Let A be a strictly diagonally dominant matrix. Suppose we use Gauss Seidel method to solve $Ax=b$, a sequence of vectors {$x_{0},x_{1},...,x_{k},...$} is obtained (where $x_{0}$ is the initial guess) Let $x^{*}$ is the solution. Show that $C_{1} \|x_{k}-x^{*}\|_{\infty} \leq \|x_{k}-x_{k-1}\|_{\infty}\leq C_{2} \|x_{k}-x^{*}\|_{\infty}$ for some positive constant $C_{1}$ and $C_{2}$.
I prove the former inequality by showing $\|x_{k}-x^{*}\|_{\infty} \leq \|A^{-1}U\|\|x_{k}-x_{k-1}\|_{\infty}$ where $U$ is the upper triangular part of $A$.(not include the diagonal) Actually, I have show that $A(x_{k}-x^{*})=U(x_{k}-x_{k-1})$. But I don't know how to show the later inequality since I don't know if $U$ is invertible.
Another way of writing gauss seidel is:
$x_{k+1}=(D+L)^{-1}(D+L-A)x_k+(D+L)^{-1}b$, thus
$x_{k+1}-x_{k}=-(D+L)^{-1}Ax_k+(D+L)^{-1}Ax^*=(D+L)^{-1}A(x^*-x_k)$, so:
$\|x_{k+1}-x_{k}\|\leq\|(D+L)^{-1}\|\|A\|\|x^*-x_k\|$
Also as $A$ is strictly diagonaally dominant, is diagonal entries are non zero, so the determinant of (D+L) is the product of all the diagonal entries of $A$, which is non zero so (D+L) is invertible.