= Solution
For homogeneous error boundary values, the tensor-product sine vectors $U_{j,k}=\sin(r\pi jh)\sin(s\pi kh)$ form a basis. The sine addition formula gives $U_{j+1,k}+U_{j-1,k}=2\cos(r\pi h)U_{j,k}$, and similarly in the other coordinate. Hence the five-point matrix has
$$
\boxed{\lambda_{r,s}=4-2\cos(r\pi h)-2\cos(s\pi h),\qquad r,s=1,\ldots,n.}
$$
There is a printing error in the displayed iteration. Taken literally, its error multiplier is $1/20-(2/5)(\cos(r\pi h)+\cos(s\pi h))$, and its fixed-point equation is $(19/4)U_{j,k}+\sum_{\mathrm{neighbors}}U=h^2f_{j,k}$, rather than the stated Poisson equation. Thus the printed affine update cannot be used as a consistent relaxation for this system.
The intended <weighted Jacobi iteration> with $\omega=4/5$ is
$$
\boxed{U_{j,k}^{\mathrm{new}}=\frac15U_{j,k}^{\mathrm{old}}+\frac15\sum_{\mathrm{neighbors}}U^{\mathrm{old}}+\frac15h^2f_{j,k}.}
$$
Its <iteration matrix> is $I-(\omega/4)A$, whose mode multiplier is
$$
\boxed{\mu_{r,s}=1-\frac45+\frac25\bigl(\cos(r\pi h)+\cos(s\pi h)\bigr).}
$$
If at least one of the two cosines is negative, their sum lies strictly between $-2$ and $1$, so $-3/5<\mu_{r,s}<3/5$. Thus every such high-frequency mode contracts by at most $\boxed{3/5}$ per sweep, uniformly in grid spacing. This <weighted Jacobi smoothing bound for the square Poisson problem> justifies the corrected method as a smoother. The literal printed update also damps modes, but toward the wrong fixed-point equation.
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