Uniform continuity of Hamiltonian flow

Question

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.

Answer 1

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 $\mathfrak{X}$.