In mathematics, the Borel–Carathéodory
theorem in complex analysis shows that an
analytic function may be bounded by its real
part. It is an application of the maximum
modulus principle. It is named for Émile
Borel and Constantin Carathéodory.
== Statement of the theorem ==
Let a function
f
{\displaystyle f}
be analytic on a closed disc of radius R centered
at the origin. Suppose that r < R. Then, we
have the following inequality:
‖
f
‖
r
≤
2
r
R
−
r
sup
|
z
|
≤
R
Re
⁡
f
(
z
)
+
R
+
r
R
−
r
|
f
(
0
)
|
.
{\displaystyle \|f\|_{r}\leq {\frac {2r}{R-r}}\sup
_{|z|\leq R}\operatorname {Re} f(z)+{\frac
{R+r}{R-r}}|f(0)|.}
Here, the norm on the left-hand side denotes
the maximum value of f in the closed disc:
‖
f
‖
r
=
max
|
z
|
≤
r
|
f
(
z
)
|
=
max
|
z
|
=
r
|
f
(
z
)
|
{\displaystyle \|f\|_{r}=\max _{|z|\leq r}|f(z)|=\max
_{|z|=r}|f(z)|}
(where the last equality is due to the maximum
modulus principle).
== Proof ==
Define A by
A
=
sup
|
z
|
≤
R
Re
⁡
f
(
z
)
.
{\displaystyle A=\sup _{|z|\leq R}\operatorname
{Re} f(z).}
First let f(0) = 0. Since Re f is harmonic,
we may take A>0. f maps into the half-plane
P to the left of the x=A line. Roughly, our
goal is to map this half-plane to a disk,
apply Schwarz's lemma there, and make out
the stated inequality.
w
↦
w
/
A
−
1
{\displaystyle w\mapsto w/A-1}
sends P to the standard left half-plane.
w
↦
R
(
w
+
1
)
/
(
w
−
1
)
{\displaystyle w\mapsto R(w+1)/(w-1)}
sends the left half-plane to the circle of
radius R centered at the origin. The composite,
which maps 0 to 0, is the desired map:
w
↦
R
w
w
−
2
A
.
{\displaystyle w\mapsto {\frac {Rw}{w-2A}}.}
From Schwarz's lemma applied to the composite
of this map and f, we have
|
R
f
(
z
)
|
|
f
(
z
)
−
2
A
|
≤
|
z
|
.
{\displaystyle {\frac {|Rf(z)|}{|f(z)-2A|}}\leq
|z|.}
Take |z| ≤ r. The above becomes
R
|
f
(
z
)
|
≤
r
|
f
(
z
)
−
2
A
|
≤
r
|
f
(
z
)
|
+
2
A
r
{\displaystyle R|f(z)|\leq r|f(z)-2A|\leq
r|f(z)|+2Ar}
so
|
f
(
z
)
|
≤
2
A
r
R
−
r
{\displaystyle |f(z)|\leq {\frac {2Ar}{R-r}}}
,as claimed. In the general case, we may apply
the above to f(z)-f(0):
|
f
(
z
)
|
−
|
f
(
0
)
|
≤
|
f
(
z
)
−
f
(
0
)
|
≤
2
r
R
−
r
sup
|
w
|
≤
R
Re
⁡
(
f
(
w
)
−
f
(
0
)
)
≤
2
r
R
−
r
(
sup
|
w
|
≤
R
Re
⁡
f
(
w
)
+
|
f
(
0
)
|
)
,
{\displaystyle {\begin{aligned}|f(z)|-|f(0)|&\leq
|f(z)-f(0)|\leq {\frac {2r}{R-r}}\sup _{|w|\leq
R}\operatorname {Re} (f(w)-f(0))\\&\leq {\frac
{2r}{R-r}}\left(\sup _{|w|\leq R}\operatorname
{Re} f(w)+|f(0)|\right),\end{aligned}}}
which, when rearranged, gives the claim
