(i) Let $P_{r}\left(e^{i \theta}\right)$ be the real part of $\frac{1+r e^{i \theta}}{1-r e^{i \theta}}$. Establish the following properties of $P_{r}$ for $0 \leqslant r<1$ : (a) $0<P_{r}\left(e^{i \theta}\right)=P_{r}\left(e^{-i \theta}\right) \leqslant \frac{1+r}{1-r}$; (b) $P_{r}\left(e^{i \theta}\right) \leqslant P_{r}\left(e^{i \delta}\right)$ for $0<\delta \leqslant|\theta| \leqslant \pi$; (c) $P_{r}\left(e^{i \theta}\right) \rightarrow 0$, uniformly on $0<\delta \leqslant|\theta| \leqslant \pi$, as $r$ increases to 1 .

(ii) Suppose that $f \in L^{1}(\mathbf{T})$, where $\mathbf{T}$ is the unit circle $\left\{e^{i \theta}:-\pi \leqslant \theta \leqslant \pi\right\}$. By definition, $\|f\|_{1}=\frac{1}{2 \pi} \int_{-\pi}^{\pi}\left|f\left(e^{i \theta}\right)\right| d \theta$. Let

$$P_{r}(f)\left(e^{i \theta}\right)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}\left(e^{i(\theta-t)}\right) f\left(e^{i t}\right) d t$$

Show that $P_{r}(f)$ is a continuous function on $\mathbf{T}$, and that $\left\|P_{r}(f)\right\|_{1} \leqslant\|f\|_{1}$.

[You may assume without proof that $\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}\left(e^{i \theta}\right) d \theta=1$.]

Show that $P_{r}(f) \rightarrow f$, uniformly on $\mathbf{T}$ as $r$ increases to 1 , if and only if $f$ is a continuous function on $\mathbf{T}$.

Show that $\left\|P_{r}(f)-f\right\|_{1} \rightarrow 0$ as $r$ increases to 1 .