= Solution
Divide the supplied scheme by $\omega$ and take
$$
M=\frac D\omega+L,\qquad N=\left(\frac1\omega-1\right)D-U.
$$
Then $A=M-N$, and $M$ is invertible because it is lower triangular with positive diagonal $D/\omega$. Symmetry gives $U=L^T$, so
$$
\boxed{W=M+M^T-A=\left(\frac2\omega-1\right)D>0\quad(0<\omega<2).}
$$
The <Householder-John theorem> therefore gives convergence.
For a direct spectral proof, let $B=M^{-1}N=I-M^{-1}A$. Expansion verifies $A-B^*AB=(I-B)^*W(I-B)$. If $Bv=\lambda v$ with $v\ne0$, then
$$
(1-|\lambda|^2)v^*Av=|1-\lambda|^2v^*Wv.
$$
The value $\lambda=1$ is impossible since $(I-B)v=M^{-1}Av$ and $A$ is invertible. Both quadratic forms are positive, hence $|\lambda|<1$. The error obeys $e^{(k+1)}=Be^{(k)}$ and tends to zero, proving \b[convergence to the unique solution $A^{-1}b$ for every initial iterate]. This is the successive-over-relaxation iteration, with no assertion of convergence at the excluded endpoints.
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