We fix an integer $d \in \mathbb{N}^*$, and we equip $\mathbb{R}^d$ with the usual inner product denoted $\langle \cdot, \cdot \rangle$ and the associated Euclidean norm $\|\cdot\|$. We denote $C := \{x \in \mathbb{R}^d \mid \|x\| \leq 1\}$ the closed unit ball of $\mathbb{R}^d$, and we are given $f \in \mathcal{C}^1(\mathbb{R}^d)$. Let $x_*$ be a minimizer of $f$ on $C$. Suppose in this question that $\|x_*\| = 1$. The objective is to show that $$\exists \lambda \geq 0,\, \nabla f(x_*) = -\lambda x_*.$$ a) Let $x, y \in \mathbb{R}^d$ such that $x \neq y$ and $\|x\| = \|y\| = 1$. Show that $\langle x, v \rangle > 0$ and $\langle y, v \rangle < 0$, where $v := x - y$. b) Suppose by contradiction that (7) is not satisfied. Show that there exists $v \in \mathbb{R}^d$ such that $\langle v, \nabla f(x_*) \rangle > 0$ and $\langle v, x_* \rangle > 0$. Deduce a contradiction and conclude. Hint: consider the quantities $f(x_* - tv)$ and $\|x_* - tv\|^2$, in the limit $t \rightarrow 0^+$.
We fix an integer $d \in \mathbb{N}^*$, and we equip $\mathbb{R}^d$ with the usual inner product denoted $\langle \cdot, \cdot \rangle$ and the associated Euclidean norm $\|\cdot\|$. We denote $C := \{x \in \mathbb{R}^d \mid \|x\| \leq 1\}$ the closed unit ball of $\mathbb{R}^d$, and we are given $f \in \mathcal{C}^1(\mathbb{R}^d)$. Let $x_*$ be a minimizer of $f$ on $C$.\\
Suppose in this question that $\|x_*\| = 1$. The objective is to show that
$$\exists \lambda \geq 0,\, \nabla f(x_*) = -\lambda x_*.$$
a) Let $x, y \in \mathbb{R}^d$ such that $x \neq y$ and $\|x\| = \|y\| = 1$. Show that $\langle x, v \rangle > 0$ and $\langle y, v \rangle < 0$, where $v := x - y$.\\
b) Suppose by contradiction that (7) is not satisfied. Show that there exists $v \in \mathbb{R}^d$ such that $\langle v, \nabla f(x_*) \rangle > 0$ and $\langle v, x_* \rangle > 0$. Deduce a contradiction and conclude.\\
Hint: consider the quantities $f(x_* - tv)$ and $\|x_* - tv\|^2$, in the limit $t \rightarrow 0^+$.