Periodic Soultions - ODE Let $f: \mathbb R^m \rightarrow \mathbb R^m$ with $f \in C^1$ en $u$ such that
$$
 u'(t) = f(u(t)); \qquad u(T) = u(0)\text{ for some $T > 0$ }
$$
I have to show that $u$ is a periodic solution.
Eventhough this problem seems quite simple I am a bit unsure how to proceed. My first idea is
Let $h(t):= t + T$ Then we must show $\forall t \in \mathbb R: u(t) = u(h(t))$. If we say $v(t) := u(h(t))$ then we must show $u = v$ and we know $u(0) = v(0)$. I further know that the oribts $\{u(t) : t \in \mathbb R\}$ and $\{v(t) \mid t \in \mathbb R\}$ are equal because $u,v$ are solutions of our problem.
How can I deduce that $u = v$ ?
Does this follow from the local uniqueness theorem of picard ?
 A: All the ingredients are here; now you have to put them together correctly.
(a) Given a differential equation $$\dot x= f(x)\qquad\qquad(1)$$ for some $C^1$ function $f:\ {\mathbb R}^m\to{\mathbb R}^m$, a $t_0\in{\mathbb R}$, and a point $x_0\in{\mathbb R}^m$ there is an   open interval $\tilde J$ containing $t_0$ and a function $\tilde u:\ \tilde J\to{\mathbb R}^m$ such that $\tilde u$ is a solution of the initial problem
$$\dot x=f(x),\quad x(t_0)=x_0\ ;\qquad\qquad(2)$$
and any solution $u:\ J\to{\mathbb R}^m$ of $(2)$ in an open interval $J$ containing $t_0$ is a restriction of $\tilde u$.
(b) If $\tilde u$ is as in (a) then for any $h\in{\mathbb R}$ the maximal solution $v$ of the initial problem
$$\dot x=f(x),\quad x(t_0+h)=x_0$$
is given by
$$v(t)=\tilde u(t-h)\qquad(t\in J+h)\ .$$
(Starting from the local existence and uniqueness theorem for ODEs some work is needed to prove (a), but (b) is verified easily.)
Assume now that $u:\ J\to{\mathbb R}^m$ is the maximal solution of the initial problem
$$\dot x=f(x),\quad x(0)=x_0\ , $$
and it so happens that $u(T)=x_0$ for some $T>0$. Then $u$ is also the maximal solution of the initial problem
$$\dot x=f(x),\qquad x(T)=x_0\ .$$
Using (b) we conclude that $J+h=J$ and therefore $J={\mathbb R}$, and secondly that $u(t)=u(t-T)$ for all $t$.
A: To show global uniqueness, consider the set $A=\{t:u(t)=v(t)\}$. Since $u$ and $v$ are continuous, $A$ is closed. By the local uniqueness (Picard), $A$ is open. Thus, $A$ is either empty or coincides with the entire interval on which solution exists, $\mathbb R$ in this case. 
