To show that the set $S = \{ x \mid |x| \leq 2 \}$ is a convex set, we need to verify that for any two points $x_1$ and $x_2$ in $S$, the line segment connecting these two points also lies entirely within $S$.

Definition of Convexity

A set $S$ is convex if for any two points $x_1, x_2 \in S$, and for any $\lambda \in [0,1]$, the point $x_\lambda = \lambda x_1 + (1 - \lambda) x_2$ is also in $S$. In other words, the line segment between $x_1$ and $x_2$ lies within the set $S$.

The Set $S = \{ x \mid |x| \leq 2 \}$

The set $S$ consists of all points $x$ such that the Euclidean norm (or absolute value in the one-dimensional case) of $x$ is less than or equal to 2. Geometrically, this represents a disk (or ball) of radius 2 centered at the origin in any Euclidean space.

Proof of Convexity

Let $x_1, x_2 \in S$. Then we know that $|x_1| \leq 2 \quad \text{and} \quad |x_2| \leq 2.$ We need to check whether the point $x_\lambda = \lambda x_1 + (1 - \lambda) x_2$ also satisfies $|x_\lambda| \leq 2$ for $\lambda \in [0,1]$.

By the triangle inequality, we have: $|x_\lambda| = |\lambda x_1 + (1 - \lambda) x_2| \leq |\lambda x_1| + |(1 - \lambda) x_2|.$ Using the fact that norms are positively homogeneous (i.e., $|c x| = |c| |x|$ for a scalar $c$), we get: $|x_\lambda| \leq \lambda |x_1| + (1 - \lambda) |x_2|.$ Since $|x_1| \leq 2$ and $|x_2| \leq 2$, we substitute these into the inequality: $|x_\lambda| \leq \lambda \cdot 2 + (1 - \lambda) \cdot 2 = 2(\lambda + 1 - \lambda) = 2.$

Thus, $|x_\lambda| \leq 2$, which implies that $x_\lambda \in S$.

Conclusion

Since for any $x_1, x_2 \in S$ and any $\lambda \in [0,1]$, the point $x_\lambda$ also belongs to $S$, we conclude that the set $S = \{ x \mid |x| \leq 2 \}$ is convex.

Would you like to dive deeper into this proof or have any specific questions about convex sets?

5 Related Questions:

1. What is the geometrical interpretation of convexity in higher dimensions?
2. Can the union of two convex sets be convex? Why or why not?
3. How would you check if a set defined by inequalities is convex?
4. How does the convexity of a function differ from the convexity of a set?
5. Can a set be convex if it is unbounded? Provide examples.

Tip:

Remember, convex sets are closely related to optimization problems. Many optimization problems have easier solutions when the feasible region (the set of possible solutions) is convex.