Exercise 13.4: Star-Shaped Sets and Convexity
A set \(\mathcal{C}\) is star-shaped around \(x^\star\) if \(\mathcal{C} - x^\star\) is star-shaped, i.e., for any \(x \in \mathcal{C} - x^\star\) and \(\alpha \in [0,1]\), we also have \(\alpha x \in \mathcal{C} - x^\star\).
(a)
If \(\alpha x + (1 - \alpha) x^\star = x^\star + \alpha (x - x^\star)\) belongs to \(\mathcal{C}\), then \(\alpha(x - x^\star) \in \mathcal{C} - x^\star\) by definition. In the opposite direction, if \(z \in \mathcal{C} - x^\star\), then \(x^\star + \alpha z \in \mathcal{C}\) for any \(\alpha \in [0, 1]\). Since any \(z \in \mathcal{C} - x^\star\) is equal to \(x - x^\star\) for some \(x \in \mathcal{C}\) by definition, this implies \(\alpha x + (1 - \alpha) x^\star \in \mathcal{C}\) for any \(x \in \mathcal{C}\) and \(\alpha \in [0, 1]\).
(b)
If \(\mathcal{C}\) is convex, then \(\alpha x + (1 - \alpha) x^\star \in \mathcal{C}\) for any \(x, x^\star \in \mathcal{C}\) and \(\alpha \in [0,1]\). Hence by (a), \(\mathcal{C}\) is star-shaped around all its points. In the opposite direction, for any \(x^\star \in \mathcal{C}\) around which \(\mathcal{C}\) is star-shaped, we have \(\alpha x + (1 - \alpha) x^\star \in \mathcal{C}\) for all \(\alpha \in [0,1]\) and \(x \in \mathcal{C}\) by (a). Hence \(\mathcal{C}\) is convex.