# Ngô Quốc Anh

## June 7, 2009

### Schwarz Reflection Principle and several applications

Suppose that $f$ is an analytic function which is defined in the upper half-disk $\{|z|^2<1,I[z]>0\}$. Further suppose that $f$ extends to a continuous function on the real axis, and takes on real values on the real axis. Then $f$ can be extended to an analytic function on the whole disk by the formula

$\displaystyle f(\overline z)=\overline{f(z)}$

and the values for $z$ reflected across the real axis are the reflections of $f(z)$ across the real axis. It is easy to check that the above function is complex differentiable in the interior of the lower half-disk. What is remarkable is that the resulting function must be analytic along the real axis as well, despite no assumptions of differentiability.

This is called the Schwarz reflection principle, and is sometimes also known as Schwarz’s symmetric principle (Needham 2000, p. 257). The diagram above shows the reflection principle applied to a function f defined for the upper half-disk (left figure; red) and its image (right figure; red). The function is real on the real axis, so it is possible to extend the function to the reflected domain (left and right figures; pink).

For the reflected function to be continuous, it is necessary for the values at the boundary to be continuous and to fall on the line being reflected. The reflection principle also applies in the generality of reflecting along any line, not just the real axis, in which case the function $f$ has to take values along a line in the range. In fact, any arc which has a neighborhood biholomorphic to a straight line can be reflected across. The basic example is the boundary of the unit circle which is mapped to the real axis by $z \mapsto \frac{z+1}{z+i}$.

The reflection principle can also be used to reflect a harmonic function which extends continuously to the zero function on its boundary. In this case, for negative $y$, defining

$\displaystyle v(x,y)=-v(x,-y)$

extends $v$ to a harmonic function on the reflected domain. Again note that it is necessary for  $v(x,0)=0$. This result provides a way of extending a harmonic function from a given open set to a larger open set (Krantz 1999, p. 95).

Source:

Several applications

• Let $f$ be a function meromorphic in the closed unit disk $\{ |z| \leq 1\}$. Suppose that $|f|$ is constant on the unit circle $|z|=1$.  Prove that $f$ must be a rational function.

Proof. If $|f| = 0$ on $\partial\mathbb{D}$, then $f = 0$ and the statement is true. W.L.O.G suppose $|f| = 1$ on $\partial\mathbb{D}$, i.e. $f(\partial\mathbb{D}) \subset \partial\mathbb{D}$ and suppose that $f \neq \textrm{const.}$ $f$ is holomorphic as a function on $\mathbb{D}$ to the Riemann sphere $\hat{\mathbb{C}}$, and $f: \overline{\mathbb{D}}\to\hat{\mathbb{C}}$ is continuous. Since $\overline{\mathbb{D}}$ is compact, so is $f(\overline{\mathbb{D}})$ and thus closed in $\hat{\mathbb{C}}$. In particular $f(\overline{\mathbb{D}}) = \overline{f(\mathbb{D})}$. Furthermore, since non-constant holomorphic mappings are open, $f(\mathbb{D})$ is open. In conclusion

$\displaystyle\partial(f(\mathbb{D})) =\overline{f(\mathbb{D})}\,\backslash\,f(\mathbb{D}) = f(\partial\mathbb{D})\subset\partial\mathbb{D}$

Therefore either $f(\mathbb{D}) = \mathbb{D}$ or $f(\mathbb{D}) = \hat{\mathbb{C}}\,\backslash\,\overline{\mathbb{D}}$ and, in both cases, $f(\partial\mathbb{D}) = \partial\mathbb{D}$.

Suppose $f(\mathbb{D}) = \mathbb{D}$ (the other case is treated similarly). Now by

$\displaystyle f(z) : = \overline{\frac {1}{f\left(\frac {1}{\overline{z}}\right)}},\quad |z| > 1$,

i.e. reflection along $\partial\mathbb{D}$, we have a, in fact, the holomorphic continuation of $f$ to $\hat{\mathbb{C}}$ by the Schwarz reflection principle (if you’re not familiar with this: The idea is that this continuation is clearly holomorphic in $\hat{\mathbb{C}}\,\backslash\,\overline{\mathbb{D}}$; on $\partial\mathbb{D}$ use Morera’s theorem). We know not that $f: \hat{\mathbb{C}}\to\hat{\mathbb{C}}$ is holomorphic. Thus $f$ has to be a rational function.

Comments: In the above solution, we can construct

$\displaystyle f(z) : = \overline{\frac {1}{f\left(\frac {1}{\overline{z}}\right)}},\quad |z| > 1$,

to be a holomorphic function by mean of the Schwarz Reflection Principle. Roughtly speaking, we first go back to the domain $\{|z| <1\}$ by the mapping $\frac{1}{\overline z}$. Then we compute its image under the function $f$. Then we go forward by the mapping

$\displaystyle\frac{1}{\overline{f(\frac{1}{\overline z})}}$

which equals to

$\displaystyle\overline{\frac {1}{f\left(\frac {1}{\overline{z}}\right)}}$.

The above complex conjugate operation reflects the word “reflection ” in this principle. Similar to the above question, we have the following

• Let $f$ be holomorphic in the unit disk $\{ |z| < 1\}$, continuous in $\{ |z| \leq 1\}$ and $|f(z)|=1$ whenever $|z|=1$. Prove that $f$ is a rational function.

Proof. If $f$ has infinite many zeros, by the isolatedness of the zeros of holomorphis functions, the zeros must have limit points on the boundary of the unit disk, otherwise, $f \equiv 0$. But it will violate the fact that $f$ is continuous in $\{|z| \leq 1 \}$ and $|f(z)|=1$ whenever $|z|=1$. Hence $f$ has only finite zeros in the unit disk. Denote all these zeros by $z_1, ..., z_n$, multiple zeros being repeated, and define,

$\displaystyle F\left( z\right) =\frac{{f\left( z\right)}}{{\prod\limits_{k = 1}^{n}{\left({\frac{{z-{z_{k}}}}{{1-\overline{{z_{k}}}z}}}\right)}}}$.

Then $F$ is holomorphic in $\{|z|<1\}$, continuous in $\{ |z| \leq 1\}$ and $|F(z)|=1$ whenever $|z|=1$. By Maximum Modulus Principle, $|F(z)| \leq 1$ in $\{|z| <1\}$. Since $F(z)$ has no zero in $\{|z|<1\}$, $\frac{1}{F(z)}$ is also holomorphic in $\{|z|<1\}$, continuous in $\{ |z| \leq 1\}$ and $\left|\frac{1}{F(z)}\right|=1$ whenever $|z|=1$. Application of the Maximum Modulus Principle to $\frac{1}{F(z)}$ yields $|F(z)| \geq 1$ in $\{|z|\leq 1\}$. Hence $|F(z)|=1$ holds in $\{|z| \leq 1\}$, which implies $F(z)=e^{i\alpha}$ with $\alpha$ a real number. So we obtain

$\displaystyle f\left( z \right) = {e^{i\alpha }}\prod\limits_{k = 1}^n {\left( {\frac{{z - {z_k}}} {{1 - \overline {{z_k}} z}}} \right)}$

which completes the proof.

Comments: It doesn’t have to just sound like one. Basically, if $f$ has a pole in $b$ you should multiply by the conformal disc automorphism $\frac {z - b}{1 - \overline{b}z}$ and divide by it if $f$ has a zero in $b$. The motivation is basically just that I knew it . The standard way to construct an analytic function in the disc with zeros in $(b_n)$ is to multiply together infinitely many factors

$\displaystyle\frac {z - b_n}{1 - \overline{b_n}z}$.

This is called a Blaschke product. There are factorization theorems saying that analytic functions belonging to certain growth classes (the Smirnov class, for example) can be factored into products of a Blaschke product (describing the zeros), a singular inner function (describing the angle of $f(z)$), and an outer function (describing the growth). With the very nice function described in the problem I knew that it would have no growth and no inner factor either (they can’t be continuous on the circle), so essentially I just knew the solution to the problem. Although I guess the solution can be easily guessed if products of Möbius transforms of the type above are familiar to you as exactly the rational functions that have constant modulus on the unit circle.