I need to solve this problem, and I'm having some trouble.
Let $f(t,x)$ satisfy the hypothesis of Picard's Theorem in the strip $-\infty<t<\infty$, $a \le x \le b$, and suppose $f$ is periodic int $t$ with period $T$. If
$f(t,a)>0$, and $f(t,b)<0$, $\forall t \in \mathbb R$,
show that this implies that the equation $\dot {x} = f(t,x)$ has a periodic solution of period $T$.
Hint: First that any solution $x(t)$ satisfying $x(0)=x_0$, $a\le x_0\le b$ is defined for $0 \le t \le T$ and takes on values betwenn $a$ and $b$. Then show there exists such a solution satisfying $x(0)=x(T)$ and that its periodic is the required solution.
I know that if $x(t)$ is between $a$ and $b$, then the domain of $x$ is $\mathbb R$ (since the graphic of $x$ is not contained in any compact). But I cannot prove that $a\le x(t) \le b$.
And I could not prove the existence of a $x_0$ satisfying $x(0)=x(T)$.
Suppose $x(0)=x_0\in(a,b)$. Then $a<x(t)<b$ for all $t>0$. Suppose that $x(t)>b$ for some $t>0$. Then there exists $s\in(0,t)$ such that $x(s)=b$. Let $\tau$ be the infimum of all such $s$. Then, since $x(t)<b$ for $t<\tau$, we have $x'(\tau)\ge0$. On the other hand, $x'(\tau)=f(\tau,x(\tau))=f(\tau,b)<0$, arriving to a contradiction. Similarly, we see that $x(t)>a$. If $x(0)=b$,since $x'(0)=f(0,b)<0$, $x(t)<b$ for$t$ close to $0$ and the previous argument applies.
Continuity of the map $x_0\to x(T)$ follows from the continuous dependence of solutions on initial data. This map takes $[a,b]$ into itself, and the intermediate value theorem proves that it has a fixed point.