Let $v\_1 =\lambda\_1 \zeta\_1$ and $v\_2 = \lambda\_2 \zeta\_2$ with $\zeta\_1 = \frac{4\pi i\omega}{3}$ and $\zeta\_2 = \frac{4\pi i\omega^2}{3}$ where $\omega = e^{2\pi i/3}$ is the third root of unity and $\lambda\_1,\lambda\_2$ some positive integers.


I would like to ask if there is an entire function $f$ such that
$$ f(z+ v\_1) = f(z) e^{2\pi i \lambda\_1 Cz/3 }$$
and
$$ f(z+ v\_2) = f(z) e^{2\pi i \lambda\_2 Cz/3 }$$
where $C \in \mathbb R$ is a constant such that $\frac{\lambda\_1 \lambda\_2 C (\zeta\_2 -\zeta\_1)}{3}=1$. This looks a bit like periodic boundary conditions, but since $z \in \mathbb C$ the modulus of these boundary conditions has of course a growing/decaying direction.


This looks pretty similar to something related to theta functions, but I don't quite get it together, as the underlying lattice looks rather different.



Entire function with almost periodic boundary condition?
The answer seems to be negative. Suppose that an entire function $f$
satisfies $f(z+v\_i)=e^{A\_iz}f(z)$, where $v\_1$ and $v\_2$ generate a lattice. Let $\Pi$ be the fundamental parallelogram of this lattice and integrate $f'/f$ over $\partial \Pi$. You obtain the ``Legendre's relation'':
$$v\_2A\_1-v\_1A\_2=2\pi in,$$
where $n$ is the number of zeros of $f$ in $\Pi$.


Substituting your values, we see that $n=1$.
Now $(f'/f)'$ is doubly periodic with respect to our lattice,
having a single double pole per parallelogram. So we may
assume (by shifting a pole to the origin) that
$(f'/f)'=\wp+c,$ and two integrations integrations give
$$f(z)=e^{P(z)}\sigma(z),$$
where $\sigma$ is the Weierstrass sigma function and $P$ is a polynomial of degree at most $2$. This is the general form of your $f$ (modulo a shift of the origin), if it exists.


Now let us try to find $P$. Sigma satisfies
$$\sigma(z+v\_j)=-e^{\eta\_j(z+v\_j)}\sigma(z),$$
where $\eta\_j=\zeta(\omega\_j)$, and $\zeta$ is the Weierstrass zeta function ($\zeta'=-\wp$),
which gives
$$P(z+v\_j)=P(z)-\eta\_j(z+v\_j)+\pi i, \quad j=1,2.$$
Trying to find such a polynomial with your data, we
just set $P(z)=az^2+bz+c$, and try to find $a,b,c$.
Equation for $a$ is satisfied in view of Legendre's relation, and we have
$$a=(A\_j-\eta\_j)/(2v\_j), \quad j=1,2,$$
but for $b$ we obtain
$$av\_j^2+bv\_j+\eta\_jv\_j=\pi i,\quad i=1,2.$$
These two equations with one variable must be consistent,
which is very unlikely, certainly not for all $\lambda\_1,\lambda\_2$. To check this one has to compute $\eta\_j$ for your lattices.


