Let M be a compact manifold of dimension $d\geqslant 1$ and $\mathscr{A}$ a finite atlas composed of $C^1$ charts.
Prove a version of Picard theorem for $C^1$ vector fields in the manifold M.
My attempt:
Let $M$ be a compact manifold of dimension $n$ such that ${(U_i,\psi_i)}$ forms an atlas. $$\psi_i:M\to\psi_i(U_i)\subset\mathbb{R}^n$$ are $C^r$ and $\bigcup_{i\in I} U_i=M$. Let $X:M\to TM$ where $TM$ is the tangent space of $M$ and $X$ defines a vector field on $M$. Then I have the following Cauchy problem:
$$\begin{cases} \dot{x}=X(x(t)) \\ x(0)=z\end{cases}$$
Making the following composition(since $\psi_i$ is an diffeomorphism): $$\psi_i'\circ X:M\to U_i\subset \mathbb{R}^n.$$
So I have for an arbitrary $i\in I$:
$$\begin{cases} \psi'_i\dot{x}=\psi'_i(X(x(t)) \\ \psi_i(x(0))=\psi_i(z)\end{cases}$$
which is in $\mathbb{R}^n$.
If I prove $\psi_i'\circ X$ is Lipschitz I can apply Picard Theorem and conclude there is a unique solution.
By the fact the composition of two continuous functions is continuous then $\psi'_i(X(x(t))$ is bounded on $M$, since $M$ is compact. If $\phi_1$ and $\phi_2$ are solutions then:
$|\psi'_i(\phi_1(x(t)) -\psi'_i(X(\phi_2(t)) |\leqslant K|||\phi_1-\phi_2|||$
Hence applying Picard theorem we get a solution in $\mathbb{R}^n$ that is unique and it is also unique via $\psi_i^{-1}$ in $M$. Since this holds for any $i\in I$, the assertion is proved.
Question:
Is my proof right? If not. How should I solve the problem?
Thanks in advance!
You have that the vector field $X$ is $C^1$, which means that its image under the chart maps is $C^1$ so that the differential equation on the charts has a $C^1$ right side, which means that it is locally Lipschitz. Then indeed you can lift the local solution on the chart with its uniqueness to a local solution on the manifold. Further, each local solution has a unique continuation to a maximal solution and as the manifold is compact, and by omission without a boundary where the solution could end, the maximal domain of the solution is $\Bbb R$.
In the details, you have some doubtful formulations.
That the atlas is $C^r$ means in general just that the chart transition maps are $C^r$. If $M$ is an embedded manifold, then the chart maps should go in the other direction, $\psi_i:U_i\to M\subset \Bbb R^N$. But let's stay with the chart maps as coordinate maps.
The push forward of the tangent vectors to the chart depends on the basis point, in your notation $v(\psi_i(p))=\psi_i'(p;X(p))$, which is a uniquely defined function on $\psi_i(U_i)$. By the assumption that $X$ is a $C^1$ vector field, $v$ is $C^1$ by definition. As such, it is locally Lipschitz
The first object you obtain is the solution $\xi(t)$ of $\dot \xi(t)=v(\xi(t))$. You need to pull that back to $x(t)=\psi_i^{-1}(\xi(t))$ to get a curve in the manifold and then check that it is a solution and plays nice with chart transitions.
Your last equation is incomprehensible. Probably you want to apply a Grönwall argument where $K$ is a local Lipschitz constant. But the parts of the formula seem simply wrong.