I have a research work concerning the equation: $$x '(t) + g (x (t)) = f (t),\quad \forall t\in \mathbb R$$
f and g are defined and continuous in $\mathbb R$ and with values ​​in $\mathbb R$.
Furthermore f is assumed to be T periodic (There is no initial condition)


**First question:**


Assume that there is a periodic solution. Using the mean formula, to show that $\frac 1T \int\_0 ^ T f (t) dt \in g (\mathbb R) $


**Second question**


Show that this condition : $\frac 1T \int\_0 ^ T f (t) dt \in g (\mathbb R) $ is sufficient for the existence of a periodic solution


***For first question:***


$ g \circ x $ being continuous, there is a $ c $
between $ 0 $ and $ T $ such as $ \frac 1T\int\_0 ^ T
 g (x (s)) ds = g (x (c)) $. but $ x $ is a solution
$ T $ periodic, then $ x (T) = x (0) $, and we
will have:


$$\frac1T \int\_0 ^ T f (s) ds = \frac 1T \int\_0 ^ T
 (g (x (s)) + x '(s)) ds = \frac1T \int\_0 ^ T
 g (x (s)) ds = g (x (c)) \in g (\mathbb R). $$


***For second question:***


I need help how to use some one of fixed point theorem or any other way to prove that the condition $\frac 1T \int\_0 ^ T f (t) dt \in g (\mathbb R) $ is sufficient



$x '(t) + g (x (t)) = f (t),\quad \forall t\in \mathbb R$ have periodic solution $\iff\; \frac 1T \int_0 ^ T f (t) dt \in g (\mathbb R) $
The condition is not sufficient. Take for example the equation
$$x'=x^2+\sin t$$ with $g(x)=-x^2$ in your notation. Then
$$\int\_0^{2\pi} \sin t\, dt=0=g(0).$$ If $x$ were a $2\pi$-periodic solution, then
$$
0=x(2\pi)-x(0)=\int\_0^{2\pi} x'(t)\, dt=\int\_0^{2\pi}(x^2(t)+\sin t)\, dt= \int\_0^{2\pi}x^2(t)\, dt$$
would imply $x\equiv 0$ in $[0, 2\pi]$.


