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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.

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    $\begingroup$ The corresponding Hamiltonian flow is Lipschitz continuous and hence uniformly continuous. One way to prove this is by applying Grönwall’s inequality to the squared norm of the difference between two copies of the Hamiltonian dynamics started at two different initial conditions in phase space. $\endgroup$ Sep 23 at 8:52
  • $\begingroup$ @NawafBou-Rabee, nice observation. Why don't you post it as an answer? I'll surely upvote it. $\endgroup$ Sep 23 at 9:27
  • $\begingroup$ @DanieleTampieri Sure thing; done. $\endgroup$ Sep 23 at 10:09

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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, $$ |\varphi_t(x) - \varphi_t(y)|^2 \le \exp(2 L t) |x-y|^2 \;. $$ 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 of the Hamiltonian vector field $\mathfrak{X}$.

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