if $f$ is holomorphic and $f(z)=f(\overline{z})$ then $f$ is constant

Question

Prove that if $f$ is holomorphic at some point and also $f(z)=f(\overline{z})$ then $f$ must be constant.

Letting $f=u+iv$, and using Cauchy-Riemann, I was able to prove that $u(x,0)$ and $v(x,0)$ are constant (i.e. along the real axis), but do not know how to prove that this implies $u$ and $v$ are constant everywhere.

Will appreciate any help. Thank you

• Tveltzel: Try proving $f'(z)$ is 0ドル$
• Dr. John: It is indeed zero but only along the real axis. How can I continue from there? All we learned so far is Cauchy-Riemann and the definition of differentiability/holomorphicity/harmonic functions
• Malady: Do you have the identity principle?
• Tveltzel: Perhaps try the Taylor series expansion of $f(z),ドル and take particular care of what it means for all $f^{(n)}(x)=0$ and the coefficients of the expansion, where $x\in \mathbb{R}$
• peter a g: Try the chain rule on the RHS.

Answer 1

Let $f(z)$ be a function that is holomorphic in an open, connected domain $D \subset \mathbb{C}$ and satisfies $f(z) = f(\overline{z})$ for all $z \in D$. Then $f(z)$ is a constant function. Let $f(z) = u(x, y) + i v(x, y)$, where $z = x + iy$. Let $I = D \cap \mathbb{R}$ be the non-empty open segment of the real axis contained in $D$. The given condition $f(z) = f(\overline{z})$ implies: $$ u(x, y) + i v(x, y) = u(x, -y) + i v(x, -y) $$ Equating the real and imaginary parts shows that $u$ and $v$ are even functions of $y$: $$ u(x, y) = u(x, -y) \quad \text{and} \quad v(x, y) = v(x, -y) $$ Differentiating these relations with respect to $y$ and setting $y=0$ (along the real axis $I$): $$ \frac{\partial u}{\partial y}(x, 0) = -\frac{\partial u}{\partial y}(x, 0) \quad \implies \quad \frac{\partial u}{\partial y}(x, 0) = 0 $$ $$ \frac{\partial v}{\partial y}(x, 0) = -\frac{\partial v}{\partial y}(x, 0) \quad \implies \quad \frac{\partial v}{\partial y}(x, 0) = 0 $$ Now, we apply the Cauchy-Riemann equations, $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$: \begin{align*} \frac{\partial u}{\partial x}(x, 0) &= \frac{\partial v}{\partial y}(x, 0) = 0 \\ \frac{\partial v}{\partial x}(x, 0) &= -\frac{\partial u}{\partial y}(x, 0) = 0 \end{align*} The derivative of $f(z)$ along the real axis $I$ must vanish: $$ \mathbf{f'(x)} = \frac{\partial u}{\partial x}(x, 0) + i \frac{\partial v}{\partial x}(x, 0) = 0 + i(0) = \mathbf{0} \quad \text{for all } x \in I $$ Since $f(z)$ is holomorphic in $D$, its derivative $f'(z)$ is also holomorphic in $D$. We use the fact that holomorphicity implies analyticity. See here

Let $x_0 \in I$. The Taylor series expansion of $f(z)$ around $x_0$ is: $$ f(z) = \sum_{n=0}^{\infty} c_n (z - x_0)^n, \quad \text{where } c_n = \frac{f^{(n)}(x_0)}{n!} $$ Since $f'(x) = 0$ on the open segment $I$, all successive derivatives of $f(z)$ must also be zero when restricted to $I$: $$ f^{(n)}(x) = \frac{d^{n-1}}{dx^{n-1}} \left( f'(x) \right) = \frac{d^{n-1}}{dx^{n-1}} (0) = 0 \quad \text{for all } n \ge 1 $$ Thus, evaluating at $x_0$, we find $\mathbf{f^{(n)}(x_0) = 0}$ for all $n \ge 1$. This implies all coefficients $c_n$ for $n \ge 1$ are zero.

The series reduces to the constant term: $$ f(z) = c_0 = f(x_0) $$ This proves that $f(z)$ is constant in a disk around $x_0$. Because $f(z)$ is holomorphic and $D$ is connected, the uniqueness of analytic continuation (the Identity Theorem) forces this constancy to hold throughout the entire domain $D$. $$ \therefore \mathbf{f(z) = C}, \quad \text{where } C \in \mathbb{C} \text{ is a constant.} $$

• Dr. John: Thank you for that!
• Rεaδ my bi0: Saying $f$ is constant if $f’=0$ is a consequence of your space being connected. @Dr.John

Answer 2

Let $U\subset \Bbb{C}$ be a connected open set which is symmetric under complex conjugation. Suppose $f:U\to \Bbb{C}$ is holomorphic and satisfies $f(z)= f(\overline{z})$. Let $\sigma(z)=\overline{z}$ be the complex conjugation map. So our hypothesis is that $f= f\circ \sigma$.

A standard fact is that if $f$ is holomorphic then $\sigma\circ f\circ \sigma$ is holomorphic (just write out the difference quotient definition). So with our additional assumption that $f=f\circ \sigma$, it follows that $\sigma\circ f$ is holomorphic (I used $\sigma\circ \sigma$ is the identity... i.e taking complex conjugate twice gives the same number). In other words, both $f$ and its complex conjugate $\overline{f}$ are holomorphic. It follows by Cauchy-Riemann (and connectedness of the domain) that $f$ is constant.

Answer 3

I can provide you a rather proof / intuitive proof using a different form of Cauchy - Riemann equation. In many standard books authored by "Ahlfors" or "Stein-Shakarchi" we find this definition, that $$F_\bar{z}=0 \ \ \ \ \text{is equivalent to saying C-R Equation}$$ Now, we are given here that, $$F(z)=F(\bar{z})$$ Say our function contains terms of "z" [Like say $F(z)=a+bz$, type], hence from our given form, $F(z)=F(\bar{z})$, the term $F(\bar{z})$ will contain terms of $\bar{z}$.

Now, as we know $F(z)$ being holomorphic (Analytic) it indeed staisfies C-R equation, hence $F_\bar{z}$ will necessarily have to be $=0$, but this contradicts our fact that $F(z)$ has terms of $z$ or by saying $F(\bar{z})$ has terms of $\bar{z}$, becuase if it contains $\bar{z}$, the partial deriavtive with respect to $\bar{z}$ will atleast be no zero constant.

Hence, for all conditions to satisfy all at once the only possibily is $F(z)$ is constant.

Answer 4

Assume that $f$ is holomorphic on the plane (and not just at one point - cf. counter-example in comments above).

So, by Cauchy-Riemann, $$ {\partial \over \partial \bar z} f = {1\over 2}\Big({\partial \over \partial x} - {1\over i} {\partial \over \partial y}\Big)f = 0.$$

Suppose that $g= u+iv$ is a differentiable function from the complex plane to itself, and consider $h= f\circ g$.

Then:

$$ \begin{align*}{\partial h\over \partial x} &= {\partial f\over \partial x} {\partial u\over \partial x} + {\partial f\over \partial y} {\partial v\over \partial x}\\ &= {\partial f\over \partial x} {\partial u\over \partial x} + {1\over i}{\partial f\over \partial y} {\partial i v\over \partial x}. \end{align*}$$

Now, since $f$ is holomorphic (Cauchy-Riemann!), $$ {\partial f\over \partial z}= {\partial f\over \partial x} = {1\over i}{\partial f\over \partial y}. $$

Therefore,

$$\begin{align*} {\partial h\over \partial x}&={\partial f\over \partial z}{\partial \over \partial x}(u+i v)\\ &={\partial f\over \partial z}{ \partial g\over \partial x}. \end {align*}$$ Similarly, $${\partial h\over \partial y} = {\partial f \over \partial z} {\partial g\over \partial y}.$$

Therefore $$ \begin{align*} {\partial h\over \partial z} &= {1\over 2}\Big({\partial \over \partial x} + {1\over i} {\partial \over \partial y}\Big) h \\ &={\partial f \over \partial z} {\partial g\over \partial z}. \end{align*} $$

In our case $g(z) = \bar z$, so
$$ {\partial g \over \partial z} = 0. $$ Hence ${\partial h \over \partial z} =0.$ On the other hand, by hypothesis $f = h$, so ${\partial f \over \partial z} = 0.$

Now, $$ \begin{align*} {\partial \over \partial x} &= {\partial \over \partial z} + {\partial \over \partial \bar z},\\ {1\over i}{\partial \over \partial y} &= {\partial \over \partial z} - {\partial \over \partial \bar z} \end{align*}$$

Therefore $f$ is constant:
$${\partial f\over \partial x} = {\partial f\over \partial y} = 0. $$

• peter a g: I started writing this up earlier - and now I see that Stardust has basically given the same (if word-free!) answer. And, in the same spirit, Arnad Tapabar... And... Ah well, I'm leaving it anyway.