Exercise 6.3.10

Answers

The sufficiency is easy since we may just pick $y = x$ . For the necessity, suppose now $∥ T (x) ∥ = ∥ x ∥$ . By Exercise 6.1.20, we have

\begin{array}{l} ⟨ x, y ⟩ & = \frac{1}{4} \sum_{k = 1}^{4} i^{k} ∥ x + i^{k} y ∥^{2} \\ = \frac{1}{4} \sum_{k = 1}^{4} i^{k} ∥ T (x + i^{k} y) ∥^{2} \\ = \frac{1}{4} \sum_{k = 1}^{4} i^{k} ∥ T (x) + i^{k} T (y) ∥^{2} \end{array}

⟨ T (x), T (y) ⟩

if $𝔽 = ℂ$ . However, for the case $𝔽 = ℝ$ the above argument could also work if we just pick $k$ to be $2$ and $4$ .

jephianlin

2011-06-27 00:00