38
VIII – Cauchy Theory
where 1/p + 1/q = 1 (Chap. V, § 3, no 14); the first integral is less that
f p and the second does not depend on f , hence (16). (16’) can be similarly
obtained from (12).
The reader will easily verify that relation (16) continues to hold if, instead
of the usual measure dxdy, we use a measure dμ(z) = ρ(z)dm(z) to define
f p , i.e. the formula
f p =
G
|f (z)|
p ρ(z)dxdy
1/p
,
(4.18)
where the given “ density ” ρ is continuous and has strictly positive values. It
suffices to note that the minimum m over K(r) of ρ is > 0. Hence
K(r)
|f (z)|
p dm(z) ≤
1
m
K(r)
|f (z)|
p ρ(z)dm(z)
for any function f .
From this result, it follows that, for any measure dμ(z) of the form (18),
the normed vector space H
p (G, μ) of holomorphic functions such that
|f (z)|
p dμ(z) < +∞
is complete:
27 Indeed, inequality (16) transforms every Cauchy sequence with
respect to the norm L
p into a Cauchy sequence with respect to compact
convergence. Hence we get a holomorphic limit which can easily be inferred
to also be the L
p - limit of the functions f n . This is obvious if we have the
benefit of the simplest results from Lebesgue theory.
28
In particular, H
2 (G, μ), equipped with the scalar product
(f |g) =
f (z)g(z)dμ(z) ,
is a Hilbert space which plays an important role in some questions, especially
complex Fourier transformations, the theory of modular functions, conformal
representations, etc.
In fact, as observed by Laurent Schwartz half a century ago, much more
can be proved. Let G be an open subset of C and D(G) the vector space of
27 As we will see in n
◦ 12, it can be reduced to zero if the open set G is not bounded,
especially in the trivial case when G = C since (17) then applies for all r > 0.
28 Given a Cauchy sequence (fn) in an L
p -space , if lim fn(x) = f (x) exists almost
everywhere, then f ∈ L
p and lim f − fnp = 0 (Chap. XI). In the case at hand,
the functions fn and f are continuous and the sequence converges uniformly on
every compact subset. Hence the theorem is in fact about Riemann integrals.
But a proof based on elementary arguments requires far more ingenuity than the
use of Lebesgue’s sledgehammer.
VIII – Cauchy Theory
where 1/p + 1/q = 1 (Chap. V, § 3, no 14); the first integral is less that
f p and the second does not depend on f , hence (16). (16’) can be similarly
obtained from (12).
The reader will easily verify that relation (16) continues to hold if, instead
of the usual measure dxdy, we use a measure dμ(z) = ρ(z)dm(z) to define
f p , i.e. the formula
f p =
G
|f (z)|
p ρ(z)dxdy
1/p
,
(4.18)
where the given “ density ” ρ is continuous and has strictly positive values. It
suffices to note that the minimum m over K(r) of ρ is > 0. Hence
K(r)
|f (z)|
p dm(z) ≤
1
m
K(r)
|f (z)|
p ρ(z)dm(z)
for any function f .
From this result, it follows that, for any measure dμ(z) of the form (18),
the normed vector space H
p (G, μ) of holomorphic functions such that
|f (z)|
p dμ(z) < +∞
is complete:
27 Indeed, inequality (16) transforms every Cauchy sequence with
respect to the norm L
p into a Cauchy sequence with respect to compact
convergence. Hence we get a holomorphic limit which can easily be inferred
to also be the L
p - limit of the functions f n . This is obvious if we have the
benefit of the simplest results from Lebesgue theory.
28
In particular, H
2 (G, μ), equipped with the scalar product
(f |g) =
f (z)g(z)dμ(z) ,
is a Hilbert space which plays an important role in some questions, especially
complex Fourier transformations, the theory of modular functions, conformal
representations, etc.
In fact, as observed by Laurent Schwartz half a century ago, much more
can be proved. Let G be an open subset of C and D(G) the vector space of
27 As we will see in n
◦ 12, it can be reduced to zero if the open set G is not bounded,
especially in the trivial case when G = C since (17) then applies for all r > 0.
28 Given a Cauchy sequence (fn) in an L
p -space , if lim fn(x) = f (x) exists almost
everywhere, then f ∈ L
p and lim f − fnp = 0 (Chap. XI). In the case at hand,
the functions fn and f are continuous and the sequence converges uniformly on
every compact subset. Hence the theorem is in fact about Riemann integrals.
But a proof based on elementary arguments requires far more ingenuity than the
use of Lebesgue’s sledgehammer.