Verify that the following is a continuous contraction semigroups on $C(\TT)$.
$$
P_tf(x)\colon=\frac1{2\pi}\int_0^{2\pi} P(e^{-t},x-y)f(y)\,dy
\quad\mbox{where}\quad
P(r,x)\colon=\frac{1-r^2}{1-2r\cos x+r^2}
=\Re\Big(\frac{1+re^{ix}}{1-re^{ix}}\Big)~.
$$
This is called the Poisson semigroup on the torus $\TT=S^1$. 2. Show verify that $P_te_m=e^{-t|m|}e_m$ for $e_m(x)=e^{imx}$, $m\in\Z$. 3. Compute the generator of $P_t$. Remark: Use the fact that for $f\in C(\TT)$ and $z=e^{-t}e^{ix}$, then function $z\mapsto P_tf(x)$ is the harmonic extension of $f$ into the unit disc $D\colon=[|z| < 1]$.
1. The function $z=e^{-t}e^{ix}\mapsto P_tf(x)$ is the harmonic extension of $f$ and the harmonic extension of $g(x)\colon=P_tf(x)$ is $z=e^{-s}e^{ix}\mapsto P_sg(x)$. Thus the latter is also the harmonic extension of $f$ at $z=e^{-s}e^{-t}e^{ix}$, i.e.
$$
P_sP_tf(x)=P_{s+t}f(x)~.
$$
2. 3. For $m\geq0$ the function $e_m$ is the boundary function of the analytic (and hence harmonic) function $z\mapsto z^m$ and thus by the remark: $P_te_m=(e^{-t}e^{ix})^m=e^{-mt}e_m(x)$. For $m < 0$ the function $e_m$ is the boundary function of the anti analytic (and hence harmonic) function $z\mapsto\bar z^{-m}$. Therefore $P_te_m=(e^{-t}e^{-ix})^{-m}=e^{mt}e_m(x)$. It follows that the generator $-H$ of $P_t$ is defined on the complex trigonometric polynomials $p=\sum_{j\in J} c_je_j$, $J\sbe\Z$ finite, and
$$
\forall m\in\Z:\quad
-He_m=-|m|e_m~.
$$