I'm sorry if this question is similar to a one already asked, I'm not aware of many basic facts about elliptic curves.
Let $g$ be a meromorphic function on $\mathbb{C}$, and $\Lambda$ be a lattice in $\mathbb{C}$. Suppose that for any $\lambda\in \Lambda$, there are constants $c_{\lambda}$ such that for $z\in \mathbb{C}$ (out of poles) the following holds: $$ g(z+k\lambda)=g(z)+kc_{\lambda}$$ Is it true that $g(z)=\alpha z+R(\mathcal{S})(z)$ for some elliptic function $\mathcal{S}$, polynomial $R$ and constant $\alpha$ ?
My ideas are as follow:
The field of meromorphic functions on $\mathbb{C}/\Lambda$ is $\mathbb{C}(\mathcal{S},\mathcal{S}')$ for some elliptic function $\mathcal{S}$ (for example the Weirestrass function) and $(\mathcal{S}')^2 = Q(\mathcal{S})$ for some polynomial $Q$
The assumption on $g$ implies that $h(z):=g'(z)$ is elliptic, that is $h(z)=P(\mathcal{S},\mathcal{S}')(z)$ by 1.
"Locally" (that is for $z$ near enough from a zero $z_0$ of $\mathcal{S}$, in order to find a determination of $\sqrt{Q(\mathcal{S})}$ ) : $$g(z)-g(z_0)=\int_{[z_0,z]} h(u) du = \int_{[z_0,z]} \mathcal{S}'(u) \frac{P(\mathcal{S},\mathcal{S}')(u)}{\sqrt{Q(\mathcal{S}(u))}} du$$
We may decompose the last integral into an integral with residu, giving $\alpha z$, and the remainder would be a function of $\mathcal{S}$ by an apporpriate change of variable.
I'm not sure because I have difficulties to write formally the statement 4. Is there a counterexample or do you know a classical proof otherwise ?
Thank you very much, best regards
The counter-example is the primitive of $\wp_\Lambda(z)$, called the Weierstrass zeta function $\zeta_\Lambda(z)$.
The derivative of $\zeta_\Lambda(z+\lambda)-\zeta_\Lambda(z)$ is identically $0$ so $\zeta_\Lambda(z+\lambda)-\zeta_\Lambda(z)$ is constant.
$\zeta_\Lambda(z)$ has only one (simple) pole on a fundamental parallelogram $P$.
So it can't be that $\zeta_\Lambda(z)=\alpha z+f(z)$ with $f$ an elliptic function, as $\int_{\partial P} \zeta_\Lambda(z)dz= 2i\pi Res(\zeta_\Lambda(z),0)=2i\pi$ whereas $\int_{\partial P} f(z)dz=0$ (the opposite sides of the parallelogram are traversed in opposite direction so they cancel each other) and $\int_{\partial P}\alpha zdz = 0$.
You'll get that any $g$ satisfying your condition is of the form $$g(z)=b \zeta_\Lambda(z)+cz + f(z)$$ with $f$ elliptic.