TIFR CSE 2016 | Part B | Question: 6

Question

A subset $X$ of $\mathbb{R}^n$ is convex if for all $x, y \in X$ and all $\lambda \in (0, 1)$, we have $\lambda x + (1- \lambda)y \in X$. If $X$ is a convex set, which of the following statements is necessarily TRUE?

• A. For every $ x \in X$ there exist $y, z \in X -\{x\}$ and $\lambda \in (0, 1)$ so that $x= \lambda y+ (1-\lambda ) z $
• B. If $x, y \in X$ and $\lambda \geq 0$, then $\lambda x + (1-\lambda)y  \in X$
• C. If $x_1, \dots , x_n \in X (n \geq 1)$, then $(x_1+ \dots + x_n)/n \in X$
• D. If $x \in X$, then $\lambda x \in X$ for all scalars $\lambda$
• E. If $x, y \in X$, then $x-y \in X$

Answer 1

A set  is convex if for any two points x,y∈X : for any λ∈(0,1)\ --> λx+(1−λ)y∈X.

This is just the “mixture” or “average” of x and y.

(c) If x₁, …, xₙ ∈ X, then (x₁ + ⋯ + xₙ)/n ∈ X

This is the average of the points.

The statement follows directly, the definition of convexity, applied repeatedly.

Note that the bracket representation that lambda belongs to gives the range between 0 and 1 but not them.