Question: top-left: “Optimization for Data Science” and “Fall 2026”.
center: “Unit 03 Assignment” with subtitle “Convex Sets and Convex Functions”, then “Lessons 05–06 Version 1.0”.
left-center: “1. Convex combinations. Let” with vectors $x = \begin{bmatrix}1 \\ 2\end{bmatrix}$ and $y = \begin{bmatrix}5 \\ -2\end{bmatrix}$, then “Compute” and $z = \theta x + (1 - \theta) y$ for $\theta = \frac{1}{4}$, followed by “Explain briefly why $z$ lies on the line segment joining $x$ and $y$.”
center-left: “2. Classifying sets. For each set below, state whether it is convex. Give a brief reason.”
(a) $C_1 = \{x \in \mathbb{R}^2 : x_1 + x_2 \leq 4\}$
(b) $C_2 = \{x \in \mathbb{R}^2 : \|x\|_2 \leq 3\}$
(c) $C_3 = \{x \in \mathbb{R}^2 : \|x\|_2 \geq 3\}$
(d) $C_4 = \{x \in \mathbb{R}^2 : x_1 = 2x_2 + 1\}$
center: “3. Proving or disproving convexity. Consider the set”
$C = \{x \in \mathbb{R}^2 : x_1 + x_2 \leq 5\}$.
Then “Use the definition of a convex set to prove that $C$ is convex. Your proof should begin with arbitrary $x, y \in C$ and $0 \leq \theta \leq 1$.”
center-left: “4. Feasible regions and active constraints. Consider the constraints”
$x_1 \geq 0$, $x_2 \geq 0$, $x_1 + x_2 \leq 4$.
(a) Sketch the feasible region.
(b) Which constraints are active at $(0,4)$?
(c) Which constraints are active at $(1,1)$?
(d) Is the feasible region convex? Explain briefly.
center-left: “5. Convex functions. For each function, determine whether it is convex. Use the simplest valid justification.”
(a) $f(x) = 3x + 2$
(b) $f(x) = x^2 + 4x + 1$
(c) $f(x) = -x^2 + 4x$
(d) $f(x_1, x_2) = x_1^2 + 5x_2^2$
center-left: “6. First-order condition and global optimality. Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is differentiable and convex, and suppose”
$\nabla f(x^*) = 0$.
Then “Using the first-order inequality”
$f(y) \geq f(x) + \nabla f(x)^T(y - x)$,
bottom-center: page number “1”.Optimization for Data Science Fall 2026
show that $x^*$ is a global minimizer of $f$.
7. Least squares is convex. Consider
$f(x) = \frac{1}{2}\|Ax - b\|_2^2$.
From class,
$\nabla^2 f(x) = A^T A$.
For an arbitrary vector $v$, complete the argument
$v^T A^T A v = (Av)^T(Av) = __________ \geq 0$.
Then explain why this proves that $f$ is convex.
Reminder: Use the definitions and tests developed in Lessons 05 and 06. Clear, concise reasoning is sufficient;
lengthy proofs are not required unless specifically requested.
1. **Problem Statement:** Compute $z = \theta x + (1 - \theta) y$ for $\theta = \frac{1}{4}$ where $x = \begin{bmatrix}1 \\ 2\end{bmatrix}$ and $y = \begin{bmatrix}5 \\ -2\end{bmatrix}$. Explain why $z$ lies on the line segment joining $x$ and $y$.
2. **Formula and Rules:** A convex combination of two points $x$ and $y$ is given by $z = \theta x + (1 - \theta) y$ where $0 \leq \theta \leq 1$. Such $z$ always lies on the line segment joining $x$ and $y$.
3. **Calculation:**
$$
z = \frac{1}{4} \begin{bmatrix}1 \\ 2\end{bmatrix} + \left(1 - \frac{1}{4}\right) \begin{bmatrix}5 \\ -2\end{bmatrix} = \frac{1}{4} \begin{bmatrix}1 \\ 2\end{bmatrix} + \frac{3}{4} \begin{bmatrix}5 \\ -2\end{bmatrix}
$$
$$
z = \begin{bmatrix}\frac{1}{4} \\ \frac{2}{4}\end{bmatrix} + \begin{bmatrix}\frac{15}{4} \\ -\frac{6}{4}\end{bmatrix} = \begin{bmatrix}\frac{1}{4} + \frac{15}{4} \\ \frac{2}{4} - \frac{6}{4}\end{bmatrix} = \begin{bmatrix}4 \\ -1\end{bmatrix}
$$
4. **Explanation:** Since $\theta = \frac{1}{4}$ lies between 0 and 1, $z$ is a convex combination of $x$ and $y$, so $z$ lies on the line segment joining $x$ and $y$.
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**Answer:**
$$z = \begin{bmatrix}4 \\ -1\end{bmatrix}$$
This completes the first question.