[proofplan] The proof is local on $T^*X \setminus 0$, so we compute in canonical cotangent coordinates $(x,\xi)$. In those coordinates the Hamilton vector field differentiates a function by pairing the $\xi$-derivatives of $p$ with the $x$-derivatives of that function and subtracting the opposite pairing. Applying this formula to the function $p$ itself gives two identical scalar sums with opposite signs, hence $H_p p=0$. Along an integral curve $\gamma$ of $H_p$, the derivative of $p \circ \gamma$ is exactly $(H_p p)\circ \gamma$, so $p$ is constant and the zero level set is invariant. [/proofplan]

[step:Compute the Hamilton vector field applied to its own Hamiltonian in canonical coordinates]Let $\rho \in T^*X \setminus 0$ be arbitrary. Choose a coordinate chart $(U,\varphi)$ on $X$ with $\pi(\rho) \in U$, where $\pi: T^*X \to X$ is the cotangent projection. The induced canonical coordinates on $T^*U$ are denoted by $(x,\xi) = (x_1,\dots,x_n,\xi_1,\dots,\xi_n)$, where $x_i$ are base coordinates and $\xi_i$ are the corresponding fiber coordinates. In these coordinates, the Hamilton vector field of $p$ is the smooth vector field \begin{align*} H_p = \sum_{j=1}^n \frac{\partial p}{\partial \xi_j}\frac{\partial}{\partial x_j} - \frac{\partial p}{\partial x_j}\frac{\partial}{\partial \xi_j}. \end{align*} Applying this first-order differential operator to the function $p: T^*X \setminus 0 \to \mathbb{R}$ gives, at every point of $T^*U \cap (T^*X \setminus 0)$, \begin{align*} (H_p p)(x,\xi)=\sum_{j=1}^n \left(\frac{\partial p}{\partial \xi_j}(x,\xi)\frac{\partial p}{\partial x_j}(x,\xi)-\frac{\partial p}{\partial x_j}(x,\xi)\frac{\partial p}{\partial \xi_j}(x,\xi)\right). \end{align*} For each index $j \in \{1,\dots,n\}$, the two factors are [real numbers](/page/Real%20Numbers) and commute under multiplication, so \begin{align*} \frac{\partial p}{\partial \xi_j}(x,\xi)\frac{\partial p}{\partial x_j}(x,\xi)-\frac{\partial p}{\partial x_j}(x,\xi)\frac{\partial p}{\partial \xi_j}(x,\xi)=0. \end{align*} Therefore $(H_p p)(x,\xi)=0$ in this coordinate neighbourhood. Since $\rho$ was arbitrary, $H_p p=0$ on all of $T^*X \setminus 0$.[/step]

[guided]We prove the identity pointwise, because both $H_p p$ and the zero function are smooth real-valued functions on $T^*X \setminus 0$. Fix a point $\rho \in T^*X \setminus 0$. Choose a coordinate chart $(U,\varphi)$ on $X$ containing the base point $\pi(\rho)$, where $\pi: T^*X \to X$ is the cotangent projection. This chart induces canonical coordinates $(x,\xi) = (x_1,\dots,x_n,\xi_1,\dots,\xi_n)$ on $T^*U$: the $x_i$ record the base point and the $\xi_i$ record the covector components in the dual coordinate basis. In these canonical coordinates, the Hamilton vector field associated to the smooth real-valued function $p: T^*X \setminus 0 \to \mathbb{R}$ is \begin{align*} H_p = \sum_{j=1}^n \frac{\partial p}{\partial \xi_j}\frac{\partial}{\partial x_j} - \frac{\partial p}{\partial x_j}\frac{\partial}{\partial \xi_j}. \end{align*} This formula says that $H_p$ is a first-order differential operator acting on smooth functions. We now apply it to the same function $p$. At a point $(x,\xi) \in T^*U \cap (T^*X \setminus 0)$, this gives \begin{align*} (H_p p)(x,\xi)=\sum_{j=1}^n \left(\frac{\partial p}{\partial \xi_j}(x,\xi)\frac{\partial p}{\partial x_j}(x,\xi)-\frac{\partial p}{\partial x_j}(x,\xi)\frac{\partial p}{\partial \xi_j}(x,\xi)\right). \end{align*} The reason this expression vanishes is not an estimate or a limiting argument; it is exact cancellation. For each fixed $j$, the two quantities $\frac{\partial p}{\partial \xi_j}(x,\xi)$ and $\frac{\partial p}{\partial x_j}(x,\xi)$ are real numbers. Real multiplication is commutative, hence \begin{align*} \frac{\partial p}{\partial \xi_j}(x,\xi)\frac{\partial p}{\partial x_j}(x,\xi)=\frac{\partial p}{\partial x_j}(x,\xi)\frac{\partial p}{\partial \xi_j}(x,\xi). \end{align*} Subtracting equal quantities gives \begin{align*} \frac{\partial p}{\partial \xi_j}(x,\xi)\frac{\partial p}{\partial x_j}(x,\xi)-\frac{\partial p}{\partial x_j}(x,\xi)\frac{\partial p}{\partial \xi_j}(x,\xi)=0. \end{align*} Every summand is zero, so the whole finite sum is zero: \begin{align*} (H_p p)(x,\xi)=0. \end{align*} Because the original point $\rho$ was arbitrary and the computation is valid in a canonical coordinate neighbourhood of any point of $T^*X \setminus 0$, the identity holds globally: \begin{align*} H_p p=0. \end{align*}[/guided]

[step:Differentiate the Hamiltonian along an integral curve] Let $I \subset \mathbb{R}$ be an interval, and let $\gamma: I \to T^*X \setminus 0$ be an integral curve of $H_p$. By definition of integral curve, for every $s \in I$, \begin{align*} \gamma'(s)=H_p(\gamma(s)). \end{align*} The composition $p \circ \gamma: I \to \mathbb{R}$ is smooth. Differentiating this composition along the curve gives \begin{align*} \frac{d}{ds}p(\gamma(s))=(H_p p)(\gamma(s)). \end{align*} By the identity already proved, $(H_p p)(\gamma(s))=0$ for every $s \in I$. Therefore \begin{align*} \frac{d}{ds}p(\gamma(s))=0 \end{align*} for every $s \in I$. Hence $p \circ \gamma$ is constant on $I$. [/step]

[step:Conclude that the characteristic set is invariant] Let $s_0 \in I$ and suppose $\gamma(s_0) \in \{p=0\}$. Since $p \circ \gamma$ is constant on $I$, for every $s \in I$ we have \begin{align*} p(\gamma(s))=p(\gamma(s_0))=0. \end{align*} Thus $\gamma(s) \in \{p=0\}$ for every $s \in I$. This proves that every integral curve of $H_p$ beginning in the characteristic set remains in the characteristic set for as long as it is defined. [/step]