We will prove that $r_k \rightarrow 0$ when $k \rightarrow +\infty$.
Each step of the TSIRM algorithm \\
+
+Let $\operatorname{span}(S) = \left \{ {\sum_{i=1}^k \lambda_i v_i \Big| k \in \mathbb{N}, v_i \in S, \lambda _i \in \mathbb{R}} \right \}$ be the linear span of a set of vectors $S$. So,\\
$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
$\begin{array}{ll}
-& = \min_{x \in Vect\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
-& = \min_{x \in Vect\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
-& \leqslant \min_{x \in Vect\left( x_{k-1} \right)} ||b-Ax ||_2\\
-& \leqslant ||b-Ax_{k-1}||
+& = \min_{x \in span\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k-1} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\
+& \leqslant ||b-Ax_{k-1}||_2 .
\end{array}$
\end{proof}
% that's all folks
\end{document}
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