S^1}\dfrac{1-r^2}{1-2r\cos(2\pi(x-y))+r^2}f(y) dy-f\Big\|_{\infty}.$$
From here, I do not know how to continue. I have no idea how to use the hint they gave me. Sure, for $f=1$, and $x=0$, we have
$$\int_{S^1}\dfrac{1-r^2}{1-2r\cos(2\pi(x-y))+r^2} dy=\sum\hat{1}(n)r^{|n|}e_n(0)=1,$$
and so $\|\sum\hat{f}(n)r^{|n|}e_n-f\|_{\infty}\to 0$ as $r\to 1$.

But it doesn't feel like I have proven anything by doing this, since this was just for a particular function, namely $f=1$. What if I pick other functions? Have the above helped me in any way to prove it for any continuous function?
Is there something I am missing? Should I use this hint in a later stage of the proof? Should i rewrite the expression as something similar to this and use triangle inequality:
$$\Big\|\sum\hat{f}(n)r^{|n|}e_n-f\Big\|_{\infty}=\Big\|\int_{S^1}\dfrac{1-r^2}{1-2r\cos(2\pi(x-y))+r^2}f(y) dy+1-1-f\Big\|_{\infty},$$
and then apply the hint?
I would be very grateful if someone could help me solve this problem, give me some hint or help me understand the hint they gave me, in the book, better. Thanks! :)
 A: Denote 
$$P_r(y):=\dfrac{1-r^2}{1-2r\cos(2\pi y)+r^2}.$$
Observe that $P_r$ is non-negative. Once you obtain 
$$\int_{S^1}P_r(x-y)dy=1,$$
and use the fact that convolution is commutative, you may see that
$$\int_{S^1}P_r(x-y)f(y)dy-f(x)=\int_{S^1}P_r(y)f(x-y)dy-f(x)=\int_{S^1}P_r(y)(f(x-y)-f(x))dy.$$
Thus
$$\left\|\int_{S^1}P_r(\cdot-y)f(y)dy-f(\cdot)\right\|_\infty\leq \int_{S^1}|P_r(y)|\|f(\cdot-y)-f(\cdot)\|_\infty dy.$$
Since $f$ is continuous on $S^1$ and $S^1$ is compact, we see that $f$ is uniformly continuous on $S^1$. Thus $\forall\epsilon>0$, $\exists0<\delta<1$ such that
$$\|f(\cdot-y)-f(\cdot)\|_\infty<\epsilon,\quad\forall |y|<\delta.$$
Split $S^1$ into two parts:
$$S^1=(S^1\setminus\{y:|y|<\delta\})\cup\{y:|y|<\delta\}.$$
We have
$$\int_{\{y:|y|<\delta\}}|P_r(y)|\|f(\cdot-y)-f(\cdot)\|_\infty dy\leq\epsilon\int_{S^1}P_r(y)dy=\epsilon.$$
And
$$\begin{aligned}\int_{S^1\setminus\{y:|y|<\delta\}}|P_r(y)|\|f(\cdot-y)-f(\cdot)\|_\infty dy&\leq2\|f\|_\infty\int_{S^1\setminus\{y:|y|<\delta\}}P_r(y)dy\\
&\leq2\|f\|_\infty\int_{S^1\setminus\{y:|y|<\delta\}}\dfrac{1-r^2}{1-2r\cos(2\pi \delta)+r^2}\\
&\leq 2\|f\|_\infty\dfrac{1-r^2}{1-2r\cos(2\pi \delta)+r^2}\to 0,
\end{aligned}$$
as $r\to 1^-$. Thus we see that
$$\limsup_{r\to 1^-}\left\|\int_{S^1}P_r(\cdot-y)f(y)dy-f(\cdot)\right\|_\infty\leq\epsilon.$$
By letting $\epsilon\to 0^+$, we obtain the desired result.
