# Real-analytic $f(z)=f(\sqrt z) + f(-\sqrt z)$?

Are there nonconstant real-analytic functions $f(z)$ such that

$$f(z)=f(\sqrt z) + f(-\sqrt z)$$

is satisfied near the real line ?

Also can such functions be entire ? And/Or can they be periodic with a real period $p>0$ ?

Does the set of equations

$$f(z)=f(\sqrt z) + f(-\sqrt z)$$

$$f(z)=f(z+p)$$

$$f ' (0) > 0$$

imply that $f(z)= 0 + a_1 z + a_2 z^2 + a_3 z^3 + ...$ where more than 50 % of the nonzero signs of the $a_n$ are positive ?

Related :

Real-analytic periodic $f(z)$ that has more than 50 % of the derivatives positive?

• If a function is analytic and real-valued, it must be constant from the Cauchy--Riemann equations. – Alex Schiff Jun 19 '14 at 19:50
• real-analytic means mapping the reals to some reals and being complex differentiable on an near the real line. Your " and " interpretation seems flawed. – mick Jun 19 '14 at 19:52
• – mick Jun 19 '14 at 19:53
• I'm not sure what you mean by "mapping the reals to some reals." Do you mean that $f(\Bbb R)\subset \Bbb R$? – Alex Schiff Jun 19 '14 at 19:56
• Yes the range is a subset of the reals. – mick Jun 19 '14 at 19:56

As pointed out by Zarrax, $$f(z^2)=f(z)+f(-z)\tag{1}$$ implies $$\sum_{k=0}^\infty a_kz^{2k}=2\sum_{k=0}^\infty a_{2k}z^{2k}\tag{2}$$ therefore, $$a_k=2a_{2k}\tag{3}$$ Obviously, $a_0=0$. Given $a_k$ for odd $k$, $(3)$ allows us to compute all $a_k$.

Functions with power series that satisfy $(3)$ are \begin{align} f_n(z) &=-\log(1-z^{2n+1})\\[6pt] &=\sum_{k=1}^\infty\frac{z^{k(2n+1)}}{k}\tag{4} \end{align} and linear combinations of the $f_n$. Note that the $f_n$ satisfy $(1)$. In fact, \begin{align} \sum_{n=0}^\infty\frac{\mu(2n+1)}{2n+1}f_n(z) &=\sum_{n=0}^\infty\sum_{k=1}^\infty\frac{\mu(2n+1)}{2n+1}\frac{z^{k(2n+1)}}{k}\\[9pt] &=\sum_{n=1}^\infty\sum_{\substack{d\mid n\\d\text{ odd}}}\mu(d)\frac{z^n}{n}\\ &=\sum_{n=0}^\infty\frac{z^{2^n}}{2^n}\tag{5} \end{align} where $\mu$ is the Möbius function. The last equation in $(5)$ follows from $$\sum_{\substack{d\mid n\\d\text{ odd}}}\mu(d) =\left\{\begin{array}{} 1&\text{if n is a power of 2}\\ 0&\text{otherwise} \end{array}\right.\tag{6}$$ We can use \begin{align} f_n\left(z^{2k+1}\right) &=-\log\left(1-z^{(2k+1)(2n+1)}\right)\\ &=f_{2kn+k+n}(z)\tag{7} \end{align} to write all possible functions satisfying $(1)$ as linear combinations of the $f_n$. That is, use $(5)$ then $(7)$ to get \begin{align} \sum_{k=0}^\infty\sum_{n=0}^\infty a_{2k+1}\frac{z^{(2k+1)2^n}}{2^n} &=\sum_{k=0}^\infty\sum_{n=0}^\infty a_{2k+1}\frac{\mu(2n+1)}{2n+1}f_n\left(z^{2k+1}\right)\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty a_{2k+1}\frac{\mu(2n+1)}{2n+1}f_{2kn+k+n}(z)\tag{8} \end{align} Therefore, all $f$ that satisfy $(1)$ are given by linear combinations of the $f_n$.

For example, we can write the example given by mjqxxxx as $$\log(1+z+z^2)=f_0(z)-f_1(z)$$

The simplest example of such a function that is real-analytic over the entire real line is $$f(z)=\log\left(1+z+z^2\right).$$ We have $$f(\sqrt{z})+f(-\sqrt{z})=\log\left(1+\sqrt{z}+z\right)+\log\left(1-\sqrt{z}+z\right)\\=\log\left((1+z)^2-z\right)\\=\log\left(1+z+z^2\right)\\=f(z),$$ as desired. It's not entire, but is analytic in the neighborhood of the real line, since its branch cuts start at $z=-1/2\pm i\sqrt{3}/2$.

• I have used your example in my answer (+1) – robjohn Jun 23 '14 at 6:45

Write $f(z) = \sum_n a_nz^n$ as a power series near $z = 0$. Then the equation $f(z^2) = f(z) + f(-z)$ implies that $$\sum_{n = 0}^{\infty} a_n z^{2n} = \sum_{n=0}^{\infty} a_nz^n + a_n (-z)^n$$ $$= \sum_{n \, even} 2a_n z^n$$ $$= \sum_{n = 0}^{\infty} 2a_{2n} z^{2n}$$ So you have to have $a_{2n} = {1 \over 2}a_n$ for all $n$. Conversely if this holds for all $n$ then the power series for $f(z)$ will satisfy $f(z^2) = f(z) + f(-z)$. So the desired equation $f(z) = f(\sqrt{z}) + f(-\sqrt{z})$ will hold for $z \geq 0$. So for example ${\displaystyle \sum_{k = 0}^{\infty} {z^{2^k} \over 2^k}}$ works. None of these functions can be made entire... if $a_n \neq 0$ then ${\displaystyle \sum_{k=0}^{\infty} a_{2^{k}n}z^{2^{k}n} = \sum_{k=0}^{\infty} a_{n}{z^{2^{k}n} \over 2^k}}$ diverges for $z > 1$.

I'm not sure what $f(z) = f(\sqrt{z}) + f(-\sqrt{z})$ is supposed to mean when $z$ is not a nonnegative real number. But if you fix a definition for $\sqrt{z}$ these functions will satisfy $f(z) = f(\sqrt{z}) + f(-\sqrt{z})$ too.

• Im not yet convinced by the idea that the function cannot be made entire. What about negative signs ? Also : Can the function have a real period ? – mick Jun 20 '14 at 18:49
• Hmm it seems the radius is 1 , so not entire. But maybe periodic after analytic continuation ? – mick Jun 20 '14 at 18:53

Note that if $f(z)$ is an odd function $f(-z) = -f(z)$, then we obtain

$$f(z) = f(\sqrt{z}) + f(-\sqrt{z}) = f(\sqrt{z}) - f(\sqrt{z}) = 0$$

Note that if $f(z)$ is an even function $f(-z) = f(z)$, then we obtain

$$f(z) = f(\sqrt{z}) + f(-\sqrt{z}) = f(\sqrt{z}) + f(\sqrt{z}) = 2 f(\sqrt{z})$$

So we look for a even function such that

$$f(z^2) = 2 f(z)$$

or more general

$$f(z^n) = n f(z)$$

The function

$$\ln(|z|^k)$$

has this property, so

$$f(z) = \ln\Big(|z|^k\Big)$$

Note that

$$f\big(\pm\sqrt{-z}\big) = \ln\Big(\big|\pm\sqrt{-z}\big|^k\Big) = \ln\Big(\sqrt{|z|}^k\Big)$$

and is well defined.

• Im sorry but $f(z)=0$ is ruled out by the OP : a nonconstant function. Also $\ln(|x|)$ is invalid : it is not real-analytic. A logaritm is not analytic at 0. Im aware of these ideas but they are too trivial and do not give an answer to the OP. – mick Jun 20 '14 at 18:41
• That is why $f(z)$ is excluded and therefore $f(z)$ MUST be an even function. That is why you can only have even powers, thus $x^2k$ or powers of $|x|$. And $\ln(|x|)$ is real analytic - it can local be written as a power serie. – johannesvalks Jun 20 '14 at 21:34
• But $f(-z)$ can only be defined IF $f(z)$ contains $|z|$ in the function, thus $f(z) = g(|z|)$. – johannesvalks Jun 20 '14 at 21:47
• $ln(|x|)$ is not analytic at $x=0$, it does not even converge there !? – mick Jun 22 '14 at 20:34
• If $f(z)$ is not even nor odd, we can always write $$f(z) = f_\textrm{even} (z) + f_\textrm{odd}(z)$$ where $$f_\textrm{even} (z) = \frac{f(z) + f(-z) }{2}$$ and $$f_\textrm{odd} (z) = \frac{f(z) - f(-z) }{2}$$ The same properties hold and from that $$f_\textrm{odd} (z) = 0$$ whence $f(z)$ can only be even. – johannesvalks Jun 22 '14 at 22:39