90
VIII – Cauchy Theory
indeed follows. This is a product of convolutions on R, and since
P y (u)du =
1, it proves that
|u f (x, y)| ≤ ≤f ,
(11.10)
where the norm is a uniform one on R. On the other hand, for all a ∈ R,
|u f (x, y) − f (a)| ≤
P y (u) |f (x − u) − f (a)| du .
For all r > 0, the extended integral over {|u| ≥ r} is, up to a factor 2f ,
bounded above by the extended integral of P y over the same set, est hence,
as seen above, is ≤ ε for all x given any sufficiently small y . In the interval
|u| ≤ r, |x − u − a| ≤ |x − a| + r, and so |f (x − u) − f (a)| ≤ ε for any u
in this interval provided |x − a| and r are sufficiently small. This integral
is, therefore, ≤ ε since the value of the total integral of P y is 1. Finally,
|u f (x, y) − f (a)| ≤ 2ε for sufficiently small y and |x − a|, qed.
Exercise. If f (x) is uniformly continuous on R, the function x → u f (x, y)
converges uniformly to f (x) on R as y −→ +0.
The function u f , therefore, solves the Dirichlet problem for the half-plane
and for continuous and bounded functions on R: find a harmonic function on
an open set G with given values on the boundary. More precisely, it is one of
the possible solutions since any function of the form u f (x, y) + ay, where a
is a constant, is also a solution of the problem. Uniqueness holds in the case
of the unit disc D considered in Chapter VII, § 5 (Theorem 22) because, if a
continuous function on the closed disc and harmonic on the interior is zero
on the boundary, then a compactness argument and the maximum principle
show that it is identically zero. But the half-plane is not compact. In fact,
the map
z −→ ζ = (z − i)/(z + i) ,
is a conformal representation of the half-plane Im(z) > 0 on the unit disc D :
|ζ| < 1; it, therefore, transforms every holomorphic (resp. harmonic) function
on the former to a holomorphic (resp. harmonic) function on the latter. The
real axis y = 0 is homeomorphically mapped onto the boundary |ζ| = 1 with
the point 1 removed, which could be obtained by making z approach infinity.
Hence, for continuous functions f (x) on R, the Dirichlet problem in the halfplane, translated into the language of the unit disc amounts to finding a
continuous function on D − {1}, harmonic on D and with given values on
T − {1}, which allows every kind of behaviour in the neighbourhood of 1. In
fact,
y = Im(z) = (1 − |ζ|
2 )/|1 − ζ|
2
is easily computed to be the Poisson disc functional; it is harmonic on D,
continuous on D − {1} and zero on T − {1}, but takes every positive value in
VIII – Cauchy Theory
indeed follows. This is a product of convolutions on R, and since
P y (u)du =
1, it proves that
|u f (x, y)| ≤ ≤f ,
(11.10)
where the norm is a uniform one on R. On the other hand, for all a ∈ R,
|u f (x, y) − f (a)| ≤
P y (u) |f (x − u) − f (a)| du .
For all r > 0, the extended integral over {|u| ≥ r} is, up to a factor 2f ,
bounded above by the extended integral of P y over the same set, est hence,
as seen above, is ≤ ε for all x given any sufficiently small y . In the interval
|u| ≤ r, |x − u − a| ≤ |x − a| + r, and so |f (x − u) − f (a)| ≤ ε for any u
in this interval provided |x − a| and r are sufficiently small. This integral
is, therefore, ≤ ε since the value of the total integral of P y is 1. Finally,
|u f (x, y) − f (a)| ≤ 2ε for sufficiently small y and |x − a|, qed.
Exercise. If f (x) is uniformly continuous on R, the function x → u f (x, y)
converges uniformly to f (x) on R as y −→ +0.
The function u f , therefore, solves the Dirichlet problem for the half-plane
and for continuous and bounded functions on R: find a harmonic function on
an open set G with given values on the boundary. More precisely, it is one of
the possible solutions since any function of the form u f (x, y) + ay, where a
is a constant, is also a solution of the problem. Uniqueness holds in the case
of the unit disc D considered in Chapter VII, § 5 (Theorem 22) because, if a
continuous function on the closed disc and harmonic on the interior is zero
on the boundary, then a compactness argument and the maximum principle
show that it is identically zero. But the half-plane is not compact. In fact,
the map
z −→ ζ = (z − i)/(z + i) ,
is a conformal representation of the half-plane Im(z) > 0 on the unit disc D :
|ζ| < 1; it, therefore, transforms every holomorphic (resp. harmonic) function
on the former to a holomorphic (resp. harmonic) function on the latter. The
real axis y = 0 is homeomorphically mapped onto the boundary |ζ| = 1 with
the point 1 removed, which could be obtained by making z approach infinity.
Hence, for continuous functions f (x) on R, the Dirichlet problem in the halfplane, translated into the language of the unit disc amounts to finding a
continuous function on D − {1}, harmonic on D and with given values on
T − {1}, which allows every kind of behaviour in the neighbourhood of 1. In
fact,
y = Im(z) = (1 − |ζ|
2 )/|1 − ζ|
2
is easily computed to be the Poisson disc functional; it is harmonic on D,
continuous on D − {1} and zero on T − {1}, but takes every positive value in