Exercise 6.5.6

Answers

If $T$ is unitary, we must have

0 = ∥ T (f) ∥^{2} - ∥ f ∥^{2} = \int_{0}^{1} | h |^{2} | f |^{2} dt - \int_{0}^{1} | f |^{2} dt

= \int_{0}^{1} (1 - | h |^{2}) | f |^{2} dt

for all $f \in V$ . Pick $f = {(1 - | h |^{2})}^{\frac{1}{2}}$ and get $1 - | h |^{2} = 0$ and so $| h | = 1$ . Conversely, if $| h | = 1$ , we have

∥ T (f) ∥^{2} - ∥ f ∥^{2} = \int_{0}^{1} | h |^{2} | f |^{2} dt - \int_{0}^{1} | f |^{2} dt

= \int_{0}^{1} (1 - | h |^{2}) | f |^{2} dt = 0

and so $T$ is unitary.

jephianlin

2011-06-27 00:00