Question on isometries

Dear Professor,

Firstly thanks for taking the time out to reply to our messages, we really appreciate the help and guidance from yourself. I wanted to ask the following short question about the 2201 notes:

On page 56 Thm 5.4.150: after showing that (i)\Rightarrow (ii) and (ii) \Rightarrow (iii), can’t we simply show (iii) \Rightarrow  (i) by saying:

||T^*Tu||=||u||

and proceeding exactly as the last line of the proof at the bottom?

Thanks in advance.

——————————————————————

Reply:

Not quite. Although with a bit more argument it might be OK. In `the last line of the proof at the bottom,’ note that an unwritten step in the argument says

\langle (T^*T-I)u, v \rangle=0

for *all* v, and then one sets v= (T^*T-I)u. This wouldn’t be admissible if we just had \langle (T^*T-I)u, u\rangle=0. Try it out to see clearly for yourself what I mean.

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