[proofplan] We prove involution by establishing an index-shifting identity: $\{H_n, H_m\}_1 = \{H_{n+1}, H_{m-1}\}_1$ for all $m \geq 1$, using the bi-Hamiltonian recurrence $\mathcal{J}_1\, \delta H_{m} = \mathcal{J}_0\, \delta H_{m-1}$ and the antisymmetry of both Hamiltonian operators. Iterating this identity shifts the second index down to $n$, producing $\{H_n, H_m\}_1 = \{H_{n + (m-n)}, H_n\}_1 = \{H_m, H_n\}_1$. Antisymmetry of the Poisson bracket then forces $\{H_n, H_m\}_1 = 0$. Conservation of each $H_n$ follows from the [Evolution of Functionals Along Flow](/theorems/1346) theorem and the involution result. [/proofplan]

[step:Establish the index-shifting identity $\{H_n, H_m\}_1 = \{H_{n+1}, H_{m-1}\}_1$] Fix $n \geq 0$ and $m \geq 1$. We compute, starting from the definition of the Poisson bracket induced by $\mathcal{J}_1$: \begin{align*} \{H_n, H_m\}_1 &= \langle \delta H_n, \mathcal{J}_1\, \delta H_m \rangle. \end{align*} The bi-Hamiltonian recurrence relation states $\mathcal{J}_1\, \delta H_m = \mathcal{J}_0\, \delta H_{m-1}$. Substituting: \begin{align*} \{H_n, H_m\}_1 &= \langle \delta H_n, \mathcal{J}_0\, \delta H_{m-1} \rangle. \end{align*} Since $\mathcal{J}_0$ is a skew-adjoint operator (being a Hamiltonian operator), we have $\langle f, \mathcal{J}_0\, g \rangle = -\langle \mathcal{J}_0\, f, g \rangle$ for all admissible $f, g$. Applying this with $f = \delta H_n$ and $g = \delta H_{m-1}$: \begin{align*} \{H_n, H_m\}_1 &= -\langle \mathcal{J}_0\, \delta H_n, \delta H_{m-1} \rangle. \end{align*} Now apply the recurrence in the other direction: $\mathcal{J}_0\, \delta H_n = \mathcal{J}_1\, \delta H_{n+1}$. Substituting: \begin{align*} \{H_n, H_m\}_1 &= -\langle \mathcal{J}_1\, \delta H_{n+1}, \delta H_{m-1} \rangle. \end{align*} Since $\mathcal{J}_1$ is also skew-adjoint, $\langle \mathcal{J}_1\, f, g \rangle = -\langle f, \mathcal{J}_1\, g \rangle$. Applying this: \begin{align*} \{H_n, H_m\}_1 &= \langle \delta H_{n+1}, \mathcal{J}_1\, \delta H_{m-1} \rangle = \{H_{n+1}, H_{m-1}\}_1. \end{align*} [/step]

[step:Iterate the shifting identity to conclude $\{H_n, H_m\}_1 = 0$] Assume WLOG that $m \geq n$ (the case $m < n$ follows by antisymmetry of the bracket). Applying the index-shifting identity from the previous step $m - n$ times (each application is valid provided the second index is at least $1$, which holds until the second index reaches $n$): \begin{align*} \{H_n, H_m\}_1 = \{H_{n+1}, H_{m-1}\}_1 = \{H_{n+2}, H_{m-2}\}_1 = \cdots = \{H_{n + (m-n)}, H_{m - (m-n)}\}_1 = \{H_m, H_n\}_1. \end{align*} But the Poisson bracket $\{\cdot, \cdot\}_1$ is antisymmetric (this is one of the defining properties of a Poisson bracket, following from the skew-adjointness of $\mathcal{J}_1$): \begin{align*} \{H_m, H_n\}_1 = -\{H_n, H_m\}_1. \end{align*} Combining the two equalities: \begin{align*} \{H_n, H_m\}_1 = -\{H_n, H_m\}_1, \end{align*} which implies $2\{H_n, H_m\}_1 = 0$, and therefore \begin{align*} \{H_n, H_m\}_1 = 0 \qquad \text{for all } n, m \geq 0. \end{align*} [/step]

[step:Deduce conservation of each $H_n$ from involution and the evolution formula] By the [Evolution of Functionals Along Flow](/theorems/1346) theorem, the time evolution of any functional $H_n$ along solutions of the Hamiltonian PDE $\partial_t u = \mathcal{J}_1\, \delta H_1$ is \begin{align*} \frac{dH_n}{dt} = \{H_n, H_1\}_1. \end{align*} Since we have just shown $\{H_n, H_m\}_1 = 0$ for all $n, m$, taking $m = 1$ gives \begin{align*} \frac{dH_n}{dt} = \{H_n, H_1\}_1 = 0 \qquad \text{for all } n \geq 0. \end{align*} Therefore each $H_n$ is conserved along solutions. Since the $H_n$ are pairwise in involution and individually conserved, they form an infinite family of commuting conserved quantities. [/step]