= Solution
Use the <Fourier amplification symbol> for the semidiscrete generator. At a mode $e^{i(k\xi+l\eta)}$, the two stencil symbols are
$$
m(\xi,\eta)=\frac{4+\cos\xi+\cos\eta}{6},\qquad
\ell(\xi,\eta)=\frac{-10+4(\cos\xi+\cos\eta)+2\cos\xi\cos\eta}{3h^2}.
$$
The mass symbol satisfies $1/3\leq m\leq1$. The numerator of $\ell$ is increasing in each cosine, since each partial <derivative> is at least two, and its maximum is zero at $(1,1)$. Hence $\ell\leq0$ and the generator $\lambda_h=\ell/m$ is real and nonpositive.
Every exact-time mode is multiplied by $e^{t\lambda_h}$, of modulus at most one. <Parseval identity> therefore proves
$$
\boxed{\|U(t)\|_{\ell_h^2}\leq\|U(0)\|_{\ell_h^2},\qquad
\|U\|_{\ell_h^2}^2=h^2\sum_{k,l}|U_{k,l}|^2.}
$$
The method is stable, uniformly in the spatial mesh. This is a semidiscrete conclusion: it imposes no time-step condition until a particular time integrator is applied. For the stated merely square-integrable initial data, cell averages give a well-defined stable initialization, with $\|U(0)\|_{\ell_h^2}\leq\|v\|_{L^2}$ by Cauchy–Schwarz. Arbitrary point sampling of an L2 equivalence class is not required.
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