1. **Problem Statement:** Discuss whether the set defined by the input requirement $V(y)$ is convex. 2. **Definition of Convex Set:** A set $S$ in a vector space is convex if for any two points $x_1, x_2 \in S$, the line segment joining them is also in $S$. Formally, for all $\lambda \in [0,1]$, $$\lambda x_1 + (1-\lambda) x_2 \in S.$$ 3. **Approach:** To determine if $V(y)$ is convex, we need to check if for any $y_1, y_2 \in V(y)$ and any $\lambda \in [0,1]$, the combination $\lambda y_1 + (1-\lambda) y_2$ also belongs to $V(y)$. 4. **Important Rules:** - Convexity depends on the nature of $V(y)$, which is typically defined by inequalities or functional constraints. - If $V(y)$ is defined by linear inequalities or convex functions, it is convex. - If $V(y)$ involves non-convex constraints, it may not be convex. 5. **Intermediate Work:** - Without explicit form of $V(y)$, assume $V(y)$ is defined by $g_i(y) \leq 0$ for convex functions $g_i$. - Then for $y_1, y_2 \in V(y)$, $g_i(y_1) \leq 0$ and $g_i(y_2) \leq 0$. - By convexity of $g_i$, $$g_i(\lambda y_1 + (1-\lambda) y_2) \leq \lambda g_i(y_1) + (1-\lambda) g_i(y_2) \leq 0,$$ so $\lambda y_1 + (1-\lambda) y_2 \in V(y)$. 6. **Conclusion:** If $V(y)$ is defined by convex constraints, it is a convex set. Otherwise, it may not be convex. More information about $V(y)$ is needed for a definitive answer.