Holomorphic function preserving real and imaginary axis

Question

Let $f$ be a holomorphic function such that $f(\mathbb{R})\subseteq \mathbb{R}$ and $f(i\mathbb{R})\subseteq i\mathbb{R}$. Prove that $f(-z)=-f(z)$.

We want to prove $g(z)=f(-z)+f(z)$ is zero. Notice that $g(0)=2f(0)=0$, because $f(0)\in \mathbb{R}\cap i\mathbb{R}$. If we prove there is an accumulation point of zeros of $g$ converging to $0$, we are done by the identity theorem.

Let me write $f(z)=u(x,y)+iv(x,y)$. In this case, $$g(z)=(u(x,y)+u(-x,-y))+i(v(x,y)+v(-x,-y))$$

$g(x)=u(x,0)+u(-x,0)$ and also $g(iy)=i(v(0,y)+v(0,-y))$. If we prove $g'(x)$ to be constant, then we have our function $g$ being zero in all of $\mathbb{R}$ and we are done as this clearly implies $0$ has an accumulation point of zeros and is itself a zero! By the Cauchy Riemman equations:

$$g'(x)=u_x(x,0)-u_x(-x,0)=v_y (x,0)-v_y(-x,0)$$

We notice $v(x,0)=0$ for any $x$ because of $f(\mathbb{R})\subseteq \mathbb{R}$. This doesn't seem very promissing.

Answer 1

Say $f(z) = \sum_{n=0}^\infty a_n z^n$ is the Taylor Series to $f$ around $0$. $f(\mathbb R) \subseteq \mathbb R$ implies that all $f^{(n)}(0)$ and hence all $a_n$ are real.

Now consider $f(z) +f(-z) = \sum_{n \text{ even}} a_n z^n$. By assumption this function evaluated on $i\mathbb R$ has values in $i\mathbb R$; but also, since all powers are even and all $a_n \in \mathbb R$, its values are in $\mathbb R$, so they are all $0$. Since $i\mathbb R$ has accumulation points, by the identity theorem, $f(-z)=-f(z)$ for all $z$.

Answer 2

A solution that doesn't use Schwartz reflection principle - write $f(z)=\sum_{n=0}^\infty a_nz^n$. By our assumption, for every $x\in\mathbb{R}$: $$f(ix)=\sum_{n=0}^\infty a_ni^n x^n=\sum_{n=0}^\infty (-1)^na_{2n}x^{2n}+i\sum_{n=0}^n(-1)^na_{2n+1}x^{2n+1}\in i\mathbb{R}$$

First of all, notice that all derivatives of $f$ sastisfy the same condition. Why? well note that for every real $x_0$: $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\in\mathbb{R}$$ as this is a limit of real numbers. Therefore, $f^{(n)}(\mathbb{R})\subset\mathbb{R}$ and hence each $a_n$ is real. Therefore: $$\sum_{n=0}^\infty (-1)^na_{2n}x^{2n}\in\mathbb{R},\,i\sum_{n=0}^n(-1)^na_{2n+1}x^{2n+1}\in i\mathbb{R}$$ So, since $f(ix)\in i\mathbb{R}$, this must mean that: $$\sum_{n=0}^\infty (-1)^na_{2n}x^{2n}$$ For all $x\in\mathbb{R}$. By the uniqueness theorem, this implies that $a_{2n}=0$ for all $n$, hence $f$ is odd.

Edit - solution using the Schwarz reflection principle. It basically says that if you have $f$ that is defined on the upper half plane and is real on the real axis, it can be extended analytically to the lower half plane and it must satisfy $f(z)=\overline{f(\overline{z})}$. Now use the condition that $f(ix)\in i\mathbb{R}$ to see that for every $x\in\mathbb{R}$: $$f(ix)=\overline{f(-ix)}=-f(-ix)$$ As this is true on a set with an accumulation point, $f(z)=f(-z)$ for all $z\in\mathbb{C}$ by the uniqueness theorem.