A case when $fg=0$ as well as $f\bar{g}=0$

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I wonder how the second last sentence can be deduced. That is, $(f-Pf)\cdot Pg=0$. Here, $e_\alpha(t) = e^{-i\alpha t}.$ I don't believe that such things as whether a complex vector space is closed under complex conjugation would be used. Thus, I thought that some simple algebraical manipulation such as $\overline{Pg} =Pg_r - iPg_i $ would be applied to some line in the below to deduce the statement, but I could not find where the point would be hidden. Also, I guess that some other fact might be to be used.

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