Let $h \in C^2\_{\mathrm{ub}}(\mathbb{R}^{2n})$, where $C\_{\mathrm{ub}}^k$ consists of $C^k$-functions that are bounded and uniformly continuous along with their derivatives up to $k$th-order.


It is clear that the Hamiltonian vector field $X$ is $C^1$ and globally Lipschitz, hence the Hamiltonian flow $\Phi\_t$ exists for all times.


Since $X$ is $C^1$, so is the flow. Can one say something similar regarding the uniform continuity? Is the flow of a function $h$ as described above uniformly continuous?


Any help would be much appreciated.



Uniform continuity of Hamiltonian flow
Denote by $L$ the Lipschitz constant of the Hamiltonian vector field $\mathfrak{X}$ and by $\varphi\_t$ the flow generated by $\mathfrak{X}$. Then for any $x, y \in \mathbb{R}^{2n}$, by the chain rule, \begin{align\*}
& |\varphi\_t(x) - \varphi\_t(y)|^2 = \\
& \quad |x - y|^2 + 2 \int\_0^t \langle \mathfrak{X}(\varphi\_s(x)) - \mathfrak{X}(\varphi\_s(y)), \varphi\_s(x) - \varphi\_s(y) \rangle ds \\
& \le |x - y|^2 + 2 L \int\_0^t | \varphi\_s(x) - \varphi\_s(y) |^2 ds
\end{align\*}
By [Grönwall's inequality](https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality#Integral_form_for_continuous_functions),
$$
 |\varphi\_t(x) - \varphi\_t(y)|^2 \le \exp(2 L t) |x-y|^2 \;.
$$
Thus, the Hamiltonian flow is Lipschitz continuous and hence uniformly continuous. $\Box$


Note that all we really need in this proof is the [one-sided Lipschitz property](https://en.wikipedia.org/wiki/Lipschitz_continuity#One-sided_Lipschitz) of the Hamiltonian vector field $\mathfrak{X}$.


