Exercise 6.2.18

Answers

Let $f$ be an odd function. Then for every even funcion $g$ we have $fg$ is an odd function since

fg (t) = f (- t) g (- t) = - f (t) g (t) .

So the inner product of $f$ and $g$ is zero. This means $W_{e}^{⊥} \supset W_{o}$ .

Conversely, for every function $h$ , we could write $h = f + g$ , where

f (t) = \frac{1}{2} (h (t) + h (- t))

and

g (t) = \frac{1}{2} (h (t) - h (- t)) .

If now $h$ is an element in $W_{e}^{⊥}$ , we have

0 = ⟨ h, f ⟩ = ⟨ f, f ⟩ + ⟨ g, f ⟩ = ∥ f ∥^{2}

since $f$ is a even function. This means that $f = 0$ and $h = g$ , an element in $W_{o}$ .

jephianlin

2011-06-27 00:00