Exercise 6.1.25

Answers

By Exercise 6.1.20 we know that if there is an inner product such that $∥ x ∥^{2} = ⟨ x, x ⟩$ for all $x \in ℝ^{2}$ , then we have

⟨ x, y ⟩ = \frac{1}{4} ∥ x + y ∥^{2} - \frac{1}{4} ∥ x - y ∥ .

Let $x = (2, 0)$ and $y = (1, 3)$ . Thus we have

⟨ x, y ⟩ = \frac{1}{4} ∥ (3, 3) ∥^{2} - \frac{1}{4} ∥ (1, - 3) ∥^{2} = 0

and

⟨ 2 x, y ⟩ = \frac{1}{4} ∥ (5, 3) ∥^{2} - \frac{1}{4} ∥ (3, - 3) ∥^{2} = \frac{1}{4} (25 - 9) = 4 .

This means this function is not linear in the first component.

jephianlin

2011-06-27 00:00