[proofplan] Differentiate the eigenvalue equation $L\psi = \lambda\psi$ with respect to $t$, substitute the Lax equation $L_t = [L, A]$, and rearrange to show $\dot\lambda\psi = (L - \lambda)(\psi_t + A\psi)$. Self-adjointness of $L$ then forces $\dot\lambda = 0$ via an inner product argument. The second claim follows immediately. [/proofplan] [step:Differentiate the eigenvalue equation in $t$] Suppose $L(t)\psi(\cdot, t) = \lambda(t)\psi(\cdot, t)$. Differentiating both sides with respect to $t$: \begin{align*} L_t\psi + L\psi_t = \dot\lambda\psi + \lambda\psi_t. \end{align*} [/step] [step:Substitute the Lax equation and isolate $\dot\lambda$] The Lax equation gives $L_t = [L, A] = LA - AL$. Substituting: \begin{align*} (LA - AL)\psi + L\psi_t &= \dot\lambda\psi + \lambda\psi_t. \end{align*} Since $L\psi = \lambda\psi$, the term $LA\psi$ can be left as is, while $AL\psi = A(\lambda\psi) = \lambda A\psi$. Rearranging: \begin{align*} LA\psi - \lambda A\psi + L\psi_t - \lambda\psi_t &= \dot\lambda\psi, \\ (L - \lambda)(A\psi + \psi_t) &= \dot\lambda\psi. \end{align*} Define $\tilde\psi = \psi_t + A\psi$. The equation becomes \begin{align*} (L - \lambda)\tilde\psi = \dot\lambda\psi. \end{align*} [/step] [step:Use self-adjointness to show $\dot\lambda = 0$] Take the $L^2$ inner product of both sides with $\psi$: \begin{align*} \dot\lambda\|\psi\|^2 = \langle\psi, (L - \lambda)\tilde\psi\rangle. \end{align*} Since $L$ is self-adjoint, $\langle\psi, L\tilde\psi\rangle = \langle L\psi, \tilde\psi\rangle = \langle\lambda\psi, \tilde\psi\rangle = \lambda\langle\psi, \tilde\psi\rangle$. Therefore \begin{align*} \langle\psi, (L - \lambda)\tilde\psi\rangle = \langle\psi, L\tilde\psi\rangle - \lambda\langle\psi, \tilde\psi\rangle = \lambda\langle\psi, \tilde\psi\rangle - \lambda\langle\psi, \tilde\psi\rangle = 0. \end{align*} Since $\|\psi\|^2 > 0$ (eigenfunctions are nonzero), we conclude $\dot\lambda = 0$. [/step] [step:Conclude that $\tilde\psi = \psi_t + A\psi$ is an eigenfunction] Now that $\dot\lambda = 0$, the equation $(L - \lambda)\tilde\psi = \dot\lambda\psi = 0$ gives \begin{align*} L\tilde\psi = \lambda\tilde\psi. \end{align*} So $\tilde\psi = \psi_t + A\psi$ is an eigenfunction of $L$ with the same eigenvalue $\lambda$. [/step]