# Holomorphic function preserving real and imaginary axis

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.

• Are you familiar with Schwartz reflection principle? Commented May 1 at 18:42
• I am not. :(. But I will study it. Is it a corollary of this principle? Commented May 1 at 18:42
• I will try to post an answer that doesn't use it, but it is easier to use it Commented May 1 at 18:43

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 $$z$$.
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.