Exam

Verify that the following is a continuous contraction semigroups on

C (T)

.

P_{t} f (x) : = \frac{1}{2 π} \int_{0}^{2 π} P (e^{- t}, x - y) f (y) d y where P (r, x) : = \frac{1 - r^{2}}{1 - 2 r \cos x + r^{2}} = ℜ (\frac{1 + r e^{i x}}{1 - r e^{i x}}) .

This is called the Poisson semigroup on the torus

T = S^{1}

. 2. Show verify that

P_{t} e_{m} = e^{- t | m |} e_{m}

for

e_{m} (x) = e^{i m x}

,

m \in Z

. 3. Compute the generator of

P_{t}

. Remark: Use the fact that for

f \in C (T)

and

z = e^{- t} e^{i x}

, then function

z \mapsto P_{t} f (x)

is the harmonic extension of

f

into the unit disc

D : = [| z | < 1]

.

1. The function

z = e^{- t} e^{i x} \mapsto P_{t} f (x)

is the harmonic extension of

f

and the harmonic extension of

g (x) : = P_{t} f (x)

is

z = e^{- s} e^{i x} \mapsto P_{s} g (x)

. Thus the latter is also the harmonic extension of

f

at

z = e^{- s} e^{- t} e^{i x}

, i.e.

P_{s} P_{t} f (x) = P_{s + t} f (x) .

2. 3. For

m \geq 0

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_{t} e_{m} = (e^{- t} e^{i x})^{m} = e^{- m t} 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_{t} e_{m} = (e^{- t} e^{- i x})^{- m} = e^{m t} 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_{j} e_{j}

,

J \subseteq Z

finite, and

\forall m \in Z : - H e_{m} = - | m | e_{m} .