Question

A function :ℝ→ℝ is strictly concave if for all x ∈ ℝ and y ∈ ℝ, where x ≠ y, and for all scalar ∈ (0, 1),

(x + (1 ― )y) > (x) + (1 ― )(y).

Using this definition, show that the function (x) ≡ 1 ― xTx is strictly concave.

HINTS: For clarity and brevity, I recommend you use, as well as , the notation ≡ 1 ― . If you cannot manage (b) for general , attempt to work it through for the special (scalar) case = 1, where (x) ≡ 1 ― x^2

Answer 1

To demonstrate that the function f(x) ≡ 1 - xTx is strictly concave, we compare f(λx + (1-λ)y) to λf(x) + (1-λ)f(y) for any distinct points x, y and any scalar λ in (0, 1), proving the inequality that defines strict concavity.

Step-by-step explanation:

To show that the function f(x) ≡ 1 - xTx is strictly concave, we need to verify the definition of strict concavity. Given any two distinct points x and y in ℝ, and a scalar λ within (0, 1), we examine the value of the function at the point λx + (1-λ)y and compare it to the weighted average λf(x) + (1-λ)f(y).

For the special case λ = 1/2, the function becomes f(x) ≡ 1 - x2. Consider f(λx + (1-λ)y) and λf(x) + (1-λ)f(y):

1. f(λx + (1-λ)y) = 1 - (λx + (1-λ)y)2.
2. λf(x) + (1-λ)f(y) = λ(1 - x2) + (1-λ)(1 - y2).

By expanding the squares and simplifying, we can show that the value of the function at the point λx + (1-λ)y is greater than the weighted average, proving that f(x) is strictly concave.