Rudin's Real and Complex Analysis, Section 9.16

Question

In Section of 9.16 from Rudin's RCA, it says

Let $\hat{M}$ be the image of a closed translation-invariant subspace $M \subset L^2$, nder the Fourier transfrom. Let $P$ be the orthogonal projection of $L^2$ onto $\hat{M}$ (Theorem 4.11): To each $f \in L^2$ there corresponds a unique $Pf \in \hat{M}$ such that $f - Pf$ is orthogonal to $\hat{M}$. Hence $$f - Pf \perp Pg, \quad (f \text{ and } g \in L^2)$$ and since $\hat{M}$ is invariant under multiplication by $e_\alpha = e^{-i\alpha t}$, we also have $$ f - Pf \perp (Pg)e_\alpha \quad (f \text{ and } g \in L^2, \, \alpha \in \mathbb{R}^1)$$ If we recall how the inner product is defined in $L^2$, we see that the above statement to $$ \int_{-\infty}^\infty (f -Pf)\cdot \overline{Pg}\cdot e_{-\alpha} \,dm = 0, \quad (f \text{ and } g \in L^2, \, \alpha \in \mathbb{R}^1)$$ and this says that the Fourier transform of $$ (f-Pf)\cdot \overline{Pg}$$ is $0$.

So far so good! However, he goes onto say "This remains true if $\overline{Pg}$ is replaced by $Pg$". Why is that!? Could you help me to understand this? Thank you in advance.

Answer 1

If $a,b:\mathbb R\to\mathbb C$ are functions in $L^2$ and $a\,\bar b=0$, then this occurs pointwise a.e. If now $a,b\in\mathbb C$ and $a\bar b=0$, then either $a=0$ or $b=0$, so $ab=0$.