Borel–Carathéodory theorem

Let a function ${\displaystyle f}$ be analytic on a closed disc of radius R centered at the origin. Suppose that r < R. Then, we have the following inequality:

${\displaystyle \|f\|_{r}\leq {\frac {2r}{R-r}}\sup _{|z|\leq R}\operatorname {Re} f(z)+{\frac {R+r}{R-r}}|f(0)|.}$

Here, the norm on the left-hand side denotes the maximum value of f in the closed disc:

${\displaystyle \|f\|_{r}=\max _{|z|\leq r}|f(z)|=\max _{|z|=r}|f(z)|}$

(where the last equality is due to the maximum modulus principle).

Define A by

${\displaystyle A=\sup _{|z|\leq R}\operatorname {Re} f(z).}$

If f is constant c, the inequality follows from ${\displaystyle (R+r)|c|+2r\operatorname {Re} c\geq (R-r)|c|}$, so we may assume 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,

${\displaystyle \operatorname {Re} f(0)={\frac {1}{2\pi }}\int _{|z|=R}\operatorname {Re} f(z)dz.}$

Since f is regular and nonconstant, we have that Re f is also nonconstant. Since Re f(0) = 0, we must have Re ${\displaystyle f(z)>0}$ for some z on the circle ${\displaystyle |z|=R}$, so we may take ${\displaystyle A>0}$. Now 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.

${\displaystyle w\mapsto w/A-1}$ sends P to the standard left half-plane. ${\displaystyle w\mapsto R(w+1)/(w-1)}$ 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:

${\displaystyle w\mapsto {\frac {Rw}{w-2A}}.}$

From Schwarz's lemma applied to the composite of this map and f, we have

${\displaystyle {\frac {|Rf(z)|}{|f(z)-2A|}}\leq |z|.}$

Take |z| ≤ r. The above becomes

${\displaystyle R|f(z)|\leq r|f(z)-2A|\leq r|f(z)|+2Ar}$

so

${\displaystyle |f(z)|\leq {\frac {2Ar}{R-r}}}$,

as claimed. In the general case, we may apply the above to f(z)-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.

We start with the following result:[1]

Theorem — If ${\displaystyle f=u+iv}$ is analytic on ${\displaystyle B(0,R+\epsilon )}$ for some ${\displaystyle \epsilon >0}$, and ${\displaystyle u\leq M}$ on ${\displaystyle \partial B(0,R)}$, then ${\displaystyle \forall n\geq 1}$,

$${\displaystyle |f^{(n)}(0)|\leq {\frac {2n!}{R^{n}}}(M-u(0))}$$

and similarly if ${\displaystyle v\leq M}$.

Proof[2]

It suffices to prove the ${\displaystyle u}$ case, since the ${\displaystyle v}$ case is found by ${\displaystyle -if=v-iu}$.

WLOG, subtract a constant away, to get ${\displaystyle f(0)=0}$.

Do three contour integrals around ${\displaystyle \partial B(0,R)}$ using Cauchy integral formula:

$${\displaystyle f^{(n)}(0)/n!={\frac {1}{2\pi i}}\oint {\frac {f(z)}{z^{n+1}}}dz=\int _{0}^{1}{\frac {f(Re^{2\pi it})}{R^{n}e^{2\pi int}}}dt}$$

$${\displaystyle \int _{0}^{1}f(Re^{2\pi it})e^{2\pi int}dt={\frac {1}{2\pi iR^{n}}}\oint f(z)z^{n-1}dz=0}$$

$${\displaystyle \int _{0}^{1}f(Re^{2\pi it})dt={\frac {1}{2\pi i}}\oint f(z)z^{-1}dz=f(0)=0}$$

Pick angle ${\displaystyle \theta }$, so that ${\displaystyle e^{-i\theta }f^{(n)}(0)=|f^{(n)}(0)|}$. Then by linearly combining the three integrals, we get

$${\displaystyle \int _{0}^{1}f(Re^{2\pi it})dt(1+\cos(2\pi nt+\theta ))={\frac {1}{2}}R^{n}|f^{(n)}(0)|/n!}$$

The imaginary part vanishes, and the real part gives

$${\displaystyle \int _{0}^{1}u(Re^{2\pi it})dt(1+\cos(2\pi nt+\theta ))={\frac {1}{2}}R^{n}|f^{(n)}(0)|/n!}$$

The integral is bounded above by ${\displaystyle M\int _{0}^{1}dt(1+\cos(2\pi nt+\theta ))=M}$, so we have the result.

Corollary 1 — With the same assumptions, for all ${\displaystyle r\in (0,R)}$,

$${\displaystyle \max _{z\in \partial B(0,r)}|f(z)|\leq {\frac {2r}{R-r}}M+{\frac {R+r}{R-r}}|f(0)|}$$

Proof

It suffices to prove the case of ${\displaystyle f(0)=0}$.

By previous result, using the Taylor expansion,

$${\displaystyle |f(z)|\leq \sum _{n=0}^{\infty }{\frac {1}{n!}}|f^{(n)}(0)|\cdot |z|^{n}\leq |f(0)|+\sum _{n=1}^{\infty }2M(r/R)^{n}={\frac {2r}{R-r}}M}$$

Corollary 2 (Titchmarsh, 5.51, improved) — With the same assumptions, for all ${\displaystyle r\in (0,R)}$, and all integer ${\displaystyle n\geq 1}$

$${\displaystyle \max _{z\in \partial B(0,r)}|f^{(n)}(z)|\leq {\frac {2n!}{(R-r)^{n+1}}}R(M-u(0))}$$

Proof

It suffices to prove the case of ${\displaystyle f(0)=0}$ as well. And similarly to above,

$${\displaystyle {\begin{aligned}|f^{(n)}(z)|&\leq \sum _{k=n}^{\infty }{\frac {k\cdots (k-n+1)}{k!}}|f^{(k)}(0)|\cdot |z|^{k-n}\\&\leq {\frac {2(M-u(0))}{R^{n}}}\sum _{k=n}^{\infty }k\cdots (k-n+1)\left({\frac {r}{R}}\right)^{k-n}\\&={\frac {2n!}{(R-r)^{n+1}}}R(M-u(0))\end{aligned}}}$$

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 ${\displaystyle 0<r<r_{k}}$,

$${\displaystyle \max _{z\in \partial B(0,r)}|f^{(n+1)}(z)|\leq {\frac {4n!}{(r_{k}-r)^{n+2}}}r_{k}M}$$

where ${\displaystyle M=\max _{z\in \partial B(0,r_{k})}\Re (f(z))=o(r_{k}^{n+1})}$.

Letting ${\displaystyle r\leq {\frac {r_{k}}{2}}}$, and taking the ${\displaystyle k\nearrow \infty }$ limit:

$${\displaystyle \max _{z\in \partial B(0,r)}|f^{(n+1)}(z)|=o(1)\to 0}$$

Thus by Liouville's theorem, ${\displaystyle f^{(n+1)}}$ is a constant function, and it converges to zero, so it is zero. So ${\displaystyle f}$ is a polynomial of degree at most ${\displaystyle n}$.

• 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.