All entire functions which satisfying : $f(2z)=f(z)^{2}$

Question

I want to solve following question:

Find all entire functions satisfying the condition that $f(2z)=f(z)^{2}$ and $f(0)\ne 0.$

I know the function $f(z)=\mathbb e^{wz}$ for every $w$ is a solution. Is there another entire function by above property?

Answer 1

First, it is easy to see that $$f(z) \ne 0 \text{ for all } z \in \Bbb C \, .$$ (Assume that $f(z_0) = 0$. It follows that $f(z_0/2) = 0$ and therefore $f(z_0/2^k) = 0$ for all $k \in \Bbb N$. Then the zeros of $f$ acccumulate at $z=0$, and it follows from the Identity theorem that $f$ is identically zero.)

So $g(z) := f'(z)/f(z)$ is an entire function.

Differentiating $f(2z)=f(z)^{2}$ gives $$ 2f'(2z) = 2f(z)f'(z) \, . $$ and therefore $$ g(2z) = \frac{f'(2z)}{f(2z)} = \frac{f(z)f'(z)}{f(z)^2} = g(z) \, . $$ Now we can argue as above: For an arbitrary $z_1 \in \Bbb C$, $C := g(z_1) = g(z_1/2^k)$ for all $k \in \Bbb N$, and it follows from the identity theorem that $g \equiv C$.

So we have $f'(z) = Cf(z)$ for all $z \in \Bbb C$, and this implies $f(z) = De^{Cz}$ for some $D \in \Bbb C $.

(If that isn't obvious, define $h(z) := f(z) e^{-Cz}$. Then $h' \equiv 0$ so that $h$ is constant.)

Finally we can substitute $f(z) = De^{Cz}$ into $f(2z)=f(z)^{2}$ and conclude that $D = 1$, so that all solutions are given by $$ f(z) = e^{Cz} \,, \quad C \in \Bbb C \, . $$

Answer 2

We clime that $f(z)\neq0$ for every $z\in \mathbb C$. If $f(z)=0$ for some $z\in \mathbb C$. by the hypothesis, $(f(\frac{z}{2}))^2=f(z)=0$. So $f(\frac{z}{2})=0$. By induction, $f(\frac{z}{2^k})=0$. When $k\to \infty$, $\frac{z}{2^k}\to 0$. Since $f$ is entire, so $f(0)=0$, contradict. So, $f(z)\neq0$ for every $z\in \mathbb C$. From $f(2z)=f(z)^2$, we can get: $f(0)=1$, and: $f'(2z)=f'(z)f(z)$. If for some $z\neq 0$, $f'(z)=0$, then, for every $k\in \mathbb N$, we should have: $f'(\frac{z}{2^k})=0$ and when $k\to \infty$, we should have: $f'(0)=0$. But $f'$ is entire and the set $\{z\in \mathbb C; f'(z)=0\}$ has a limit point. So $f'(z)=0$ for every $z\in \mathbb C$. its mean $f(z)=1$ for every $z\in \mathbb C$.
Now let $f'(0)\neq 0$. define $g(z)=\frac{f(z)}{\mathbb e^{f'(0)z}}$. Then, it is easy to check that: $$ g(2z)=(g(z))^2 \,\,\,\,\,\, (1)$$ We know: $g,g'$ are entire and $\,\,$ $g(0)=1$ $\,,\,$ $g'(0)=0$.
If we prove: $g^{(n)}(0)=0$ for every $n\in \mathbb N$, then $g\equiv 1$ and we are done.
Consider that $g$ is entire. So $g$ has a power series $$g(z)=a_0+a_1z+a_2z^2+a_3z^3+...+a_nz^n+...$$ which for every $z\in \mathbb C$ is convergence and $a_n=\frac{g^{(n)}(0)}{n!}$. From $g(0)=1$ $\,,\,$ $g'(0)=0$ we get: $$g(z)=1+a_2z^2+a_3z^3+...+a_nz^n+...$$ From equation $(1)$ we should have: $$ \begin{align} 1+4a_2z^2+8a_3z^3+...+2^na_nz^n+...&=(1+a_2z^2+a_3z^3+...+a_nz^n+...)^2\\ &=1+c_2z^2+c_3z^3+...+c_nz^n+... \end{align} $$ Which $c_n=2a_n+a_2a_{n-2}+a_3a_{n-3}+...+a_{n-2}a_2$. From here: $4a_2=2a_2$ which mean: $a_2=0$. By induction, let $a_1=0,a_2=0,...,a_n=0$. We want to prove: $a_{n+1}=0$. $$2^{n+1}a_{n+1}=2a_{n+1}+\sum_{j=1}^{n-1}a_ja_{n-j}$$
But $\sum_{j=1}^{n-1}a_ja_{n-j}=0$, and it's mean $a_{n+1}=0$. So $g\equiv 1$. Thus: $f(z)=\mathbb e^{f'(0)z}$ are all of function's satisfying $f(2z)=(f(z))^2$

Answer 3

Hint: Let $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Then

$$f(2z) = \sum_{n=0}^{\infty}(2^na_n)z^n,\,\,\, \, f(z)^2 = \sum_{n=0}^{\infty}\left (\sum_{k=0}^{n}a_ka_{n-k}\right )z^n.$$

Equate coefficients for small $n$ and a nice pattern emerges, which you can then prove by induction.

Answer 4

This result really comes down to local behavior of $f$ around $0$ so I ignore that $f$ is entire and just assume some Domain including $0$. Note that $f(0)=1$. It trivially holds that $f =\exp\big( 0\cdot z\big)$ when $f$ is constant.

Now assume $f$ is non-constant and using the standard logarithm $g:=\text{log}\circ f$. In some $B(0,\delta)$ we know $g$ is analytic, $g(0)=0$ and obeys $2\cdot g(z) = g(2\cdot z)$, which looks a lot like a linear function. Confirm this by observing that $g(z) = b_1 z + \sum_{k=2}^\infty b_k z^k$ and analytic function $0 = g(2\cdot z) - 2 \cdot g(z) = \sum_{k=2}^\infty(2^k -2)\cdot b_k z^k\implies b_k=0$ for all $k\geq 2$, so $g(z) = b_1 \cdot z$, i.e. $g$ is linear. Since $f$ is non-constant and $f\big(B(0,\delta)\big)$ is an open set (open mapping theorem), and $\log$ is non-constant (check derivative or power series expansion at $1$) we know $g\big(B(0,\delta)\big)$ is an open set $\implies b_1\neq 0$.

Finally $h(z):=\frac{\log(z)}{b_1}$ is defined on the Domain $f\big(B(0,\delta)\big)$ and $h\circ f=z$ and $h\circ \exp\big(b_1 z\big)=z$ i.e. $h$ is a surjective map to $B(0,\delta)$ and both $\exp\big(b_1\cdot z\big)$ and $f$ act as right inverses. But the first derivative $h^{(1)}(0)\neq 0$ so $h$ is locally injective as well. Select small enough $r\in (0,\delta)$ and
$h:f\big(B(0,r)\big)\longrightarrow B(0,r)$ is both surjective and injective so (subject to enforcing suitable domains and co-domains)
$\exp\big(b_1 z\big)$ and $f$ are bona fide inverses of $h\implies f(z) = \exp\big(b_1 z\big)$ since inverses are unique and if two analytic functions agree when restricted to $B(0,r)$, then they are agree everywhere in the original Domain (identity theorem).