Borel–Carathéodory theorem

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

:$\backslash |f\backslash |\_r\; \backslash le\; \backslash frac\{2r\}\{R-r\}\; \backslash sup\_\{|z|\; \backslash le\; R\}\; \backslash operatorname\{Re\}\; f(z)\; +\; \backslash frac\{R+r\}\{R-r\}\; |f(0)|.$

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

:$\backslash |f\backslash |\_r\; =\; \backslash max\_\{|z|\; \backslash le\; r\}\; |f(z)|\; =\; \backslash max\_\{|z|\; =\; r\}\; |f(z)|$

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

Define A by

: $A\; =\; \backslash sup\_$

Take |z| ≤ r. The above becomes

:$R|f(z)|\; \backslash leq\; r|f(z)\; -\; 2A|\; \backslash leq\; r|f(z)|\; +\; 2Ar$

so

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

as claimed. In the general case, we may apply the above to f(z)-f(0):

:$$

\begin{align}

|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{align}

which, when rearranged, gives the claim.

We start with the following result:{{Cite web |last=Ishita Goluguri, Toyesh Jayaswal, Andrew Lee |title=The Prime Number Theorem: A PRIMES Exposition |url=https://math.mit.edu/research/highschool/primes/materials/2020/December/6-Goluguri-Jayaswal-Lee.pdf }}

{{Math theorem

| name = Theorem

| note =

| math_statement = If $f\; =\; u\; +\; iv$ is analytic on $B(0,\; R+\backslash epsilon)$ for some $\backslash epsilon\; >\; 0$, and $u\; \backslash leq\; M$ on $\backslash partial\; B(0,\; R)$, then $\backslash forall\; n\backslash geq\; 1$,

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

and similarly if $v\; \backslash leq\; M$.

}}

{{Math proof|title=Proof{{Cite web |last=Liu |first=Travor |title=Borel-Caratheodory Lemma and Its Application |date=16 February 2021 |url=https://travorlzh.github.io/2021/02/16/borel-caratheodory-lemma.html}}|proof=

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

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

Do three contour integrals around $\backslash partial\; B(0,\; R)$ using Cauchy integral formula:

$$f^\{(n)\}(0)/n!\; =\backslash frac\{1\}\{2\backslash pi\; i\}\; \backslash oint\; \backslash frac\{f(z)\}\{z^\{n+1\}\}dz\; =\; \backslash int\_0^1\; \backslash frac\{f(Re^\{2\backslash pi\; i\; t\})\}\{R^ne^\{2\backslash pi\; in\; t\}\; \}dt$$

$$\backslash int\_0^1\; f(Re^\{2\backslash pi\; i\; t\})\; e^\{2\backslash pi\; int\}dt\; =\; \backslash frac\{1\}\{2\backslash pi\; i\; R^n\}\backslash oint\; f(z)\; z^\{n-1\}dz\; =\; 0$$

$$\backslash int\_0^1\; f(Re^\{2\backslash pi\; i\; t\})dt\; =\; \backslash frac\{1\}\{2\backslash pi\; i\}\backslash oint\; f(z)\; z^\{-1\}dz\; =\; f(0)\; =\; 0$$

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

$$\backslash int\_0^1\; f(Re^\{2\backslash pi\; i\; t\})dt\; (1\; +\; \backslash cos(2\backslash pi\; nt\; +\; \backslash theta))\; =\; \backslash frac\; 12\; R^n\; |f^\{(n)\}\; (0)|/n!$$

The imaginary part vanishes, and the real part gives

$$\backslash int\_0^1\; u(Re^\{2\backslash pi\; i\; t\})dt\; (1\; +\; \backslash cos(2\backslash pi\; nt\; +\; \backslash theta))\; =\; \backslash frac\; 12\; R^n\; |f^\{(n)\}\; (0)|/n!$$

The integral is bounded above by $M\backslash int\_0^1\; dt\; (1\; +\; \backslash cos(2\backslash pi\; nt\; +\; \backslash theta))\; =\; M$, so we have the result.

}}

{{Math theorem

| name = Corollary 1

| note =

| math_statement = With the same assumptions, for all $r\backslash in\; (0,\; R)$,

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

}}

{{Math proof|title=Proof|proof=

It suffices to prove the case of $f(0)\; =\; 0$.

By previous result, using the Taylor expansion,

$$|f(z)|\; \backslash leq\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\{1\}\{n!\}|f^\{(n)\}(0)|\backslash cdot\; |z|^n\; \backslash leq\; |f(0)|\; +\; \backslash sum\_\{n=1\}^\backslash infty2M\; (r/R)^n\; =\; \backslash frac\{2r\}\{R-r\}\; M$$

}}

{{Math theorem|name = Corollary 2

| note = Titchmarsh, 5.51, improved|math_statement=

With the same assumptions, for all $r\backslash in\; (0,\; R)$, and all integer $n\backslash geq\; 1$

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

}}

{{Math proof|title=Proof|proof=

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

$$\backslash begin\{align\}$$

|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 rR\right)^{k-n} \\

&= \frac{2n!}{(R-r)^{n+1}} R(M- u(0))

\end{align}

}}

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.

{{Math theorem

| name = Liouville's theorem, improved

| note =

| math_statement = If $f$ is an entire function, such that there exists a sequence $r\_k\; \backslash nearrow\; \backslash infty$ with $\backslash max\_\{z\; \backslash in\; \backslash partial\; B(0,\; r\_k)\}\; \backslash Re(f(z))\; =\; o(r\_k^\{n+1\})$, then $f$ is a polynomial of degree at most $n$.

}}

{{Math proof|title=Proof|proof=

By Borel-Caratheodory lemma, for any ,

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

where $M\; =\; \backslash max\_\{z\backslash in\; \backslash partial\; B(0,\; r\_k)\}\; \backslash Re(f(z))\; =\; o(r\_k^\{n+1\})$.

Letting $r\; \backslash leq\; \backslash frac\{r\_k\}\{2\}$, and taking the $k\; \backslash nearrow\; \backslash infty$ limit:

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

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

}}

{{Math theorem

| name = Corollary

| note

| math_statement = If an entire function $f$ has no roots and is of finite order $\backslash rho$, then $f(z)\; =\; e^\{p(z)\}$ for some polynomial $p$ of degree $deg(p)\; \backslash leq\; \backslash rho$.

}}

{{Math proof|title=Proof|proof=

Apply the improved Liouville theorem to $g\; =\; \backslash log(f)$.

}}

{{reflist}}

• 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.

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Category:Theorems in complex analysis

Category:Articles containing proofs