We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ admits a convergent subsequence. We denote by $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ the extractor and $x_{**}$ the corresponding limit, so that $x_{\varphi(n)} \rightarrow x_{**}$ when $n \rightarrow \infty$. Hint. We may use without proof the Bolzano-Weierstrass theorem: from any bounded sequence in $\mathbb{R}$, we can extract a convergent subsequence. b) Show that $f'(x_{**}) = 0$. c) Deduce that $x_{**}$ is a minimizer of $f$, then that $\left|x_n - x_{**}\right| \rightarrow 0$ when $n \rightarrow \infty$.
We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$.\\
a) Show that the sequence $\left(x_n\right)_{n \in \mathbb{N}}$ admits a convergent subsequence. We denote by $\varphi : \mathbb{N} \rightarrow \mathbb{N}$ the extractor and $x_{**}$ the corresponding limit, so that $x_{\varphi(n)} \rightarrow x_{**}$ when $n \rightarrow \infty$.\\
Hint. We may use without proof the Bolzano-Weierstrass theorem: from any bounded sequence in $\mathbb{R}$, we can extract a convergent subsequence.\\
b) Show that $f'(x_{**}) = 0$.\\
c) Deduce that $x_{**}$ is a minimizer of $f$, then that $\left|x_n - x_{**}\right| \rightarrow 0$ when $n \rightarrow \infty$.