Let $y: \mathbb{R} \longrightarrow \mathbb{R}$ be an odd and continuous function and $T>0$. Suppose that there exists $\varphi: \mathbb{R} \longrightarrow \mathbb{R}$ a continuous function such that $$ \int_0^T \varphi(x)\; dx=0 \tag{1} $$ and, for some constant $\gamma>0$, $$ y(x+T)=y(x)+\gamma\varphi(x), \; \forall \; x \in \mathbb{R}. \tag{2} $$ Question. Is it true that $$ \int_0^T y(x) \; dx =0? \tag{*} $$
I thought of the following: integrating $(2)$ over $[0,T]$ and using $(1)$ we obtain $$ \int_0^T y(x+L) \; dx = \int_T^{2T} y(u) \; dx. \tag{3} $$
Now, I know that, since $y$ is odd then holds $$\int_{-T}^T y(x)\; dx=0. \tag{4}$$
Can I use $(4)$ in order to prove $(*)$? If so, how can I use it?
Consider $y = \sin(x)$ and $T = \frac{\pi}{2}$. We have $y(x + T) = \sin(x + \frac{\pi}{2}) = \cos(x)$. Then define $\phi(x) = \cos(x) - \sin(x)$, $\gamma = 1$. Then we have $y(x + T) = y(x) + \gamma \phi(x)$. And note that $y$ is odd.
Furthermore, note that $\int\limits_0^T \phi(x) dx = 0$. But $\int\limits_0^T y(x) dx = 1$.
So your claim is not true.