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 [ edit ] Let a function f {\displaystyle f} 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).
Define A by
A = sup | z | ≤ R Re f ( z ) . {\displaystyle A=\sup _{|z|\leq R}\operatorname {Re} f(z).} If f is constant c , the inequality follows from ( R + r ) | c | + 2 r Re c ≥ ( R − r ) | c | {\displaystyle (R+r)|c|+2r\operatorname {Re} c\geq (R-r)|c|} f is nonconstant. First let f (0) = 0. Since Re f is harmonic, Re f (0) is equal to the average of its values around any circle centered at 0. That is,
Re f ( 0 ) = ∫ 0 1 Re f ( R e 2 π i s ) d s . {\displaystyle \operatorname {Re} f(0)=\int _{0}^{1}\operatorname {Re} f(R{\rm {e}}^{2\pi {\rm {i}}s}){\,{\rm {d}}}s.} Since f is regular and nonconstant, we have that Re f is also nonconstant. Since Re f (0) = 0, we must have Re f ( z ) > 0 {\displaystyle f(z)>0} z on the circle | z | = R {\displaystyle |z|=R} A > 0 {\displaystyle 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} P to the standard left half-plane. w ↦ R ( w + 1 ) / ( w − 1 ) {\displaystyle w\mapsto R(w+1)/(w-1)} 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.
Alternative result and proof [ edit ] We start with the following result:[ 1]
Proof[ 2] It suffices to prove the u {\displaystyle u} v {\displaystyle v} − i f = v − i u {\displaystyle -if=v-iu}
WLOG, subtract a constant away, to get f ( 0 ) = 0 {\displaystyle f(0)=0}
Do three contour integrals around ∂ B ( 0 , R ) {\displaystyle \partial B(0,R)}
f ( n ) ( 0 ) / n ! = 1 2 π i ∮ f ( z ) z n + 1 d z = ∫ 0 1 f ( R e 2 π i t ) R n e 2 π i n t d t {\displaystyle f^{(n)}(0)/n!={\frac {1}{2\pi i}}\oint {\frac {f(z)}{z^{n+1}}}dz=\int _{0}^{1}{\frac {f(Re^{2\pi it})}{R^{n}e^{2\pi int}}}dt}
∫ 0 1 f ( R e 2 π i t ) e 2 π i n t d t = 1 2 π i R n ∮ f ( z ) z n − 1 d z = 0 {\displaystyle \int _{0}^{1}f(Re^{2\pi it})e^{2\pi int}dt={\frac {1}{2\pi iR^{n}}}\oint f(z)z^{n-1}dz=0}
∫ 0 1 f ( R e 2 π i t ) d t = 1 2 π i ∮ f ( z ) z − 1 d z = f ( 0 ) = 0 {\displaystyle \int _{0}^{1}f(Re^{2\pi it})dt={\frac {1}{2\pi i}}\oint f(z)z^{-1}dz=f(0)=0}
Pick angle θ {\displaystyle \theta } e − i θ f ( n ) ( 0 ) = | f ( n ) ( 0 ) | {\displaystyle e^{-i\theta }f^{(n)}(0)=|f^{(n)}(0)|}
∫ 0 1 f ( R e 2 π i t ) d t ( 1 + cos ( 2 π n t + θ ) ) = 1 2 R n | f ( n ) ( 0 ) | / n ! {\displaystyle \int _{0}^{1}f(Re^{2\pi it})dt(1+\cos(2\pi nt+\theta ))={\frac {1}{2}}R^{n}|f^{(n)}(0)|/n!}
The imaginary part vanishes, and the real part gives
∫ 0 1 u ( R e 2 π i t ) d t ( 1 + cos ( 2 π n t + θ ) ) = 1 2 R n | f ( n ) ( 0 ) | / n ! {\displaystyle \int _{0}^{1}u(Re^{2\pi it})dt(1+\cos(2\pi nt+\theta ))={\frac {1}{2}}R^{n}|f^{(n)}(0)|/n!}
The integral is bounded above by M ∫ 0 1 d t ( 1 + cos ( 2 π n t + θ ) ) = M {\displaystyle M\int _{0}^{1}dt(1+\cos(2\pi nt+\theta ))=M}
Corollary 1 — With the same assumptions, for all r ∈ ( 0 , R ) {\displaystyle r\in (0,R)}
max z ∈ ∂ B ( 0 , r ) | f ( z ) | ≤ 2 r R − r M + R + r R − r | f ( 0 ) | {\displaystyle \max _{z\in \partial B(0,r)}|f(z)|\leq {\frac {2r}{R-r}}M+{\frac {R+r}{R-r}}|f(0)|}
Proof It suffices to prove the case of f ( 0 ) = 0 {\displaystyle f(0)=0}
By previous result, using the Taylor expansion,
| f ( z ) | ≤ ∑ n = 0 ∞ 1 n ! | f ( n ) ( 0 ) | ⋅ | z | n ≤ | f ( 0 ) | + ∑ n = 1 ∞ 2 M ( r / R ) n = 2 r R − r M {\displaystyle |f(z)|\leq \sum _{n=0}^{\infty }{\frac {1}{n!}}|f^{(n)}(0)|\cdot |z|^{n}\leq |f(0)|+\sum _{n=1}^{\infty }2M(r/R)^{n}={\frac {2r}{R-r}}M}
Corollary 2 (Titchmarsh, 5.51, improved) — With the same assumptions, for all r ∈ ( 0 , R ) {\displaystyle r\in (0,R)} n ≥ 1 {\displaystyle n\geq 1}
max z ∈ ∂ B ( 0 , r ) | f ( n ) ( z ) | ≤ 2 n ! ( R − r ) n + 1 R ( M − u ( 0 ) ) {\displaystyle \max _{z\in \partial B(0,r)}|f^{(n)}(z)|\leq {\frac {2n!}{(R-r)^{n+1}}}R(M-u(0))}
Proof It suffices to prove the case of f ( 0 ) = 0 {\displaystyle f(0)=0} | f ( n ) ( z ) | ≤ ∑ k = n ∞ k ⋯ ( k − n + 1 ) k ! | f ( k ) ( 0 ) | ⋅ | z | k − n ≤ 2 ( M − u ( 0 ) ) R n ∑ k = n ∞ k ⋯ ( k − n + 1 ) ( r R ) k − n = 2 n ! ( R − r ) n + 1 R ( M − u ( 0 ) ) {\displaystyle {\begin{aligned}|f^{(n)}(z)|&\leq \sum _{k=n}^{\infty }{\frac {k\cdots (k-n+1)}{k!}}|f^{(k)}(0)|\cdot |z|^{k-n}\\&\leq {\frac {2(M-u(0))}{R^{n}}}\sum _{k=n}^{\infty }k\cdots (k-n+1)\left({\frac {r}{R}}\right)^{k-n}\\&={\frac {2n!}{(R-r)^{n+1}}}R(M-u(0))\end{aligned}}}
Borel–Carathéodory is often used to bound the logarithm of derivatives, such as in the proof of Hadamard factorization theorem .
The following example is a strengthening of Liouville's theorem .
Proof By Borel-Caratheodory lemma, for any 0 < r < r k {\displaystyle 0<r<r_{k}}
max z ∈ ∂ B ( 0 , r ) | f ( n + 1 ) ( z ) | ≤ 4 n ! ( r k − r ) n + 2 r k M {\displaystyle \max _{z\in \partial B(0,r)}|f^{(n+1)}(z)|\leq {\frac {4n!}{(r_{k}-r)^{n+2}}}r_{k}M} M = max z ∈ ∂ B ( 0 , r k ) ℜ ( f ( z ) ) = o ( r k n + 1 ) {\displaystyle M=\max _{z\in \partial B(0,r_{k})}\Re (f(z))=o(r_{k}^{n+1})}
Letting r ≤ r k 2 {\displaystyle r\leq {\frac {r_{k}}{2}}} k ↗ ∞ {\displaystyle k\nearrow \infty }
max z ∈ ∂ B ( 0 , r ) | f ( n + 1 ) ( z ) | = o ( 1 ) → 0 {\displaystyle \max _{z\in \partial B(0,r)}|f^{(n+1)}(z)|=o(1)\to 0}
Thus by Liouville's theorem, f ( n + 1 ) {\displaystyle f^{(n+1)}} f {\displaystyle f} n {\displaystyle n}
Proof Apply the improved Liouville theorem to g = log ( f ) {\displaystyle g=\log(f)}
Lang, Serge (1999). Complex Analysis (4th ed.). New York: Springer-Verlag, Inc. ISBN 0-387-98592-1 . Titchmarsh, E. C. (1938). The theory of functions. Oxford University Press. 
